Cty TNIIH MIV DVVH Khang Vigt
Cam nang 6n luy^n thi dgi hgc 18 chuy§n d§ H6a hqc - Nguyen Van Hk\
Lai
m=
gidi:
Cac p h a n u n g hoa hoc:
gf 3Cu
Cu
.Q|J
+ 8HN03{loang)
> 3Cu(N03)2 + 2 N O + 4 H 2 O
+ mN03{dac,nguoi)
> Cu(N03)2 + 2NO2 + 2H2O
Lwu y: V o i d u n g d i c h h o n h(7p
cac i o n
— 3 C u 2 *
+ 2 N p + H2O
'
A. M g
,
nj^ >
s i n h ra V l i t k h i N O (san p h a m k h u d u y nhat, 6 dktc). Gia t r i ciia V l a j t j
Lai
n Cu =
6,4
64
, ^^^^y^, D . 1,120. j -j^^i
' ,
nf^j.
Nhan xet: k h i axit H N O 3 c6 m a t d o n g thoi v o i axit H2SO4 l o a n g thi l u g n g
t r o n g d u n g d i c h l a d o 2 axit p h a n l i ra - > G i a i theo p h u o n g t r i n h i o n .
3Cu
Mol:
a09
+
<-
8H^
+ 2NO3
a24->
'v^^'
.^^^
; ^ 0*./sH Qd;^ j r i i
> 3Cu2*
a06
V = 0,06.22,4 = 1,344 l i t
+ 2NO
'
,
:D
+ 4H2O
0,06
.loW
D a p an A .
V i dv 17: Cho 3,36 g a m M g tac d y n g v o i d u n g dich H N O a d u , t h u d u g c 0,448 lit
MM(N03)2
Vi
0,02
= ^ = 0 , 1 4 mol; n ^ = ^ = a 0 2 m o l .
24
22,4
Chat khu:
M g - 2e
C h a t oxi hoa:
2N*5
+ jOe
D . 23,00.
6c,r)dc-'i;
*
^'.[:
>U
^^^^^
+ ge
;gn A p d y n g bao toan electron: 2 n ^ g = 10 n ^ j + 8 nNH4N03
6!fifi^» (..f^'r-riiih , 2 . 0 , 1 4 - 1 0 . 0 , 0 2
,
nNH4N03 =
^
=0,01 m o l .
Dap
an
D.
KV.
1
> NH4Na
xet: Cho X + H C l d u giai p h o n g H2
NMn
:i
- " f r u ri
X chua Z n - > Z n con d u . (s, i :
*::^A
> Zn(N03)2
+ 2Ag
> Zn(N03)2 + C u
>
ZnS04 + H2
^^9^
/F''
'^-^r^'h&tki
m ^i^yi:o':) <
> y r M rijl rifiq dnhl g n w j i ! ' ?
nAi = "^^^^°^ + n c u ( N 0 3 ) 2 + " H 2 = 0,04 m o l - > mzn = 2,6 g a m . . |
'
'•
"
xet: Z n tac d u n g v o i 3 chat o x i hoa la A g % Cu^^ v a
jb
Bao toan electron:
2nz„ = In
15,, j , ,
- inW m M i i X :! ub . .
'
man
;,
D . 3,90.
Phan l i n g hoa hoc:
•
i-OesH
/J ^
*
Cach 1 : nAgNOs = 0/02 m o l ; n c u ( N 0 3 )2 = 0,01 m o l ; n H j = 0,02.
Qach^i
2N^5
^
C.0,65.
-> D a p a n A .
> Mg^z
> N2;
(Zn)
Lai gidi:
^d^twriT
- - ^ ^ = ,C>MM^
= 65
B.1,30.
+ H2SO4
'e,!+
eori ixoffed:.
^,(1 ^„ „ .
A . 2,60.
Zn
Lodgidi:
>^J.:,
-'[^-^t,^'"
tricuamla:
+ Cu(NOa)2
.
M
^ u i ; hiuaf^ a u j
+ 2Ag
0,01
^
J-MJ
19: Cho m g a m Z n vao 100 m l d u n g d i c h g o m Cu(N03)2 0,1M va A g N O a
Zn
\'
> M(NOa)2
-»
= ^
m u o i k h a n . Gia t r i ciia m la
C. 22,20.
,
,. .0 ihA-:> ,(§.'•/ u&A fBr!:> rifiedl < j
,
.
^..^
^^^^^
,
, .,
; ,
2AgNOa
+ 2AgN03
• '
i8n
+
0,01 < -
Zn
B. 21,52.
D . Zn^,^^^^
A g N O a p h a n u n g he't.i-ii
> nAgivjo3
k h i N2 t i n h khiet (dktc). Co can d u n g d i c h sau p h a n l i n g t h u dugc m gam
A. 20,71.
C. Fe
^-^^^^^ ^
tac d y n g v o i d u n g d i c h H2SO4 (loang, d u ) , t h u d u g c 0,448 l i t k h i (dktc). Gia
n^,^_ = 0,12 m o l ; n^ o_ = 0,06 m o l .
NO3
' so]
'
P h u o n g t r i n h i o n n i t gon:
j^.
0,2M. Sau k h i cac p h a n u n g xay ra hoan toan, t h u du(?c chat r a n X. Cho X
= nHNOa + 2nH2S04 = 0,12 + 2.0,06 = 0,24 m o l
T r o n g X:
—
M
Mol:
_ . , „f , ; ' : 8 f t f i | ••
B. C u
Phan u n g hoa hoc:
gidi:
•• 0,1 m o i . ' x ^ i ^
\
Lai gidi:
^
,Rb^.r«i*ir>ofm^,
N^jflM xet: T r o n g cac d a p an, k i m loai c6 k h o i l u g n g m o l I o n nhat la Z n - >
tjii g o m H N O a 1,2M v a H 2 S O 4 0,6M. Sau k h i cac p h a n u n g xay ra hoan toan,
C. 1,792.
f
d u n g d i c h A g N 0 3 0,1M cho den k h i p h a n u n g hoan toan t h u d u g c chat ran
X va d u n g d i c h Y chua 1,89 g a m m u o i . K i m loai M la
V i dv 16: Cho 6,4 g a m bpt C u tac d y n g v o i 100ml d u n g d i c h X h o n hop
B. 1,008.
•
Vx dv 18: N h i i n g m o t la k i m loai M (hoa t r i 2) c6 k h o i l u g i i g 5 g a m vao 200 m l
^DapanC.
rlrrA.1,344,
YACJ .1
NH4NO3.
muoi
> ^jJioyif,
(do H2SO4 phan li) va N O 3 (do KNOa p h a n l i ) nen c6 t i n h chat
+ 8H* + 2NO3
= 0,14.142 + 0,01.80 = 21,52 gam.
--»DapanB.
g i o h g H N O 3 loang —> hoa tan d u o c C u theo p h u o n g t r i n h :
3Cu
+ mNH4N03
lieu y: Cac k i m loai M g , A l , Z n c6 the tac d u n g v o i H N O a l o a n g tao thanh
H2SO4 loang, t r o n g d u n g dich c6 chua
K N O 3 +
mMg(N03)2
'
'^^^^^'
iJ-- .* •
+ + 2 n ^ 2+ + ln„+ = 0,02 + 0,02 + 0,04 = 0,08
nzn= 0,04 m o l
mzn= 2,6 g a m - > D a p a n A . yfcgmib :»6i W^''
'' '
181
[Ml111 MTV UVVM Khang Vi$t
Ca'm nang 6n luyQn thi djii hpc 18 chuy6n dg Hoa hpc - Nguyen Van Hk\
2. D A Y D I E N H O A C U A K I M L O A I
a. D a y d i f n hoa
,, ,
K
Na
•••• Mg
2H^^
H2
C u ^ % _Fe^^
Cu
Fe^""
Al
Ag
., , „ ^ ,
Zn
Fe
Ni ^
Sn ^
Pb ^
+
Chieu ciid phan ung oxi hoa-khu: Phan ung oxi hoa-khu dien ra theo chieu
tao thanh chat khu va chat oxi hoa yeu hon.
+ Do vay, trong day di^n hoa, chat khu cua cap dung trudc se tac dung duoc
voi chat oxi hoa ciia cap dung sau.
IT^
>
Fe^^
>^ Fe^^ + ZFe^*
.
Phuang trinh phan tii:
^"^ 1
M'
Hay:
^«
Cu + 2FeCl3
3: ^ - 2 ^ >
> CuS04 + 2FeS04, j l . '
> C u C h + 2FeCl2
> Ve^* + Ag ^'^''^^^O +
Phuong trinh phan tii: Fe(N03)2 + A g N 0 3
n
Vi dy 1: X la kim loai phan ling duQ-c voi dung djch H2SO4 loang, Y la kim loai
tac dung dugc voi dung dich Fe(NC)3)3. Trong day di^n hoa, c|ip Fe^/Fe^*
diing truoc cap AgVAg. Hai kim loai X, Y Ian lugt la
A. Fe, Cu.
B. Cu, Fe.
C. Ag, Mg.
D. Mg, Ag. ,
Loai B va C do Cu va Ag khong tac dung voi dung djch ^2804 loang.
Y la kim loai tac dung dugc vai dung dich Fe(N03)3 - > Loai D.
182
iV/jfl« xet: Do cap ~—>
^'"'^
~2+
Fe
Bo '
M^di .Q
+ 2FeCl3
> 3FeCl2 '^'^'^^
0 01 -> 0 02 '
Mol-
P^^"
^
voi muoi FeCb truoc.
' ^^^^^^^^
+ ^'^^^ '-^'^^ "^^"^^^^
+ CuCl2
° "-^
° ^^•^^
'^
'^
> FeCl2 -H CuW^^S'^^'*^^^:'^^^^^^'^^'^^^'''^^^
•H'^^/^^'^
^ T "
'
Chat ran gom: Fe du = 0,01 mol; Cu = 0,01 mol ^ m = 1,20 gant.
^DapanD.
Vi dv 3: Cho cac phan ling xay ra sau day:
A. Mn^^ H^ Fe3^ Ag\. Ag^ F e ^
C. Ag^, M n ^ H*, ¥e^.i ''",
^*
Mn^\
,'
'"
" F.(*02)S9^ + u D
'
AgN03 + Fe(N03)2
> Fe(N03)3 + Ag (1)
Mn+2HC1
>MnCl2 + H2 f - ^ S t ,
(2)
lO.Q
10,0
Odf.ro^-fnv
^'^^ J"^*'
\
D. Mn^\^ Fe^\
•'['
Laigidi:
Nhqn xet: T u (1) -> Cap
Fe
Tu (2) ^ cap
• , Mn
> Fe(N03)3 + A g
j^to,,.
D.1,20.
COM.
Laigidi:
y ntio) mod riBi s o H :?! u h
np^= 0,03 mol; npecis = 0,02 mol; ncuClz = O'Ol "^ol.
j ,Qg;|.| Vi^j^ y n u l
v^^v
^ Tinh khii Fe^^ > A g ; tinh oxi hoa Ag^ > Fe^*
—> Co xay ra phan ling: Fe^* + Ag^
:\ luv't
Day cac ion dugc sap xep theo chieu tang dan tinh oxi hoa la
> Cu^* + ZFe^*
Cu + Fe2(S04)3
> FeS04 ++..««22
> Cu(N03)2 + 2Fe(N03)2
CU + 2Fe(Na)3
' dV 2: Cho 1,68 gam hot Fe vao 100ml dung dich chxia FeCb 0,2M va CuCh 0,1M.
Mol: aoi <- aoi
, Fe + Fe2(S04)3
> 3FeS04; Fe + 2Fe(N03)3
> 3Fe(N03)2 j ^ ' ! ^
Cv^+ Fe^*
Vt du 2: — — > — — -» Tinh khii Cu > Fe^^; h'nh oxi hoa Fe^* > Cu^\ '
Co xay ra phan ling: Cu + 2Fe3*
+ H2SO4
•
^, . . . . „
_
_ .
-> Tinh khii Fe > Fe^*; tinh oxi hoa Fe^* > Fe^*
F?
Co xay ra phan ling: Fe + 2Fe^*
Phuong trinh phan tii:
Pe
pg
Fe
ran. Gia tri ciia m la
^0,56.
B.1,12.
Tir trai sang phai, tinh khu ciia cac kim loai giant dan, con tinh oxi hoa ciia
cac ion kim loai tang dan.
c. S u dung day d i f n hoa
^'^
'uui
jt.j ^ a u i i . 4 M A n.,:
Fe^*
Dap an A. Cac phan ung hoa hgc:
iChuay deu tai phan ling hoan toan thu dugc dung dich X va m gam chat
^•
- -.-'.o--nfiriq i r i i n9b !ifb;MI,0(iOH§A r b i b
b. Y nghia cua day di^n hoa
Vtdu 1:
^
diing truoc cap — ^ .
Ag
diing truoc cap
'
^ J
4?-H2 , ^ v( 63 cOVIttifXfi irlM :iVt
Thu tu trong day di?n hoa: - ^ > - ^ ^
-> Tinh oxi hoa tang dan theo thii ty: Mn^* <
^
< Fe^" < Ag*.
'
V
'
Dap an A.
'
.
dy 4: y j tri ciia mgt so cap oxi hoa-khii dugc sap xep nhu sau: Fe^^/Fe,
Cu^VCu, Fe3VFe2^ AgVAg, CI2/2CI-. Trong cac chat sau: Cu, AgN03, CI2,
chat nao tac dung voi dung dich Fe(N03)2:
A. Cu, AgNOs, CI2.
B.AgN03,Cl2.
C CI2.
D. Cu, AgNOs.
183
dm
__^[yih_«a^iai
nang On luyjn thi dgi hqc IS cliuyen ae M O J
LOT ^J'ai;
Lww y: Fe(N03)2 c6 the tac d\ing vai ca cha't khvr va cha't oxi hoa. Dieu ki^n Q
day la chat k h u diing truoc Fe^VFe; chat oxi hoa dung sau Fe^^/Fe^*.
, Theo bai cho: Khong c6 cap nao dung truoc Fe^VFe va c6 hai cap dung sa^
}, Fe3"/Fe2* —> Dap an B.
Cac phan ung hoa hoc: Fe(N03)2
6Fe(N03)2
V i du 5:
dung
dugc
' khan.
+ AgN03
> Fe(N03)3 + A g
> 4Fe(N03)3 + 2FeCl3
+ 3Cl2
Hoa tan hoan toan 9,92 gam h6n hgp bot Fe304 va Cu trong 160ml
dich H2SO4 I M (loang). Sau khi cac phan ling xay ra hoan toan, thy
dung dich X (khong con axit du). Co can X thu dugc m gam muoj
Gia tri ciia m la
A . 19,68.
B. 16,64.
C. 10,72.
D. 23,36.
"
'
Laigiduqr
.
.^ij'
G9isomol:Fe304 = a;Cu = b.Theobai:232a + 64b = 9,92. 1 ^ . fQ^O i.
Nhan xet: Chi c6 Fe304 phan ung true tiep v6i axit. Cac phan ling hoa hgc:
Fe304 + 4H2SO4
>
FeS04 + Fe2(S04)3 + 2H2O
Mol: a04 < - 0,16
^
0,04
a = 0,04 ^ b = 0,01mol.
Cu
;
- > 0,01
„
->
0,01
0,02
'•I/^nSf^'rOM;:
m = 0,01.160 + 0,03.400 + 0,04.152 = 19,68 gam - > Dap an A .
{jg
r1
V i d v 6: Hoa tan hoan toan 1,12 gam bgt Fe t r o n g 100ml dung dich gom
HNO3 I M va H2SO4 0,3M thu dugc dung dich X. Dung dich X c6 the tac
dung voi toi da m gam bgt Cu. Biet k h i N O la san pham k h u duy nha't ciia
N*^ trong cac qua trinh tren. Gia trj cua m la
A . 1,92.
B.2,56.
C. a64.
'
3Cu
+
'Mol: a03 ^
Cu +
8H*
'
+ 2NO;
0,08 ^
> 3Cu2* + 2NO + 4H2O
0,02"'^' ^ '
2Fe3*
Mol: 0,01 < - 0,02
0,03
''^•^
""^^^
T ^ ' ^ - -^-^^-''^
> Cu^* + 2Fe2*
,
s n u b IrnnCS. oftv Ife rtirsH} iDrn -^miuV-A
ri>^>
m = 0,04.64 = 2,56 gam
Dap an B.
Vi dV 7: Cho 2,24 gam Fe vao 200 ml dung dich gom NaN03 0,3M va H2SO4
0,6M, thu dugc dung dich X. Cho mot thanh Cu vao X, sau Idii cac phan
^ g hoan toan, thay khoi lugng thanh Cu giam m gam. Gia tri ciia m la
A. 3,2.
B.6,4.
C.4,8.
D. 8,0.
,
Laigidi:
'
..
.
,
,
*
f j B r.u':> n i i;>t,; ' i .
n„+=0,24mol;n^^_=0,04mol..
Fe
Mol: 0,04
+ 4H^ + NO30,16
' ' ar^O ^3
"
> Fe3^+ N O + 2H2O
0,04
•
3Cu
Mol:
0,03
+ 8H* + 2 N O i
0,08
Cu + 2Fe3^
Mol: 0,02
0,04
> 3Cu2* + 2NO + 4H2O
0,02
>
'
0,04
X CO chua: n^ 3+ = 0,04; n„+ = 0,08; n^,^_ = 0,02; Na^ va SO^''''**^ '
iu,u;i<,
^
..^ , . . a ^ j j 061 f:
.anBC}fcf'
> CuS04 + 2FeS04 -Xft 'M(f^ei^5 oriS :1E
+ Fe2(S04)3
Mol: 0,01
0,04
r
Cty TNHH MTV D W H Khang Vigt
^1
Cu2* + 2Fe2^
•
'
itHOsV.
£10',./,'
-
.H
3"^'^- "^'"-J''
- - •-
'
-> ncu = 0,03 + 0,02 = 0,05 mol - » mcu = 0,05.64 = 3,2 gam ^ Dap an A.
...y^
3. B A I T A P O N L U Y E N
airiqr'^iiJ.'ib**.•uec^ .M i .u. (UH^:/. •.
Bai 1: Dot chay hoan toan 7,2 gam kim loai M (hoa tri 2) bang mot lugng vira
dii la 0,25 mol hon hgp khi CI2 va O2, thu dugc 23 gam chat ran. K i m loai M
la
D. 3,20.
A. Be.
B. Cu.
C. Ca.
'»T->rr^D. M g .
Bai 2: Dot nong 1,40 gam hon hgp X gom Cu, Zn va M g trong khi oxi, thu dugc
nFe=
56
=0,02 mol.
- -
*^. , f
' \..U:v^,:
Nhqn xet: Khi axit HNO3 c6 mat dong thoi vai axit H2SO4 loang thi lugng
H * trong dung djch la do 2 axit phan l i ra -> Giai theo phuang trinh ion.
V
= nRNOg + 2nH2S04
= 0,1 mol; n
n
i i.x^j'irtntf •<
TIRIJ
Phuang trinh ion nit ggn:
Mol:
Fe
+
4H*
0,02
<-
0,08
./^
+ NO;
0,02
^ Fe^
->
+ N O + 2H2O ,,
0,02
K h i d i o X tac dung vai Cu: 4 5 | , D . i . d D , d ^ ^ , ,
djch H2SO4 2M. Gia tri cua V la
A. 40.
B.30.
_
C.20.
„,
D. 50.
^ai 3: Cho 1,68 gam kim loai R (hoa tri n khong doi) tac dyng voi 0,025 mol O2,
thu dugc chat rMn X. Hoa tan het X trong dung djch HCl, thu dugc 0,448 lit
= 0,1 + 2.0,03 = 0,16 mol
2- = 0,03 mol.
2,04 gam hon hgp Y gom 3 oxit. De hoa tan het Y can toi thieu V m l dung
•*
,i
j^Su
''
%iinilv
,,JA,IJ3.
khi H 2 (dktc). K i m loai R la
A. Ca.
B. Zn.
cU,u ,, j , .
C. A l .
D. M g .
4: Hoa tan het 20,88 gam mgt oxit sat bang dung dich H2SO4 dac, nong thu
^irgc dung dich X va 3,248 lit khi SO2 (san pham k h u duy nhat, 6 dktc). Co
X, thu dugc m gam muoi sunfat khan. Gia trj cua m la
A. 52,2.
B.54,0.
'
C.58,0.
D. 48,4.
''
185
Cim
nang On luy^n thi d?i hqc 18 chuy6n <3i H6a hpc - Nguygn Van HSi
Cty TNHH MTV D W H Khang V i j t
Bai 5: N g a m m o t d i n h sat t r o n g 200ml d u n g d i c h C u S 0 4 a mol/1. Sau k h i
p h a n l i n g ke't thiic, lay d i n h sat ra k h o i d u n g d i c h , rira sach, say k h o v a
d e m can, thay k h o i l u g n g d i n h sat tang 0,8 gam. Gia t r i cua a la
A . 0,50.
B. 0,25.
C. 0,30.
l^han xet: Bao toan k h o i l u g n g : m o j = 2,04 - 1,40 = 0,64 g a m
Vay: n o = ^
D . 0,20.
Bai 6: N h i i n g m o t thanh sat vao 200ml d u n g d i c h g o m Cu(N03)2 0,2M va
A g N O a 0,1M. Sau m o t t h o i gian lay thanh k i m loai ra, riia sach l a m kho,
B. 1,40 gam.
C. 1,68 g a m .
D . 1,96 gam.
Bai 7: Cho h o n h g p hot g o m 1,3 gam Z n va 2,8 gam Fe vao 400ml d u n g dich
CuS04 a mol/1 sau p h a n u n g hoan toan t h u d u g c 4,4 g a m chat r a n X g o m hai
k i m loai. Gia t r j cua a la
A . 0,20.
'^"^
trong axit tao t h a n h H 2 O theo p h u o n g t r i n h :
+
02M o l : 0,04
2H0,08
- » nH2S04 = ^ " H *
. S0,0
>H20
"
tO,0 :lo„M
Ou-fj*->^.i,:,^,.f-,viv=t'^''i..^^^^^^
= 0-04 m o l
^
VH2SO4 = ^
' c^q
= 0,02 l i t = 20 m l
^
''^^'^
B.0,30.
C.0,15.
" c O i / b . aiO.
" 'ir
Bai 8: Cho h o n h o p X g o m Fe203, Z n O va C u tac d u n g v o i d u n g dich H C l (du)
t h u d u g c d u n g d i c h Y v a phan k h o n g tan Z. Cho Y tac d y n g v o i d u n g dich
N a O H (loang, d u ) t h u d u g c ket tiia
A . Fe(OH)2, C u ( O H ) 2 va Z n ( O H ) 2 .
B. Fe(OH)3 va Z n ( O H ) 2 .
C. Fe(OH)2 va Cu(OH)2. )MS +
D . Fe(OH)3.
- 'HH + uDt
t h u d u g c d u n g d i c h Y chua m gam m u o i va k h i N O (san p h a m k h u duy
nhat). Gia t r i ciia m la (bie't cap Fe^VFe d u n g t r u d c cap Fe^VFe^*) .
B.24,2.
C. 29,04.
_> D a p an C
^_
j
g ^ j „ eO,0.de • ( B + f0,0).d5 • \^^jta ' -
Nhan xet: R Ian l u g t tac d u n g voi 2 chat oxi hoa la O 2 va H C l (H*).
—> n , = 4 n o 2 + n H c i = 4 n o 2 + 2nH2 = ^'^ ^ ^'04 = ai4.
1,68
.
Bai 9: Cho 0,12 m o l Fe tac d u n g v o i 0,4 m o l axit H N O 3 t o i p h a n u n g hoan toan,
A . 25,32.
n ^ 2 - = a04 m o l .
,.,)./,„,An
K h i cho o x i t k i m loai tac d u n g v o i axit, i o n O^" t r o n g o x i t se ket h g p v o i H *
thay khoi l u g n g thanh sat tang 1,76 gam. K h o i l u ^ n g sat da p h a n u n g la
A . 1,12 g a m .
= 0,04 m o l ^
;
,, _
,^yj.
,^. ,
n = 0,14
R = 12n. V a y n = 2 va R = 24 ( M g ) ^ D a p an D . ^ ^^^'i^
\!iii:t .
^ uJ
^0
Bai 4:
Nhqn xet: O x i t sSt FexOy tac d u n g v a i H 2 S O 4 d^c, n o n e
SO2
t h i o x i t la FeO
hoac Fe304.
Cac e m lu'u y r a n g 1 m o l F e O hoac Fe304 deu chua 1 m o l Fe^* nen d e u c6
D . 10,80. '
Bai 10: Cho h o n h g p bgt g o m 0,27 gam A l v a 0,56 g a m Fe vao 550 m l dung
d i c h A g N O s 0,1M. Sau k h i cac phan u n g xay ra hoan toan, t h u d u g c m
kha nang n h u o n g 1 m o l electron, d o do:
3 248
- n F e , O y = 2nso2 ^
' ^0,0»j^oa./:)« V
Hp^^Qy = 2 - - ^ = 0,145 m o l
qsG <
.3
g a m chat ran. Gia t r i ciia m la (bie't cap Fe^^/Fe^* d u n g t r u a c A g V A g ) : •
{ A . 5,94.
B. 6,48.
C. 3,24.
4. H I / 6 N G D A N - L O I G I A I
-BD ^
B^j 1.
ffV vr*i r^'?'--'' Y nrirf
"D.
M F e , O y = 7 r - ^ = 72 (FeO).
5,40.
.9^ >
" r V f r n E § Ot^,| § n 6 n i o Q :S'
Nhqn xet: Bao toan khoi l u g n g : 7,2 + m o j + n i c i j = 23
M'
0,05
+
M o l : 0,2
Ch
0,2
g i
'" > M a 2
4h^nu^^nMih'i'mo
,fis:.a
,>t>{t>) , H id
.o./
i^^^^^
- > i i M = 0,2 + 0,1 = 0,3 m o l ^ M = ^ - 24 ( M g ) .
^ ^ i ^ 8wb,t>9^j?
. . . . . j - i
...J)>?.
,
.u^,^ ; s •ygxih udl
m,
-> D a p an D .
186
j '
.Q.8e.3
Xl,^ M
~-
> Fe2(S04)3 x - 20,0)d5 = x.m :DBrW
mFe2(so4)3= 0,145.^.400 = 58 gam ^ D a p an C.
IG} '
Onh^hQ
<^ ^
^ ^P^^ "Of^ ^^J^
^ 32x + 71y = 15,8.
M a t k h a c : x + y = a 2 5 x = a05; y = a20.
- ( " ^ ^ .
i^;^^:
,s.t
„
.:„,. .,.
fib{i)n rsol mix nih^ O i i , ! o i r O ' t f
.jj,,...
2 M + 0 2 - ^ ^^o^fy-WnBii^m
-''^''^^
M o l : 0,1
B a o t o a n n g u y e n t o F e : 2FeO
"
.£,£6./
^ach 1: Giai theo p h u o n g t r i n h hoa hgc:
Fe
+
- -^.^
> FeS04 + C u
CuS04
^
...v. ,r>,u a2a
M o l : 0,2a < - 0,2a
,,..3
j^pjc
,
^
O'lmol
> ImolCu
t
- > K h o i l u g n g tang 8 g a m .
~* " C U S O 4 " F e = 0,1 m o l - > 0,2a = 0,1
. j
.
.... _
^
,
^yj^^^
,
"^n 2: T a n g g i a m k h o i l u o n g :
ImolFe
,^
^^^^ ^ i l u 3 : l . u .
^ 6 1 l u g u g d i n h sat tang = 64.0,2a - 56.0,2a = 1,6a
1,6a = 0 , 8 a = 0,5 ^ D a p an A. ,
+ fO-
OHd
,
gmih
tang as gam.
a = 0,5 ^ D a p an A .
,
i oi1
xivui'J
^ qj.<-f <
Cty T N H H M I V
Ca'm nang On luy^n thi dgi hpc 18 chuy6n dg H6a hqc - Nguygn V a n H&\
D V V H Khang V i j t
Bai6:
nAgN03 = 0,02 m o l ; ncu(N03)2 =
• >l
mol.
xet: T i n h o x i hoa A g * > Cu^* —> A g N O s p h a n l i n g t r u o c r o i mod de'ti
NMn
Cu(N03)2. \
Phan l i n g hoa hQC:
•'\„^,„^,:''
• ; •;,
^„
,
O /loi Jixr, r r i , j^nub 'ifiJ ihOl mix' ii/o orJ'j i ,
M o l : 0,01 < - 0,02
Fe
Mol: a
<—
a
a
mtsng
Theo bai: m Fe wng = 1,76 gam - > 1,6 + 8a = 1,76
= (64-56)a = 8a gam.
Bai7:
„
Nzn = 0,02 m o l ;
npe
"
. ^
.
, $^1
M o l : 0,02
^ '
1
Fe
Mol:
^
^
X
+
0,02
mx =
Cu
.,
X
mFedu+ mcu
^
X
^,
,
0,4a = 0,06
nFe(pu)
+
> A1(N03)3
^ ^, ^ , ^ ^ ^ ^ ^
,
x = 0,04; a = 0,15.
2+
= 2n
> Fe(N03)2 + 2 A g
0,01
0,02
> xay ra p h a n u n g :
> Fe(N03)3
+ AgN03
0,005
^
Cu
> 2FeCl3 + 3H2O
6HC1
> ZnCh
^
+ 2FeCl3
> CuCh
o
i
+ H2O
nr> " r
.iieiO;:
^.
an
C.
+ Ag
Q^Q05
(3)
,M.0Om:flA
'
"''^***'^'
-£nS <"H noi •
'
^
'
'J I * - ififi
(r)
9£ - n \
• ii'j ijii
- 'H .
•~;6dri n^ib riom njl B I VGX '§b n?i>! uyrb lit,
"Hb Ififb hvfl. ri-jtb j^nub JOfirt g r f i b o^v gnurfn ivjv*
.jfcol miA nbm ah j^jidrii rf')'- •
) = 66,;
FeCls he't.
"'^ ''"^'^ <^^^^'^
Hai ket tua la Fe(OH)2 va Cu(OH)2.'"
Lieu y 2: Zn(OH)2 luong tinh nen da tan trong N a O H du.
Dap
(a-ifoo^OBiH
r j ^ u p qert nfiij^ 'jfiori) u s d n i ^ v q ^ h Dt^it 3uy qaiJ (£) ^ucff^ ohn >
I'.
+ 2FeCl2
Y gom: F e C h , C u C h va Z n C h .
C h o Y + dung dich N a O H d u ^
—>
—
;--):»tiMV
ibiV
»
"', •
tac dung dugc voi muoi FeCla vua tao thanh:
Chat ran Z la C u con d u
^
(2)
i
Lm< y; Q u a n t r o n g nhat la cac e m can viet d u o c p h a n u n g (3).
nio1
^^j^^ ^^
V _
+ 2HC1
Lieu y l:Cu
M^if^ffOitt^Q^mi
0,03 ^^n^^ni^^if^'mm-m
N/ian xet: K h i cho X + H C l d u t h i CO 2 chat tac dving true tiep:
ZnO
nj.
-> 0,02 + x = 0,4a.
-> Dap an C.
+
X)^ md mm
+ 3Ag
^
Chat dn la A g = 0,055 m o l ^ m = 5,94 gam.
= x m o l . Bao toan electron: 2n2;n + 2npg(pf,^
M a t khac: mx = 56(0,05 - x) + 64.0,2a = 4,4
Fe203
m^ji noni n/
^Xffit>
.r.oii r •••
3AgN03
Mol:
a = 0,15 m o l .
fifjiiH
-> D a p an A.
-> 2.0,02 + 2x = 2.0,4a
Bai 8:
, kith
.
:m\\ a
Fe(N03)2
x = 0,04 m o l .
-> D a p an C.
Cach 2: G p i :
r^jp srisi/
j
M o l : 0,01 - > 0,03
•
= (2,8-56x) + 64.(0,02 + x) = 4,4
ncuS04 = 0/02 + X = 0,06 ^
'
0,06
Fe'^"^
Ae'*'
Nhdn xet: D o cap — j — > —
Fe^""
Ag
Cu
—>
0,04
VIOMM^,
«mf i n mtiA' .f,
u u . i m f i eriq v >
>3Fe(N03)2
Fe + 2 A g N 0 3
M o l : aOl
0,02
0,02
> FeS04 +
CuS04
<—
+
2Fe(N03)3
_ > m = 0,06.242 + 0,06.180 = 25,32 gam.
Al
> ZnS04
+
M o l : 0,02
= 0,05 m o l .
CuS04
A O J M m
n A i = 0 , 0 1 m o l ; nFg=0,01 m o l ; nAgNOs = 0,055 m o l . •
„
g o m C u va Fe (du) - > C u S 0 4 het.
+
0,1
Bai 10:
'
Cach 1: Nhan xet: T i n h k h i i Z n > Fe - > Z n p h a n u n g t r u o c r o i m o i den Fe —>• X
Zn
0,4
_->. Fe d u = 0,02 m o l v a tiep tuc p h a n u n g v o i Fe(N03)3:
—> D a p an A.
a = 0,02 m o l .
= 56.(0,01 + a) = 56.0,03 = 1,68 gam -)• D a p an C.
mFe(pu)
MA
> Fe(N03)3 + N O + 2H2O
Sau p h a n u n g : Fe(N03)3 d u = 0,06 m o l ; Fe(N03)2 = 0,06 m o l .
Fe(N03)2 + C u
^'^'^^
'
+ 4HN03
M o l : 0,1 < Fe
0,02 ->m.ang = 2,16-0,56 = 1,6 gam.
Cu(N03)2 — >
+
Fe
> Fe(N03)2 + iZAg »
Fe ' + 2 A g N 0 3
->
Cac p h a n l i n g hoa hoc:
r i
iriq j\w
,
,t,A..<.M:>^
'Ml'.'!"
^
grtorf-t fn*.- ,5
! nri>l
'•>'' '
Ca'm nang 6n luyCn thi dai hpc 18 chuy6n dS H6a hpc - Nguygn Van HSi
Cty TNHH MTV DWH
C h u y e n de 8
+
AN M(m
V i d v , de bao v? v 6 tau bien bang thep, n g u d i ta gan vao v6 tau (phan c h i m
^-^^
g r m nsffq oyt qsii 6v lorn S0;0 = t i h •
'
Phuang phdp dien hoa:
No'i k i m loai can bao ve v o i 1 tarn k i m loai khac c6 t i n h k h u m a n h h o n .
K I M LOAI - IHltU €Hlt K I M LOAI
1. A N M O N K I M L O A I
a. K h a i n i ? m
-'
Khang Vijt
t r o n g n u o c bien) n h i i n g tarn kem. K h i tau hoat dong, tarn k e m b i an m o n
dan, v 6 tau d u p e bao v^. Sau m p t t h o i gian n g u a i ta thay cac tarn k e m khac.
Su pha h u y k i m loai hoac h o p k i m do tac d u n g hoa hoc cua m o i truorig
x u n g q u a n h goi la s u an m o n k i m loai.
b. Phan loai
+
VI D U
^-''("OVi'v-r^
-.vrHf
rtsrfr:
MAU
V i dvi 1: K h i d i e u che h i d r o t u Z n va d u n g d i c h H2SO4 loang, neu t h e m vao
An m o n k i m loai d u g c chia t h a n h 2 loai chinh: A n m o n hoa hoc va an m6n
m p t v a i gipt d u n g d i c h CuS04 t h i tha'y k h i H2 thoat ra m g n h h o n . Ban chat
d i ^ n hoa.
cua h i ^ n t u p n g t r e n la
•„-•!•>
• ^ .i s-:
•
- f/.'^
-
>5
^
ff'^v*^ 4'^'^
An mon hoa hoc:
A. A n m o n dien hoa.
A n m o n hoa hpc la sy pha h u y k i m loai do k i m loai p h a n u n g hoa hoc v6i
C. A n m o n k i m loai.
^^"t"'
^' • ' '
'
^'
B. A n m o n hoa hpc.
'i'i - -
D. A n m o n hpp k i m .
axit hoac cac chat k h i (hoi) 6 nhiet dp cao.
Dac d i e m ciia an m o n hoa hoc la qua t r i n h oxi hoa-khu, electron dupe chuyen
Vidu\
, ,.s;Vt«nf
Fe +
H2S04(io3ng)
3Fe
+
+ 2O2
.^'-S'iffW
•
+€^q
-
'v .,gA^ ^ " ^ ^
'" > Fe304
H2
.£0.0 •<-^-10,0 ;{o:
> FeS04 + H2
—> T i n h o x i hoa Cu^* > H *
Cu
nen k h i t h e m CuS04 se xay ra ngay qua t r i n h :
Zn
' ' "
+ CuS04
>
*
^ x i ii^
ZnS04 + C u
-
nkqkQ
^
C u thoat ra b a m tren hat Z n nen thoa m a n 3 d i e u kien ve qua t r i n h an m o n
An mon dim hod:
d i f n hoa: Co h a i k i m loai khac n h a u ( Z n - Cu), tie'p xuc true tie'p v o i nhau
A n m o n d i ^ n hoa la s u pha h u y k i m loai do k i m loai tie'p xiic v o i d u n g dich
(Cu b a m vao Z n ) va c u n g n h u n g t r o n g d u n g dich chat dien l i (H2SO4 loang).
chat d i e n l i tao nen d o n g electron chuyen d a i t u cue a m sang cue d u o n g va
D o do xay ra qua t r i n h an m o n dien hoa —> D a p an A.
phat sinh d o n g d i f n.
Co che an m o n d i e n hoa:
Vi
•.^_.,f^y.y..:M.._y.^;_,^^
v \. <» i-^
. ., ., ...^ ^
.
2: T r u o n g h p p nao sau day xay ra an m o n di§n hoa:
B. K e m de t r o n g d u n g d i c h H2SO4 loang.
C. Sat chay t r o n g k h o n g k h i .
> Zn^*
D. V 6 tau bien bSng thep n g a m t r o n g nuoc b i e n . ^
Tai cue d u o n g (eatot), i o n H"^ chuyen den be m a t C u de n h a n electron t u cue
a m chuyen t o i va b i o x i hoa tao thanh k h i H2:
2 H * + 2e
>
H2
Ba d i e u k i ^ n de xay ra an m o n d i f n hoa:
8""*^
'
^mi>nkdaoBoo.r
Laigidi:
Loai A, C: T h e p de t r o n g k h o n g k h i kho n o n g hoac dot chay s^t t r o n g
khong k h i - »
(1) Co hai k i m loai c6 ban chat
^- '
,
Tai cue a m (anot), Z n ( k i m loai m a n h hon) b i o x i hoa:
,
"'^"^
A. Thep de t r o n g k h o n g k h i kho, nong.
V i d u CO 2 tarn k i m loai Z n va C u dup-c n h u n g vao d u n g d i c h H2SO4.
Z n - 2e
,
> ———
Nhqn xet: T r o n g day d i e n hoa, cap
t r y c tie'p t u chat oxi hoa deh chat k h u len k h o n g phat sinh d o n g di§n.
D e u bj an m o n hoa hpc.
SOB^E
L o a i B: K e m de t r o n g axit H2SO4 loang
EJJ9>1£
K e m bj an m o n hoa hpc.
•
•
i
khac n h a u , (2) tie'p xiic t r y c tie'p v a i n h a u (hoac gian tie'p qua day dan) va (3)
—> D a p an D .
dupe n h i i n g vao c i m g m p t d u n g dich chua chat d i ^ n l i .
Luu y: V 6 tau bang thep la h p p k i m Fe - C (chua 2 d i | n cue c6 ban chat khac
c. C a c h chong an mon k i m loai:
nhau va tie'p xiic t r y e tie'p v o i nhau) va ciing tie'p xiic v a i d u n g d j c h d i ^ n l i
+
(nuae bien) nen xay ra an m o n dien hoa.
Phuomg phdp bdo ve be mat:
D i m g n h u n g chat ben v o i m o i t r u o n g p h u len be m a t k i m loai. D o la:
:,;
V i d v 3: Cho cac h p p k i m sau: Cu-Fe (I); Zn-Fe (II); Fe-C (III); Sn-Fe (IV). K h i
D u n g son cho'ng gi, vecni, dau m o , trang men, p h u h p p chat poiime.
tie'p xiic v o i d u n g djeh chat dien l i t h i cac h p p k i m m a t r o n g d o Fe b i an
M a m p t so' k i m loai ben n h u c r o m , niken, dong, kem, thie'c len be m a t kif^
m o n t r u o c la:
A. I , I I va I V .
loai can bao ve.
B.I, II vain.
C.I,IIIvaIV.
D. I I , I I I va IV.
1Q1
Cty TNHH MTV DVVH Khang Vi$t
C^m nang 6n luy^n thi dgi hpc 18 chuySn 66 Hoa hpc - Nguygn Van HSi
Lai gidi:
^
Nhan xet: Trong cac qua trinh an mon dien hoa, k i m loai m ^ h han se bi an
mon —> Fe bi an mon truoc khi no la chat k h u manh hon (dong vai tro la
anot - cue am).
Trong cac cap (I), (III), (IV) thi Fe deu manh han, rieng cap (II) thi Zn manh
hon.
•^DapanC. , .
.
,
• , .:>;;.f:,,,, , ,,•
V i di^ 4: Gang bi an mon dien hoa trong khong khi am. Qua trinh kliir xay ra tren
•n- be mat aia gang la: ,fe^,^|,.,B,M>i.#«i:
A. O2 + 2H2O + 4e
C. Fe
> 40H- '
-)1»ft»l^s*&i^.»v:t<,r
B. 2H2O + 4e
> Fe2* +2e
D. Fe
> O2 + 4 0 H -
> Fe^* +3e
Lai gidi:
'
Thep la hop k i m Fe-C (vdi mot so nguyen to'khac). K h i tie'p xiic voi khong
khi am (c6 chiia chat di^n li) se xuat hi^n qua trinh an mon di?n hoa. Fe la
cue am (bi an mon), C la cue duong se xay ra su k h u
O2 + 2H2O + 4e
> 40H-
^nfi^YGx6«<)au^frl9.flj^/l>tm-•
—> Dap an A.
Luu y: Cac em de chon riham B v i bo qua vai tro cua chat oxi hoa la oxi
khong khi.
V i du 5: Co 4 dung dich: a) H C l , b) CuCh, c) FeCb, d) H C l c6 Ian CuCh.
Nhiing vao moi dung dich mpt thanh Fe nguyen chat. So truong hop xuat
hien an mon dien hoa la:
A.O.
,
B. 1.
C.2.
D.3.
yu,<)(i!M"
Laigiai,
Nhan xet:
+ Khi nhiing thanh Fe vao cac dung dich H C l va FeCk, Fe bi an mon hoa hpc
theo cac phan ling:
gr Fe + 2HC1
Fe + 2FeCl3
> FeCk + H2
'
> SFeCk
+ K h i rihiing Fe vao dung dich CuCk se CO qua trinh:
Fe + CuCk
.:o,r;
0
> Cu + FeCk
Jfi
K i m loai Cu sinh ra bam vao Fe tao thanh cap di#n eye nhiing trong dung
•'
dich dien l i la CuCk nen xay ra an mon dien hoa.
+ Tuong t u voi dung dich H C l c6 l l n CuCk, Fe se day Cu theo phan ling:
ifl Fe + CuCk
> Cu + FeCk
m, —> Cu bam vao Fe —> Xay ra an mon dien hoa.
—> Dap an C.
192
p l E U CHE K I M LOAI
phtfcng phap nhif t Iuy?n
}Chdi niem:
IsThiet luy^n la phuang phap dieu che kim loai bang each dung cac chat khu
nhu C, CO, H2, A l de khvr oxit kim loai thanh kim loai a nhi?t dp cao.
pham vi dp dung
• '
'•'
Chat khu la CO, H2 se khvr dug-c cac oxit cua kim loai trung binh va yeu
nhu: Oxit sat, (FeO, Fe203, Fe304), oxit dong, oxit chi (khong khu dugc cac
oxit kim loai kha manh nhu: AI2O3, MgO...).
Chat khu la C: Khu dugc cac oxit cua kim loai tir Zn va tao thanh khi CO.
Chat khu la Al: khu dugc cac oxit ciia kim loai tu Cr.
. Phtfcng phap thuy luy?n
-i,
Thuy luyen la phuang phap dieu che kim loai bang each dimg kim loai
m^nh hon day kim loai yeu hon ra khoi dung dich muo'i.
hieu y: Do'i voi cac kim loai kiem va kiem tho (tru Be, Mg) khi tac dung vai
dung dich muoi, chiing se tac dung vai nuoc -> Dung dich baza, bazo tao
thanh c6 the tac dung tie'p voi muo'i:
Vi du cho N a kim loai vao dung dich CuS04 xay ra cac phuang trinh:
2Na + 2H2O
> 2 N a O H + H2
2NaOH + CuS04
Bien phan nong chay
Day la phuang phap dieu che cac kim loai manh. Cac kim loai kiem, kiem
tho thuang dieu che bang phuang phap dien phan nong chay muoi
halogenua; rieng Al dugc dieu che'bang phuang phap di§n ph?n nong chay
AkOs voi duang cue bang than chi.
MgCk _iP2£_> Mg + C k
2Ak03
o
:;iV~H^M;
''P"" > 4A1 + 3O2
Dien phan dung dich
Day la phuang phap dieu che cac kim loai trung binh va yeu tu cac dung
dich muo'i tuang ling ciia chiing.
iLuong hap 1: Goc axit tham gia di^n phan:
CuCk _ i P ^ Cu + Ck
Iruang hop 2: Goc axit khong tham gia dien phan:
CuS04 + H2O —
;W
> Na2S04 + Cu(OH)2^
Phttong phap di^n phan
u + -O2 + H2SO4
2
4AgN03 + 2H2O — ^ 2 ^ 4Ag + O2 + 4HN03
C
'
193
\jt.y
Ca'm nang 6n luyen thi dgl hoc 18 chuyen d8 H6a hpc - Nguyin Van Hi\
D i n h luat Faraday ( t i n h lup-ng chat t h u d u g c 6 cac d i | n cue): m =
m: K h o i l u g n g chat t h u dug-c a d i ^ n cue
Lm
Alt
FexOy
(gam)
n: So electron trao d o i m a ion d a n h a n de t^o t h a n h n g u y e n t u (phan tu)
t u o n g u n g 6 d i e n cue.
i
...m,-
' . . ^ 1 .
Lieu y: T u cong thue d i n h luat Faraday, so mol electron
.
v
trao d o i t r o n g qua
t r i n h d i ^ n p h a n d u o c t i n h theo eong thue: n g = — .n = —
Day la cong thue giiip giai nhanh chong nhieu bai tonffcfien
ViDVMAU
phan.
• ,^.nn
n u n g a nhiet d o eao. Sau p h a n u n g h o n hgp ran con lai la:
V , rfn-
Lai giai:
Nhan xet: D a y la each d i e u che k i m loai bang p h u a n g p h a p n h i f t luyen. 6
n h i ^ t d p cao, k h i H 2 C 6 the k h u cac oxit k i m loai d i m g sau Z n .
6 bai nay, k h i H2 k h u dupe cac oxit CuO, Fe203 t h a n h C u , Fe; k h o n g khir
V i d^ 2: Cho l u o n g k h i C O (du) d i qua 12 gam h o n h p p g o m Fe203 v a M g O
n u n g n o n g den k h i p h a n u n g hoan toan, t h u d u p e 9,6 g a m chat ran. Khoi
l u p n g Fe203 c6 t r o n g h o n h p p ban d a u la:
B. 16 gam.
a
ax
^'^ = 0 , 1 m o l
56
npe
„,,^ , ,^
2
- =
3
O
y
/;Sv;.„-v-V,
V c o = 0,15.22,4^ = 3,36 l i t - ^ D a p an D .
I
\.H:/sf
i yjnij
?A i; u i
'i;''.*
i>t;..>i
Vrno =
= 3,36
Nhan xet: XXQ ^^xit) = " c o = "002 = 0,15 -> Vco2
3,36 ->• L o a i B v a C.
p h a n u n g ke't thue, lay la k e m ra k h o i d u n g dich, r u a sach, say k h o v a dem
can, thay kho'i l u p n g la k e m g i a m 0,05 gam. Gia t r i cua a la
A. 0,30.
B. 0,25.
C. 0,50.
1 mol Zn
> 1 m o l Cu
giai:
0,05 m o l
<-
r.C/Ag D . 0,20.
'
d
d£
ioi
' • 6C\'..Q)
—> Kho'i l u p n g g i i m 1 gam.
<—
g i a m 0,05 gam.
r i C u S 0 4 = " Z n = 0'05 m o l -> 0,1a = 0,05 ^
a = 0,5 - >
^
D a p an C.
V i dv 5: K h i d i ^ n p h a n d u n g d i c h CuS04 ( v o i d i ^ n eye t r o ) t h i cac qua t r i n h
C. 24 gam.
A. O x i hoa H2O v a k h u i o n Cu^*.
B. K h i r i o n Cu^"^ v a o x i hoa H2O.
C. K h u i o n Cu^* va o x i hoa i o n SO 4 " .
D. K h u H2O va o x i hoa i o n Cu^*.
Lai
D . 12 gam.
2Fe + 3C02
Cyc a m (eatot) eo i o n Cu^* v a H2O
H2O
> Cu
>-O2
2
+ 4 H * +2e
,r
giai:
Cye d u o n g (anot) eo H2O va i o n SO 4
T a n g - g i a m k h o i l u p n g : m o (trong F e 2 0 3 ) = 12 - 9,6 = 2,4 g a m
I o n Cu^* b i d i $ n p h a n :
{Qua trinh khie)
—> H2O b i d i ^ n p h a n :
(Qua trinh oxi hoa)
80,6 " : l o i .
—> D a p an B.
2 4
n o = - ^ = 0,15 m o l
"
16
n
Ghi nha cau tha: A n h la A n o t n h u o n g e, Em la Catot n h a n e t h o i ma.
V i dy 6: D i e n p h a n 800 m l d u n g d i c h A g N O a 0,1M v o i d i e n cue t r a t r o n g t gia,
rnpejOa ~ 0/05.160 = 8 gam -> D a p an A .
V i d\ 3: K h u h o a n toan m o t oxit sat FcxOy a n h i ^ t d p cao can v i r a d u V l i t kl^'
C O (dktc), sau p h a n u n g t h u dupe 5,6 gam Fe v a 0,15 m o l k h i CO2. Cong
t h i i c cua o x i t sSt v a gia t r i cua V la
C. FeO v a 2,24.
ax = 0,1; ay = 0,15 ^
^f,;^'
Nhan xet: K h i C O c h i k h u d u p e Fe203 theo phan l i n g :
A. Fe304 va 3,36.
ay
gach2:
Cu2* + 2e
Fe203 + 3 C O
xFe + y C 0 2
xay ra 6 r y e a m v a cue d u o n g Ian l u p t la
d u p e M g O , A I 2 O 3 - > D a p an D.
A. 8 gam.
giai:
N/zflM xef: T a n g g i a m k h o i l u p n g :
D . C u , Fe, A h O a , M g O .
.i,,.,,
yCO — ^
Lai
B. C u , Fe, A l , M g .
'
viyt
V i dv 4: N g a m m o t la k e m t r o n g 100 m l d u n g d i c h CuS04 a mol/1. Sau k h i
V i dxf. 1: Cho l u o n g k h i H2 (du) qua h o n h g p cae o x i t C u O , Fe263, A I 2 O 3 , MgO
C. C u , Fe, A l , M g O .
iMiuiiy
M a t khae: n^^: TYQ = 0,5:0,15 = 2:3 -> Fe203 -> D a p an D .
_
A . C u , FeO, A I 2 O 3 , M g O .
Mol:
+
O x i t sat la Fe203
t: t h a i gian d i # n p h a n (giay)
F: H a n g so Faraday (F = 96.500)
ivi i V L ^ v v i i
ach 1: Giai theo p h u o n g t r i n h hoa hpc
nP
A : Kho'i Itfong m o l n g u y e n t u (mol.phan t u ) ciia chat t h u dug'c a d i ^ n cue.
I : C u o n g d p d o n g d i e n (ampe)
MViiii
, ..,
-s^
. ,,,
B. Fe304 v a 4,48. - . m »• EOMS
D . Fe203 va 3,36.
;
c u o n g d p d o n g di?n k h o n g d o i 1,34A (hi^u sua't qua t r i n h d i ^ n p h a n la
100%), t h u d u p e chat ran X, d u n g d i c h Y v a k h i Z. Cho m o t t h a n h Fe vao Y,
sau k h i cac p h a n u n g ke't thiic thay k h o i l u p n g t h a n h Fe tang t h e m 2,36
gam va c 6 k h i N O (san p h a m k h u d u y nhat ciia N"^"^) bay ra. Gia t r i eiia t la
A. 0,8.
B. 1,2.
.-
.
C.1,0.
D.0,3.
"
195
dm
nang On luy^n thi dgi hpc 18 chuy6n di H6a hpc - NguyJn Van HSi
Cty TNHH MTV DWH Khang Vigt
Lot gidi:
Phan ling dien phan dung dich AgNOs:
' "
.
+ H2O —^Edd_). 2Ag + ^ 0 2
2AgN03
|.
+ 2HN03
Mol:
2a
"'^
' '
' ' 2a
' 2a '
'^^'
Nhqn xet: Vi khoi lug-ng Fe tang them nen thanh Fe dupe A g bam vao-^ y
chiia A g N 0 3 - > AgNOs chua bi di^n phan het*
l«l^/;v.' Cac phan ung khi cho Fe vao Y:
Fe
Mol:
+
4HN03
0,5a
> Fe(N03)3 + N O + 2 H 2 0 , o y a n i ^ M '
2a
->
0,5a
f'
Can luu y Fe tac dung voi Fe(N03)3:
Fe
+
2Fe(N03)3
Mol: 0,25a •(- 0,5a
Fe + 2AgN03
Mol: b
<-
2b
Bai 3: Trong pin di^n hoa Zn-Cu, qua trinh khir trong pin la
A. Zn^* + 2e
/ -l^ j '
> 3Fe(N03)2
/^^v
3.BAITAP6NLUYEN
pai 1: "Sat tay" la sat dupe trang mot lop thiec mong de bao v§ sat khong b} an
mon, ten gpi eua phuong phap eho'ng an mon nay la:
, .
A. Phuong phap hoa hoe.
B. Phuong phap d i f n hoa.
, ,
C. Phuong phap bao ve be mat.
D. Phuong phap tao hpp k i m khong gi.
Bai 2: Co 4 dung dieh rieng biet: CuS04, ZnCh, FeCb, AgNOa. Nhiing vao moi
dung dich mpt thanh N i . So truong hpp xuat hif n an mon di^n hoa la
A. 2.
.B.4.
C.3.
D. 1.
rVrt',
r,-'^ " - ^
' ' ' , i » V i i i W.KWMi *
> Fe(N03)2 + 2Ag
' ' ' .ofex*
- > 'jsufTO-f
C. Cu
Vi
C. 17,20 gam.
D. 5,12 gam.
LMgidi:
2,688
nt =
„\
,
= 0,12 mol.
, „
,
Mol:
0,08
<-
> CuCh
. M o l : 0,08
Y,:.
CUSO4
y
" M o l : 0,08
+
0,08
Na2S04
^^nwib D
0,08
196
B.2.
C.4.
D.3.
"
hon hpp chat ran X. Cho X tac dung voi dung djch N a O H du, thu dupe 6,72
lit khi H2 (dktc). Hi^u suat phan ling nhi?t nhom la:
... A. 60%.
B.50%.
•
C. 75%.
D. 80%.
Bii 6: Trpn m gam bpt A l voi 16 gam bpt Fe203 thu dupe hon hpp X. Nung
0,08
0,08
+ H 2 0 — i E ^ i ^ Cu + -02
2
0,08 < - 0,04
^
B.4,48.
C. 6,72.
D. 8,96.
7: Tien hanh phan ung nhi^t nhom voi h6n hpp bpt gom A l va CuO, thu
dupe eha't ran X. Chia X thanh hai phan bang nhau.
,j i"-.
+ Phan 1 tac dung vua d i i voi 100ml d u r g dich NaOH I M .
^ Phan 2 tac dyng voi dung djch HNO3 loang, d u thu dupe V lit N O (san
; pham k h i i duy nha't, o dktc).
, , ^
Gia tri eiia V la:
+ H2SO4
u
-> Khoi lupng dung dich giam = 64.0,16 + 0,08.71 + 0,04.32 = 17,2 gam.
Dap an C.
" \
Bai 5: Trpn 32 gam Fe203 voi 13,5 gam A l roi nung 6 nhi^t dp eao, thu dupe
0: A. 2,24.
Thii t u dien phan n h u sau:
> Cu + CI2
'
thu dupe V lit N O (san pham khu duy nha't, 6 dktc). Gia tri eiia V la:
^ X gom cac muoi: CuCh = 0,08 mol; CuS04 = 0,16 mol va Na2S04 = 0,08 m o l .
CuCh
^^"^ "'"'^
dupe 21,4 gam chat ran Y. Cho Y tac dung voi dung dich HNO3 loang, d u
i*-^ v
0,16
'
(4) Cho thanh Fe tiep xiic voi thanh Cu roi nhiing vao dung dich HCl.
v^,,,
Phan ling "quy d o i " gia dinh:
2NaCl
^'
nong X trong dieu ki?n khong c6 khong khi toi phan ling hoan toan, thu
22,4
CuS04 +
,
^
So truong hpp xuat hi^n an mon di^n hoa la
tra, mang ngan xop) den khi tha'y thoat ra 6 anot 2,688 lit khi (dktc) thi ngung
B. 12,08 gam.
^
> Zn^* + 2e.
(3) Nhiing thanh Cu vao dung djch FeCb;
A.l.
di|n phan. Khoi luong dung dich sau dien phan giam d i bao nhieu gam?
D. Zn
(2) Nhiing thanh Fe vao dung dich CuS04;
^ = 0/04 mol ^ t = 2880 giay = 0,8 gio ^ Dap an A.
A. 6,96 gam.
> Cu2* + 2e.
> Cu.
(1) Nhiing thanh Fe vao dung djch FeCb;
2b
7: Di?n phan dung dich X gom 0,16 mol NaCl va 0,24 mol CuS04 (dien cue
B. Cu^* + 2e
Bai 4: Tien hanh boh thi nghi^m sau:
Theo bai: 2b.l08 - 56(0,75a + b) = 2,36.
w: a'.:
Matkhac: nAgNOa = 2a + 2b = 0,08
a =0,02 mol; b =0,02 moi. £ jo;
ne= 2a = 0,04
> Zn.
A. 2,24.
j
B.4,48.
C. 6,72.
D. 8,96.
^ai 8: Hon hpp Y gom CuO, FeO va Fe203. Hoa tan hoan toan 27,6 gam Y bang
dung dich H2SO4, thu dupe 59,6 gam muol. M§t khae, ne'u khii hoan toan
27,6 gam X can vua d i i V lit khi CO (dktc). Gia trj eua V la
A. 4,48.
B.6,72.
C. 8,96.
D. 5,60.
,.CI
.J
't.8<^,>
CJm
^
nang 6n luy^n thi dgi hpc 18 chuygn dg H6a hqc - Nguygn Van HJi
pai 9: Nhung mot la kim loai R (hoa tri 2) c6 khoi lugng 10 gam vao 200ml dung
dich CuS04 0,1M cho den khi phan ling hoan toan thu duoc chat rAn X va
Cty TNHH MTV DVVH Khang Vigt
THI/O'NG DAN-LCJI G I A I
Rai ^'
dung dich Y chua 3,22 gam muoi. K i m loai M la
Sn trang len Fe se bao phii be mat tranh cho Fe bi an mon
A. Mg. .
phvrang phap bao ve be mat —> Dap an C .
^'^
B. Be.
C . Fe.
D. Zn.
dien hoa la de bao v§ kim loai, can ghep no vol kim loai m^nh hon.
hoan toan, thu dug-c dung dich X gom hai muo'i va chat ran Y gom hai kim
2:
loai. H a i muoi trong X la
A.MgCh vaFeCh.
^
,
C . F e C h va CuCb.
AgNOa 0,2M. Sau mpt thoi gian lay thanh kim loai ra, rua sach lam kho,
thay khoi luong thanh M g tang 2,72 gam. Khoi lugng M g da phan ung la
C . 0,48 gam.
0,2M va Cu(N03)2 0,5M. Sau khi cac phan ung xay ra hoan toan, thu dugc
B.5,36.
C . 4,08.
-.''-:vr
r:-
-
Tao thanh cac cap pin di^n N i -
+ Nhung N i vao dung dich Z n C h thi khong xay ra phan ling - > N i khong bi
an mon.
+ Nhung N i vao dung dich FeCla thi N i bi an mon hoa hgc do phan ling:
Ni
+ 2FeCl3
>
NiCh
+ 2FeCl2
—>• D a p an A.
pai 12: C h o 1,3 gam hot Z n vao 100ml dung dich chua hon hgp gom AgNOa
A. 2,80.
•••:> •
C u va N i - A g —> A n mon dien hoa.
D . 0,96 gam.
dung dich X v a m gam chat ran Y . Gia tri cua m la:
if-y
cac kim loai nay lai bam tren be mat N i
D. MgCh va FeCh.
B. 0,72 gam.
.:,..,..,,.v.,i
+ Nhiing N i vao cac dung dich CuS04 hoac AgNOa thi N i se day C u v a A g ra,
B. M g C h va CuCh.
pai 11: N h u n g mot thanh M g vao 100ml dung dich gom Cu(N03)2 0,3M va
A. 1,20 gam.
day cac em de chgn nham dap an B. C a n nho rang phuang phap
luu y:
pai 10: Cho hon hgp gom M g va Fe vao dung djch C u C h deh khi cac phan virig
D a y la
Bai3:
Van dung cm tha: A n h la Anot nhuang e, E m la Catot nhan e thoi ma.
^
Trong pin Z n - C u , ta c6:
D . 2,16.
•
pai 13: Cho hon hgp bpt gom 1,3 gam Z n va 3,2 gam C u vao 200ml dung dich
Anot (cue am), Z n nhuong electron thanh Zn^* (qua trinh oxi hoa Zn):
^
AgNOa a mol/1 sau phan ung hoan toan thu dugc 12,08 gam chat ran X gom
Zn
hai kim lo^i. Gia tri cua a la
Catot (cue duang): Ion Cu^* den nhan electron (qua M n h khii ion Cu^*):
A. 0,20.
B.0,30.
T'lMj
C.0,15.
•
pai 14: Cho 4,1 gam hon hgp X gom Z n va Fe vao 200ml dung dich CuS04 0,2M.
, Sau khi cac phan ihig xay ra hoan toan, Igc thu dugc 4,24 gam chat ran Y va
.
> Zn2- + 2e.
Cu2* + 2e
D . 0,50.
"
'
y ^_..i_-_Q|i:>,,iA igrm riE/iq tj.u
> C u . ' i A ^^:tt.:
Dap an B.
'
••:>iMJ%;'
' • •
'
BM4:
dung dich Z chiia hai muoi. Khoi lugng cua Z n trong X la:
Nhan xet: 6 thi nghi^m 2 (Fe day C u v a C u bam vao Fe) v a 4 (Fe tiep xuc vol
A. 0,65 gam
Cu) -> D e u sinh ra cap di|n c\fc Fe - C u , tiep xiic tryctiep vai nhau va ciing
B. 1,3 gam
C . 1,95 gam.
D . 2,6 gam.
nhung trong mgt dung dich di|n li
pai 15: Di^n phan 200ml dung dich gom AgNOs 0,5M va Cu(N03)2 I M bang
Lieu y: 6 thi nghi?m 1 v a 3, Fe b} an mon hoa hgc theo cac phan ling:
li dong dif n mgt chieu vol cuong dg dong dif n 2,68A trong thai gian 4 gia.
|,,
Kho'i lugng kim loai thoat ra 6 catot la:
A. 20,4 gam.
B. 23,6 gam.
C . 10,2 gam.
D . 11,8 gam/'
Xuat hi|n an mon di|n hoa -> D a p an B.
-fy?^ vt;;/''^
Fe + 2FeCla
> 3FeCl2
C u + 2FeCla
> C u C k + 2FeCh
f + Xo^»^ •
pai 16: Di#n phan dung dich hon hgp chua 0,15 mol CuS04 va 0,1 mol H C l
trong thai gian 2 gio v a i dong difn c6 cuong dg la 1,34A. Biet h i f u suat ciia
J?/
= 0,5 mol; n^^^o, = ^
nAl= ^
= ^'^
"H2 ~
= 0'^
qua trinh di|n phan la 100%. Khoi lugng kim loai thoat ra tai catot la:
A. 6,4 gam.
B. 3,2 gam.
C . 12,8 gam.
2A1
D . 9,6 gam.
Bai 17: D i ? n phan dung dich X gom 0,2 mol N a C l va 0,2 mol CuS04 (di^n afc
Mol:
2a
+
<-
Fe203 — ^
a
tro, mang ngan xo'p) den khi khoi lugng dung dich giam di 17,5 gam thi
K h i cho X + N a O H du:
ngung di^n phan. The rich khi (dktc) thoat ra 6 anot la
Al
A. 1,12 lit.
B. 1,68 lit.
C . 2,24 lit.
D . 2,80 lit.
+
NaOH
+ H2O
AI2O3
i >m
•
+
2Fe
2a
- > nAidu = 0 , 5 - 2 a .
^
> NaAICh + | H 2 U S , ^ • U 0
,
,
>
199
Ca'm nang On luy^n thi dgi hpc 18 chuy§n 6i Hoa hgc - Nguyjn VSn Hai
AI2O3 +
2NaOH
3
Cty TNHH MfV DVVH Khang Vi$t
> 2NaA102 + 2H2O
_^
Mian xe^: n H , = as ^ - (a5-2a) = 0,3-> a = 0,15.
•
Bai6:
H2
'
2 ^ '
'
:
'
-
;
'^^^ •••1 /Till j^nrv^'
H = ^ ^ . 1 0 0 % = 75% ^ D a p a n C ,
0,2
^
*
m = 5,4 ^
nA,= ^
»
=
X
a2 mol.
^ " ^ ° 3 > AI3^, pe^^.
H M ^ T ? > "CUSO4 -> C u S 0 4 p h a n u n g h e t .
fr^.
;
,
r
Nhan thdy: t u tr^ng thai dau tien den trang thai cuoi ciing thi: So'oxi hoa ciia
A p dyng bao toan electron: ng=3nAi =3nNO
CuS04
^ <
,, ,^
C u C k het
.
i
>
1 ,
> MSO4 + C u
0,02
1^.1
0,02
, ,
M = 65(Zn) ^
D a p an D .
'
A , ;
Nhan xet: Y gom 2 kim l o ^ —> Y chiia Mg hoac Fe
,
Fe khong doi (luon la +3), con so oxi hoa ciia A l tang tit 0 len +3.
+
,
n; i"
- > MMSOA = - ^ = 161 ^
0,02
Bai 10:
,
^
M
Mol: 0,02 <-
So do phan ling:
Al^ FeaOg
Nhan xet: Trong cac dap an, kim loai c6 khoi lugng mol Ion nha't la Z n
Phan ling hoa hpc:
N/zfl« A:et: Truoc het can ap dung bao toan khoi lugfng: m x = m y
m + 16 = 21,4 ^
Bai 9:
- i
Mpt kim loai con d u - >
Loai B va C .
M|t khac: Do tinh khii M g > Fe —> Mg phan ling truac roi moi den Fe.
Gia sii M g con d u —> Fe con nguyen va C u bi " d a y " ra het - > Chat rSn X
HC*
gom 3 kim lo^ii —> Dieu gia sii la sai.
-> n ^ o = r»Al= 0/2 mol -> V = 0,2.22,4 = 4,48 lit -> Dap an B.
Mg phan ling het va Fe phan ling mpt phan —> Y gom Fe d u va C u .
Bai7:
-> Hai muoi trong X la M g C h va F e C h -> Dap an A .
Nhan xet:
K h i cho phan 1 + N a O H , tat ca A l va
AI2O3 deu chuyen vao hp-p chtYt
NaA102 -> A p d y n g bao toan nguyen to'Na va A l ta c6:
nNaOH="NaA102 = " A l
) A l ^ \.
Nhan thdy: T u trang thai dau tien den trang thai cuoi cung thi: So oxi hoa
ciia C u khong doi (luon la +2), con so'oxi hoa cua A l tang tu 0 len +3.
A p dyng bao toan electron:
-> n ^ o = " A r 0,1 mol ^
" A g N 0 3 = 0,02 mol; ncu(N03)2 = 0,03 mol.
Nhan xet: T i n h oxi hoa Ag* > Cu^*
Cu(N03)2.
»i>UV!iJ C/»?T S S > , •
nAi= 0,1.1 = 0,1 mol.
So do phan ling: A l , C u O — ^ X
Bai 11:
Phan ling hoa hpc:
M g + 2AgN03
Mol: 0,01 < - 0,02
Mol:
V = 0,1.22,4 = 2,24 lit.
a
<—
.
, f TlfWC{" &fo
«r--..v.
> Mg(IJ03)2 + C u
a
a
Theo bai: m Mg tang = 2,72
-> Dap an A .
AgNOa phan ling trudc roi moi den
\
iri*
JOP;
> Mg(N03)2 + 2Ag
0,02 -> mtang = 2,16-0,24 = l,92g.
M g +CU(N03)2
= Sn^i = S n ^ o
•? >
mtang = (64-24)a = 40a.
1,92 + 40a = 2,72.
-> a = 0,02 mol — > mMg (pu) = 24.(0,01 + a) = 0,72 gam.
Bai8:
Nhan xet:
-> Dap an B.
^,v,,Wj./«-i5 «ir/;,
K h i cho X + H2SO4 thi oxit chuyen thanh muoi sunfat va mot ion O^" trong
> 1 mol S O ^ ~ - > Khoi lu(?ng tang 96 - 1 6 = 80 gam.
a mol
'
= 0,02 mol; n^^j^o^
oxit dupe thay the bang mpt ion S O 4" . '
1 mol 02-
Bai 12:
'
tang 59,6-27,6 = 32 gam.
32
-> a = — =0,4 mol -> n o (x) = 0,4 mol.
80
K h i cho 27,6 gam X tac d y n g voi C O thi: no (x) = n c o = 0,4 mol.
V c o = 0,4.22,4 = 8,96 lit - > Dap an C .
^4 ;sh ,
+
^^^^^^ ^^^^^ .^^
Cu(N03)2
0,01
'
= 0,02 mol; ncu(N03)2 = 0,05 mol.
Zn
+ 2AgN03
Mol: 0,01
->
Zn
..•iii,;,;£;*.!'?.;iK
^'^'^
Phuong trinh hoa hpc:
Mol: 0,01->
'
'
> Zn(N03)2 + 2 A g
0,02
j « ; ;
> Zn(N03)2 + C u
->
0,01
- > mY = 0,02.108 + 0,01.64 = 2,80 gam - > D a p an A .
>«k;<T;V>,
...
r j c i t i ( ' • • ; c':
dm
nang On luyQn thi dgi hgc 18 chuySn dg H6a hpc - Nguygn Van HSi
Cty TNHH MTV D W H Khang Vijt
Bai 13:
= 0,02 m o l ;
nzn
ncu
= 0,05 m o l .
Cachl:
-
N/iflM xet: X g o m hai k i m loai
So m o i CuS04 phan u n g = 0,15 m o l .
.m
if, fi m i ) j
qf ^
•
^
mcntot tang = m A g + m c u
Or ^
va C u ( d u ) .
Mol: 0,02^
0,04
"
X
V -
mx =
2x
f . ,
->
„
:h-:
.
= (3,2 - 64x) + 108.(0,04 + 2x) = 12,08
" ^'^
^'^^ " ^'-^
- > a = 0,50 m o l -)• D a p an D .
- > 0,02 + x = 0,la.
mv - > Fe = 56 da p h a n u n g m p t p h a n de tao t h a n h C u = 64 —> Z gom
ZnS04 v a FeS04.
^.. - _
G o i so Z r ban d a u = a m o l ;
nFe(p.)
= b. N h a n tha'y:
'tidrt
> a m o l C u -> K h o i l u g n g g i a m (65-44)a = a.
b m o l Fe
> h mol Cu
->
gnv .m-
r^yi^^:^^^ +.
a mol Zn
Catot:
C u - + 2e
Mol:
0,05
— ^
<-0,l
^
Cu
->
j^.,^^,
0,05
r.-..^
Bai 17:
,
Phan u n g
.
,
^
t'
tang (64-56)b = 8b
~^^Zn'^
^'^Fe(pu)
= a + b.
^ a = 0,02; b = 0,02
J-
0,1
<-
>
0,2
T-/-.-,-r
«. , j ,
CuCh +
->
0,1
r>i.,tjr
K2SO4
mzn = 0,02.65 = 1,3 gam.
g
" ""^
V
- > D a p an B.
Bails:
• •
Nhan xet: D a y la bai toan d i | n phan d u n g djch chua n h i e u chat, ca.c e m nen
ap d u n g cong thuc t i n h so m o l electron trao d o i :
m
It
2,68.4.3600 ^ ,
,
n„=—.n=—=
= 0,4 m o l .
^ A
F
96500
Ta c6: So'mol electron A g * nh^n vao: 0,1.1 = 0,1 m o l .
- > So'mol electron Cu^* nh|n vao = 0,3 m o l
„
„
4-
"
0,1
"
''
T h u t u d i e n p h a n n h u sau:
Mol:
CuCl2
Cu
0,1
0,1
+
CI2
"-"•••"•^ ^-^^^^"'^
0,1
mgiam
= 0,1.(64+ 71)= 13,5 g a m .
-> K h o i l u g n g d u n g d i c h con g i a m tie'p 17,5 - 1 3 , 5 = 4 g a m .
cuso4 +
Mol:
a
H20
- i p ^
Cu +
a
io2
0,5a
+
H2SO4
, *
''•TDfeMS
^
cO:dr
64a + 32.0,5a = 4 - > a = 0,05 m o l .
,.^14 4 r;(HO)!/
Vc,2 + V02 = (ai + 0,025).22,4 = 2,8 l i t .
L a o c J i o ioinv
-» Dap anD.wi
,.
,
q u y d o i gia d m h :
CuS04 + 2KC1
Mol:
gj./j
= 4 , 2 4 - 4 , 1 =0,14 g a m 8 b - a = 0,14.
Bao toan electron: 2 n ^ 2+
-> a04
.
- > X g o m : C u C h = 0,1 m o l ; CuS04 = 0,1 m o l va K2SO4 = 0,1 m o l .
Nhan xet: T i n h k h u Z n > Fe ->• Z n p h a n u n g t r u o c r o i m a i d e n Fe. D o mx <
mt5ng
MLM ^,
anlAnyH
1,34.2.3600
,
^^^^^
=0,1 m o l .
•
- » D a p an D .
Theobai:
umUMYl
.
, , ^
It
molelecti-on: n g = - =
;^a;j i » « . v i
,4 . m . .
M a t khac: mx = 64(0,05 - x) + 108.0,2a = 12,08 - > x = 0,03; a = 0,5. ? (BH •
,
• •
-""'-^
- » m = 0,05.64 = 3,2 g a m - > D a p an B.
. , ,
Cach 2: G p i : ncu (pu) = x m o l . Bao toan electron: 2n2n + 2ncu(ph4n ijng) = "^g+
2.0,02 + 2x = 0,2a
^'
'
X chua: n^^2+ = 0,15mol; n^^. = O ,lm ol; n^+ = O ,lm ol; ng^2- = 0,15mol.
2x
n^gNOa "
'
IsJhanxet:
.
> Cu(N03)2 + 2 A g
mcudu+ mAg
- > X = 0,03 m o l - »
-
0,04
^
Bail6:
p a y la p h a n l i n g dien phan h o n h g p —> N e n d u n g cong thuc t f n h n h a n h so'
> Zn(N03)2 + 2 A g
Cu + 2 A g N 0 3
Mol:
X gom Ag
,
Zn + 2AgN03
= 0,1.108 + 0,15.64 = 20,4 gam.
D a p an A .
c6 chua Z n hoac C u —> A g N O s he't.
D o t i n h k h u Z n > C u —>• Z n phan u n g truac r o i m a i den C u
::.,,f:-./:..J.!A.f./'-:'.,-'
malBlHqlonyrtq [dmx
•
, \nnU laix w
Cty T N H H M T V DVVH Khang Vi$t
dm
nang On luygn thi dgi hpe 18 chuyfin dg H6a hgc - NguySn Van Hi\
phan img nhiet phan
Chuyen de 9
Na2C03 + CO2T + H2O -
2NaHC03
KIM LOAI KIIIM - KIltM THO - NHOM
1. K I M L O A I K I E M
.
a. KimloaiMem
+
^rv.'/-.,
Tac dung vai nude
Kim
N e u N a d u : 2Na + 2H2O
Tac dung vai dung dich muoi
;,
> 2NaOH + H 2 t
it
f
+
+
+ 2NaOH
> K2SO4 + Cu(OH)2i
M
"
"
+ H2O
Mol:
M2O
+ H2O
Ta c6:
nMOH
NaHCOs + NaOH
> NaCl + C02t + H2O
^,,,
C.Cs.,jy^|j;^XcmP'lfe)e(*Oe)::.
0,02
col' + H2O
NaHC03 + HCl
j-j
^^_^>„c,,^.a
MM.'
> M O H + ^ H 2 t ' ^ i l ^ ' 0 . H '+
0,02
x
NaHC03 + H C l
> NaCl + C 0 2 t + H2O
d. Muoi hidrocacbonat
+ Tinh chat ludng tinh
•
do
Mol:
> NaHC03 + NaCl
,
'
^,^^1^
Lcrigidi:
Phan ieng thuy phan (quy tim ~^ xanh; phenolphtalein ~^ hong) >0
Na2C03 + H C l
,
, J _^
>'•'
B.Li.
Tac dung vdi hap chat luang tinh
Cho tu tu dung dich axit HCl vao dung djch Na2C03
j,^^,,
2NOt+4H20
va 0,224 lit khi H2 (dktc). K i m loai M la
> Fe(OH)3i +3NaN03
Tac dung v&i dung dich axit
^
no vao nuoc, thu dugc 300ml dung dich chiia mQt chat tan c6 nong dp 0,2M
Fe(N03)3 + 3NaOH
HCO3 + OH-
,
.
•
Cac phuang trinh hoa hpc:
> NaAI02 + 2H2O
1,j
Vi dii 1: Hoa tan hoan toan 1,7 gam hon hop gom k i m loai kiem M va oxit ciia
A.Na.
Al(OH)3 + NaOH
c. Muoi cacbonat
^
^^^^^^^^^
^
^ 3Cu2- +
CO2 + NaOH
> NaHCOa
Tac dung vdi dung dich muoi
> 2NaA102 + H2O
+
/
VIDUMAU
> NaiCQh + H2O
AI2O3 + 2NaOH
of
^ NaHS04 + H C l t
3Cu +2NO3 + 8H*
b. Hidroxit,,. „„,„.„:„.„ rfe:)|'iWomr.O.':^
+ Tac dung vdi oxit axit
;
CO2
f •
+ Tinh oxi hod (khi c6 mat axit)
Cho Cu vao dung dich hon h(?p KNO3 va H2SO4 loang, Cu tan dan tao
thanh dung dich mau xanh theo phuong trinh:
^
> 2KOH + H2t
2KOH + CuS04
+ H2t
2KNO2 + O2
2KN03
'I.XI
Cho K k i m loai vao dung dich CuS04 xay ra cac phuong trinh:
2K + 2H2O
NaCl (ran) + H2S04(^flc)
f. Muoi nitrat
; Phan ung nhiet phan
> 2NaCl + H 2 t
* 2Na + CI2T
^ 2NaCl + 2H2O
"P""''""
>
2NaOH + C h t
+ TflcdwM^^<^2''^'*W2S04(fflc(dieuche'khiHCl)
^
Tac dung v&i dung dich axit
IVluoi clorua
phan leng dim phan: 2NaCl
• •
> 2NaOH + H 2 t
2Na + 2HC1
+
vv,;:,r
loai kiem khir mxac de dang 6 nhi?t dp thvrong, giai phong khi hidro:
2Na + 2H2O
+
>,
iff
*•
0,01
^^^^^^^^^^^^^^
^ 2MOH
f^iapi^^^
2x
= 0,3.0,2 = 0,06
* "
0,02 + 2x = 0,06 - > x = 0,02.
Mat khac: 0,02M + (2M + 16).0,02 = 1,7
M = 23 (Na) - > Dap an A.
^1 du 2: Hoa tan hoan toan 1,794 gam kim loai kiem M vao 200 m l dung dich
H2SO4 0,1M. Co can dung dich sau phan ung thu dugc 4,36 chat ran khan Y.
Kim loai kiem M la
A.Rb.
,
.
ifiui
B.K.
C.Na
Lcrigidi:
,
|oe
>•
D.Li.
f^Q
Cac phan ung hoa hoc:
2M
+ H2SO4
^ M2SO4 +H2 va M +H2O
^ MOH + ^H2
^ NaiCO^ + H2O
205
Hoa TiQc - NQuyen van Hai
Nhdn xet: Bao toan kho'i lugng -> m y
= ITIM +
Cty TNHH MTV D W H Khang Vijt
"^coZ- +
^
Al(OH)3
O H
Mol:
-> m _ , . = 4,36 - 1,794 - 0,02.96 = 0,646 g a m - > n ^ „ . = 0,038 moi. u
O H
O H
= 0,078
OH
0,04
+ 2H2O
r
0,04
_^ a = 0,08.78 + 0,1.233 = 29,54 gam
1 794
-> nM = 2nH,so.
M
H2t>U4
> AIO2
+ OH-
f-,,, r „•.*„
.
,
,
Dap an A.
yf du 5: C h o 100ml dung dich H3PO4 a mol/1 vao 100ml dung dich K O H 2 M
M= ^
= 23 (Na).
0,078
,i,
thu dugc dung dich Y c6 chua 16,42 gam hon hgp muo'i. G i a tri ciia a la
~^ D a p an C .
A. 0,9.
B. 1,0.
V i d u 3: H o a tan hoan toan 4,6 gam N a vao SOOjnl H2SO4 0,1M, thu dugc khi
C . 0,5.
Lai gidi:
D.0,8.
i,«Tr
rr, r»
,• .-
r r . . 7.
H2 va dung dich Y . Co can Y thu dugc m gam chat ran khan. G i a tri ciia m la
Hhan xet: V i Y chua hon hgp muo'i —>^ chac chan c6 chua muoi axit —>
A. 11,1.
K O H phan ung he't.
Hfga
B.7,1.
C.14,2.
D . 18,2.
Laigidi:
= 0-2 mol; nH2 = 0/1 mol; nH2S04 = 0,05 mol.
0,05
Na
+
0,05
H2O
^ NaOH
+
^
V i d ^ 4: C h o m gam hon hgp gom N a va Ba (ti 1# mol 2:1) tac d y n g voi nuoc
(du), thu dugc dung dich X v a 4,48 lit H2 (dktc). C h o 300ml dung dich
Hon hgp X gom:
.ji^^ n.3i>l
A.LivaNa.
B. N a v a K .
C . 24,86.
_
.•••M^^^'
Na
+ H2O
NaOH + i H 2
1
2a
a
>
Mol: 2a
Ba
Mol:
+ 2H2O
a
I':
Trong X:
A13^
Mol: 0,12
' .
M
^ ,,,
0,06
= 28,33
0,1
+ 30H0,36
'
fef
>BaS04^
0,1
'
•*>:•
0,12
,
, iOaMyi''
-H2t
2
t. iO'ABVi <
0,03
H a i kim loai la N a v a K - > Dap an B.
•
^
,^^,,|
dugctaorala
A. 9,40 gam.
B. 8,69 gam.
^
^
.d>I
nH, =
^
g r m nB/!q Di'
:'}H.%<3eJ4'-^^—- w(D&H + :
'
+ H2O
K
+ H2O
D. 9,90 gam.
.
: n^VT * : .
= a05mol.
Cac phan ung hoa hoc:
Na
A t " "
C . 7,98 gam.^^
Laigidi:
M m'jc>{ ifiol f
.8
,
*
dung dich X va 1,12 lit khi H2 (dktc). D u n g dich Y gom H C l 0,6M va H2SO4
; '
4H2
> Na*
+ OH" +
>
+ OH- + - H 2
2
--^J^i^J
> Al(OH)3 4
, '.
iOnmA
0,2M. Trung hoa dung dich X boi dung dich Y , tong khoi lugng cac muoi
. Q^^,
^
D.RbvaCs.
OiM
Ba + 2H2O
206
^
: ^ ^ J '
•
- aO,0 = SAc.*-
n^^2. = aimol; n ^ ^ . = 0,2mol.
Ba2- + SO^0,1
^
MOH +
= HNaOH + 2nBa(OH)2 = 0,2 + 2.0,1 = 0,4 mol
Phuong trinh ion:
Mol:
:
.>h!.•(
= 0,2 mol -> 2a = 0,2 ^ a = 0,1 mol.
C.KvaRb.
_
>
VBf;f
Vi du 7: C h o 5,35 gam hon hgp gom N a , K va Ba vao nuoc (du), thu dugc
'
OsH
a
Mol: 0,06
. -wriq
^
> Ba(OH)2 + H2
a
•
M + H2O
D . 9,34.
-tikAn \M i i i b
Ggi 2 kim loai lien tiep nhom l A la M
Al2(S04)3 0,2M vao X, thu dugc m gam ket tua. Gia tri cua m la
Voigidi:
,
y^,,, ,.
trong nhom l A ) tac dung voi nuoc d u , thu dugc 0,672 lit khi hidro (dktc).
m = mNa2S04 + rnNaOH= 142.0,1 + 40.0,1 = 11,1 gam —> D a p an A .
B. 6,24.
3 ,•
> Muoi + H2O.
Vi dv 6: C h o 1,7 gam hon hgp X gom hai kim loai (thupc hai chu ki ke tiep
-Hi
'•^•'•'••al
A. 29,54.
0,2.
—> Dap an A.
^^^^ ^
2
Mol: 0,1
nH20 = H K O H =
Bao toan khoi lugng: 0,la.98 + 0,2.56 = 16,42 + 0,2.18 - > a = 0,9.
Luu y: N a con d u se tiep tuc phan ung voi nuoc:
-
"H-iH3P04) = ^'^^
So do phan ling: H3PO4 + K O H
> Na2S04 + H 2 t
2Na + H2SO4
Mol: 0,1
. " O H - ( K O H ) = ^'2'
2
^
i f B fiu> r
''
> Ba2* + 2 0 H - + H2
207
Cty TNhH
Ca'm nang fin luy$n thi dgi hpc 18 chuy§n de 116a hqc - Nguygn Van H i !
Nhdn xet: n ^ „ . = 2 n H , = 0,1 m o l .
OH
Cac p h u o n g t r i n h p h a n l i n g k h i pha t r p n :
^
M a t khac, n o n g d o H C l gap 3 Ian H2SO4 -> T r o n g c i i n g m p t the tich thi
n H c i = 3nH2S04- G o i n H c i = 3a -> nH2S04 = a-
-> T r o n g Y: n , . = 5 a n i o l ; n
'
i_ = aniol;n
"
,
\
=3aniol/
+
Ivlol: 0,02
- > 5 a = 0,l
> H2O
+ OH"
^ a = 0,02mol.
K h o i Ivtqmg m u o i t h u d u g c = 5,35 + m^,- + ^cr,2-
SO^
,j
->DapanA.
AP*
*
0,02
''
--'i-.
•<*^'''—
So m u o i t r o n g day k h i n h i ^ t p h a n hoan toan t h i tao ra so m o l k h i I o n h o n so
B.3.
J;'Cd,i'
IciA
C.4.
^
D.5.
:d .
' i'^-^ -mddo
Laigidi:
AIO2
-
+ 2H2O
< - 0,01
:
=
V i dv 10: H o a tan hoan toan m gam h o n h o p X g o m L i , N a , K v a o nude, t h u
chat ran k h a n . Gia t r i cua m la:
B. 1,28.
t')'-«tf+
.•>:'-a.,.f:,/
M
K2Mn04 + Mn02
+02t
C. 1,62.
MOH
<-
Bao toan k h o i l u p n g : m H 2 0 +
2
-Oi't
- > D a p an A
'X
•
M.Mm)itx>iA.A
Uv
M
1
+ - H2
2
0,02
- m + mHj
- > mx = 1,62 + 0,02.2 - 18.0,04 = 0,9i g a m
CuO + 2N02t +
:>fd Iv D . 1,26.
,
Laigidi:
+ H2O
M o l : 0,04
NaN02 + - 0 2 t
2
CU(N03)2
0,02
->
Gpi cong thue c h u n g ciia cac k i m loai la M .
Cac p h a n ving nhiet phan:
NaN03
> Al(OH)34
+ O H '
A. 0,94.
n h i | t ciia cac muo'i t r o n g c h u o n g t r i n h t u lop 10 den I d p 12.
^
0,03
—> D a p an C.
Nhan xet: D a y l a cau h o i n h a m k i e m tra kien t h i i c cua hpc sinh v e d p ben
2KMn04
BaS04i
dupe 0,448 l i t k h i h i d r o (dktc) v a d u n g d i c h Y. Co can Y t h u d u p e 1,62 g a m
m o l m u o i p h a n l i n g la:
.(••A:Ab) cp.tid
, u, ,
V l y : a = 0,01.78 + 0,03.233 = 7,77 g a m
' ,
V i d\ 8: Cho day cac m u o i : K M n 0 4 , NaNCh, Cu(N03)2, CaCOs, KCIO3, AgNOs.
A . 2.
>
+ 30H"
M o l : 0,01
1, i
•,
0,03
Al(OH)3
'•'1
H2O
i.^.,,*-i
M o l : 0,02 - > 0,06
'''
=5,35 + 0,06.35,5 +0,02.96 = 9,40 gam.
"I'"
>
+ SO4"
Ba2-
'
OH"
M o l : 0,03
TrunghoaXboiY:
1TV DWH Khang Vigt
••' ..;.(,{:••'•
vlS + —
DM
^
EWS + < ) , •
• - f 3 H + rf^^
0~ ,.(y-)n'^
m.im.n
V i d\ 1 1 : Cho 5,65 g a m h o n h p p X g o m L i , N a va K v a o 100 m l H2SO4 I M ,
•:>rl .nofi, cn.fijjf ?L,c o d " } :\ .
> CaO + C02t
CaC03
KCIO3
KCI + - 0 2 t
2
•it'.M.
^ A g + N02t_+-at_^^^^^,,^^
AgN03
d u n g d i c h Y g o m Al2(S04)3 0,1M va H2SO4 0,1M, t h u d u p e a g a m ket tua
' ptj^si fMsnij!
B.4,66.
,
C. 7,77.
miiq-.:.
D.8,55.,
2+ = 0,03 m o l ; n
+ = 0,03 m o l ; n
C. 22,45.
D . 17,80. j
i ?;[
ni c n III 1., .
•
Bao
=n2nH2
0,35 ->
n x == 2n= 0,35.
A
p dtoan
u n g electron:
d i n h luat tru
g hoa=dien:
1.0,35
2- + l^ir^u-
^ •
= 0,35 - 0,2 = 0,15 m o l .
OH
= 0,09 m o l .
n
+
T r o n g Y: n ^ ,3+ = 0,02 m o l ; n ^ ^ 2 - = 0,04 m o l ; n „ + = 0,02mol.
A'
H
jd-iS. +
208
B. 15,25.
n H , = - ^ ^ — = 0,175.
,
Laigidi:
T r o n g X: n
A. 19,80.
Laigidi:
V i d\ 9: T r p n 300ml d u n g d i c h X g o m Ba(OH)2 0,1M v a N a O H 0,1M v 6 i lOOml
A. 6,99.
ran k h a n . Gia t r i ciia m la
3 92
—> D a p an B.
Giatricuaala
thoat ra 3,92 l i t k h i H2 (dktc) v a d u n g d i c h Y. Co can Y t h u d u p e m g a m chat
Bao toan k h o i l u g n g :
=m
m xx +1^502+ mgQ + n ^ o H - " ^'^^
in =
^'^"^^
0,15.17 = 17,8 g a m .
D a p an D .
209
Cty TNHH MTV DVVH Khang Vigt
CSim nang 6n luy$n thi dgi hpc 18 chuyen dg H6a hgc - NguySn v a n HSi
V i dii 12: Hoa tan hoan toan 6,65 gam hon hop muo'i clorua cua hai kim lo^ij
kiem thuoc hai chu ki ke tiep nhau vao nuoc duoc dung dich X. C h o X tg^
•
dung voi dung dich AgNOs (du), thu dugc 14,35 gam ket tua. Hai kim \Q^^
kiem tren la
"AgCl ^
B. N a v a K .
.
i r.
r^i
D. K v a R b .
mo\. G o i cong thijfc chung ciia hai muo'i la M C I .
MCI +
.
AgNCh
>
+
MNO3 + A g C l i
0,1
0,1
= 66,5
•!('>)!/
8-;:.,!Gj,) •
B.K.
CRb.
D.Li.
Laigiai:
M2CO3 + 2HC1
> 2MC1 + CO2
+ H2O
MHCO3 + H C l
> M C I + CO2
+ H2O
>
''^^
Mian x e f : n x = n c o 2 = 0,05 mol.
^
^^y.i-m
kmqk<^
'
' ' '
M H C O 3 < 97,2 < M2CO3.
- > M + 61< 97,2 < 2 M + 60 ^
18,6 < M < 36,2
^
M la N a ^
Dap an A.
ra 2,8 lit khi H2 (dktc) va dung djch Y . Co can Y thu dugc m gam chat ran
A . 27,75.
c
B. 19,05.
C . 17,20.
D . 18,05. ^
Laigidi:
nH2 = 0,125. B a o toan electron: ne=2nH2 = 0,25 -> nx = ne=0,25.
U
OH
n ^ ^ . = 0,25 - 0,2 = 0,05 mol.
Bao toan khoi l u g n g :
m = mx + m
+ m ^ „ . = 11,1 + 0,2.35,5 + 0,05.17 = 19,05 gam - > Dap an BOri
+ N2
3Mg
^ MgaNa
Ba + 2H2O
> Ba(OH)2 + H2
• r C D i o /
'
'
.
^
.
*
^ MgS04 + H 2 t
+ IOHNO3
^ 4Mg(N03)2 + N 2 0 t
+ 5H2O
, i b q g H : l .>
b. Hidroxit
Tjjnoi
+ Tac dung vai exit axit
v inV ,0
> C a C 0 3 4 + H2O
+ Ca(OH)2
.>
.(rCOfiD) idv K C *
* Ca(HC03)2
> Ba(AlC)2)2 + 4H2O
2Al(OH)3 + Ba(OH)2
c. Canxi cacbonat
•
''''''HP''TfiO-if---T^f^riVvoriAt^
+ Tac dung vai dung dich axit
CaC03 + 2HCl
„
> CaCh + C02t
1200"C
CaCOs
K
+ 2H2O
^ Ca(HC03)2
{Nuac chay da man)
O . ' - v ^ ; :''-: 't!,,?
,
•
^ C a O + CO2
Tinh chat luang tinh
Ca(HC03)2 + 2 N a O H
—
'"""'"'^
...Qki^
<
-M +
— - iDHS (
> CaCOs + Na2C03 + 2H2O '^S'-^^''
Phan ung phan huy
Ca(HC03)2
\
(Phan ung nung voi)
d. Canxi hidrocacbonat
S^"'.yy&"^'
> C a C 0 3 i + CO2 + H2O
- ' i
•
{Tao thqch nhU a hang dong, can da voi a noi hai, am dun nude)
A p dung dinh l u a t trung hoa di#n: 0,25 = l n ^ , _ + I n
^1
^ Ca(OH)2 + H2
CaCOs + CO2+ H2O
+ Phan ung nhiet phan
V i dy 14: C h o 11,1 g a m hon hgp X gom Na, K va Ba vao 200ml H C l I M , thoat
khan. G i a tri c u a m la
,
+ Tac dung vai hap chat luang tinh
>J,QQ •
-A,:..,.,
^
Ca + 2H2O
2CO2 + Ca(OH)2
^^^^
n c o 2 = - ^ = 0 , 0 5 mol. Cac phan ling hoa hQc:
—
486
Mx = 7 7 ^ = 97,2
^ 2MgO
. ,
+ Tac dung vai nuoc
Cf nhiet do thuong. Be khong phan ung voi nuoc, Mg phan ling cham. Cac
CO2
(dktc). K i m loai M la
210
Tac dung vai phi kim
4Mg
5'MV
kim loai kiem M tac dung het voi dung dich H C l (du), sinh ra 1,12 l i t khi
^
4.
Mg + H2SO4 (/oflMg)
V i du 13: C h o 4,86 gam hon hop X gom muoi cacbonat va hidrocacbonat ciia
A.Na.
,
+ Tac dung vai dung dich axit
M = 31
-> H a i k i m loai la N a va K —>^DapanB.
^
Kim loai kiem tho
kim loai con lai k h u manh nuoc tao thanh hidro:
,
4i..=K;;n
0,1
MCI = ^
mi
C.LivaNa.
Laigiai:
Phan ung hoa h o c :
Moi:
,
2Mg + O2
A.RbvaCs.
-.r-
2. K I M L O A I K I E M T H O
Nuoccung
( ;
H
.
Khai niem:
Nuoc cung la nuoc c6 chiia nhieu ion kim loai hoa trj 2 n h u Mg^*, Ca^^....
N u o c mem la nuoc khong chua ho^c c6 chua it cac ion tren.
211
dm
CtyTNHH MTV DWH Khang Vigt
nang On luy^n thi dgi hpc 18 chuy6n dg H6a hqc - NguySn V5n Hi\
Bao toan nguyen to CI:
Phan loai nude cieng
Nuoc cxing tarn thoi: Chua ion goc axit H C O 3 , khi dun soi se mat tinh
Cling do cac muo'i hidrocacbonat bi phan hiiy:
dun nong
,
,
.
'''
• > i
'
CaCO.U + CO2 + H2O
Ca(HC03)2
Nuoc Cling vTnh cuu: chua ion goc axit CI", SO 4" . De lam mem nuoc ciing
j^,, j ^.^^
vTnh cuu, thuong cho ket tiia cac ion kim loai:
CaCh + NaaCOs
^ CaCOa^ + 2NaCl
f
Nuoc Cling toan phan: Co ca tinh ciing tarn thoi va tinh ciing vinh ciiu (c6
chiia ion goc axit HCO 3 , Ch, SO 4 " ) .
88,55.C
95
36,5(100-C)
C = 10
Cac phuong trinh phan ling:
> FeS04 + H2
Nhan thay:
npe + nivi
' r :
1' /
• j,()gj;^a' +
= n^j = 0,1 mol
^
xuong?
A. Thach cao nung (CaS04.H20).
B. Da voi (CaCOs).
C. Voi song (CaO).
5
D. Thach cao song (CaS04.2H20).
Laigidi:
Nhan xet: Nhac den thach cao, cac em c6 the loai ngay B va C.
De CO the diing due tugng, bo bot khi gay xuong, nguoi ta can nung thach
cao song de tao thanh thach cao nung. Thach cao nung moi c6 kha nang hiit
nuoc va dong Cling—> Dap an C.
V i d u 2: Hoa tan hoan toan hon hgp X gom Be va Mg bang mot lugng vua dii
dung dich HCl C%, thu dugc dung dich Y. Nong do phan tram ciia BeCh va
MgCh trong Y Ian lugt la 7,18% va 4,27%. Gia trj cua C la
A. 5.
B. 10.
C.20.
D.15. '
Lai giai:
Phan ling hoa hoc:
Xet voi 100 gam dung dich Y:
,
B. Na2C03 va Na3P04.
C. Na2C03va Ca(OH)2.
D. NaCl va Ca(OH)2.
LM giai:
Nhan xet: De lam mem nuoc ciing vTnh eiiu bang phuong phap hoa hgc,
nguoi ta them vao nuoe ciing cac muo'i tan chtia goc axit c6 the ket tiia voi
ion kim loai hoa tri 2 (Mg^*, Ca^*,...).
Cac goc axit do thuong la goc cabonat, goc photphat. - > Dap an B.
V i d^ 5: Cho 2,74 gam Ba vao 100 gam dung dich CuS041,6 %, thu dugc khi X
B.3,31.
C. 5,64.
' ''* D. 3,13.
Lai giai:
r ^
; s)u/l
nB.-, = 0,02mol; ncuS04 =0,01 mol.
Ba + 2H2O
n^HjO (Y) = 100 - 7,18 - 4,27 = 88,55 gam.
Nhan xet: Lugng H2O trong Y ciing ehinh la lugng H2O c6 trong dung dich
HCl ban dau.
88,55.C
"<) l i ' H .
-> Ba(OH)2 + H2
CuS04 + Ba(OH)2
> Cu(OH)2i + BaS04i
Phan ling nhi^t phan:
BaS04
Mat khae, do nong do HCl bang C% nen
C
A. Na2C03 va HCl.
, ||
gam chat ran. Gia tri eiia m la
mBeCl2 = 7,18 gam va m^gciz = 4,27 gam
C
V i dii 4 : Hai chat dugc dung de lam mem nuoc ciing vTnh cuu la
A. 2,33.
> MgCh + H2
+ 2HC1
M=^'^=38,4
MM < 38,4 < MFe
Loai B va C. / 'i.
0,1
'
• • "
''onn '
10.(1
1,8
fit'
Mat khae: Mx >
= 22,5
Loai D
Dap an A.
0,08
va ket tua Y. Nung Y 6 nhiet do cao den khoi lugng khong doi thu dugc m
>
> BeCh + H2
Be + 2HC1
mHci
2.4,27
Yi du 3: Hoa tan hoan toan 3,84 gam hon hgp gom Fe va kim loai M (hoa tri 2)
trong dung dich H2SO4 loang du, thu dugc 2,24 lit khi H2 (dktc). Mat khae,
1,8 gam M tan hoan toan trong 160ml dung dich HCl IM. Kim loai M la
D. Be.
A.Mg.
B. Ca.
C. Zn.
Lai giai:
M + H2SO4 — > MSO4 + H2
V i d u 1: Hop chat nao ciia canxi dugc dung de due tugng, bo bgt khi gay
^
80
+"
Dap an B.
Fe + H2SO4
VI D V M A U
Mg
2.7,18
2nBeCl2 + 2nMgCl2 - " H C l
Cu(OH)2 -
-» BaS04 {khong bi nhiet phan)
CuO
+ H2O
^ = mBaso4 + i " C u O = 0,01.80 + 233.0,01 = 3,13 gam ^ Dap an D.
213
dm
Cty TMHH M i V D V V I ! Khang Vigt
nang 6n luy$n thi dgi hpc 18 chuy6n d6 H6a hpc - Nguyjn Van HSi
V i du 6: Tron 200ml dung dich X gom NaOH 0,1M va Ba(OH)2 0,1M voi lOOrril
dung dich Y gom H2SO4 0,1M va MgS04 0,2M, thu dugc m gam ket tua. Gia
tri cua m la
1
A. 5,24.
B. 5,82.
C. 4,66.
D. 6,99.
'
Lai gidi:
TrongX: n„ ,+= 0^02 m o l ; n^,+ = 0,02 m o l ; n ^ ^ _ = 0,06 mol.
TrongY: n
2+ = 0,02mol;
n
o_ = 0,03 m o l ; n^+= 0,02mol.
Cac phuong trinh phan ling khi pha trpn:
+ OH"
,
Mol:
0,02
^
0,02
iy':.*Uy.
> Mg(OH)2l
->
•">MM
0,02
v
S^VAK
4~K8e.»Y|^--M
(GiC':
B. 0,04.
C. 0,048.
nco2 =0,12mol; nBaC03 = ^ ' ^ ^ m o l . "
;
,
Cach 1: Giai theo phuong trinh
Ba(OH)2 + CO2
^ BaCOs + H2O
Mol:
0,08
0,08
^
0,08
Ba(OH)2 + 2 C O 2
^ Ba(HC03)2
0,02
C.5,68.
mol
a = 0,04 M
M^-Vi:
.v, , ^
,
^.^.
^
^
+
;())sJj + cO-
loai kiem M va k im loai kiem tho M ' bang dung dich HCl d u . Sau phan
thu dugc dung djch X va 0,448 lit khi (dktc). Co can X thu dugc m gam
muoi khan. Gia t r i cua m la
C. 2,30.
Chat oxi hoa:
2N*^ + 8e
> N2O;
2N^-'^
+ 8e
s
;
1
--i'-i
^^
rMiu\^\xn'/UK\
O d A S ' ^ " — sOE + L .
> NH4Na
Ap dyng bao toan electron: 2 n ^ g = 8 n N 2 0 + 8nNH4N03
'"^^'^
_ 2.0,08-8.0,01 _
g
=0,01.
DHd+ i
Vi dy 10: De hoa tan hoan toan 3,2 gam hon hgp X gom k i m loai R (thugc
D. 2,34.
A. Ba.
B. Be.
C. M g .
Lbigidi:
R + 2HC1
^han
>
D . Ca.
-
, ,
RCI2 + H2
5^^^!,,
^^^g^
^i^^^^^ :
ri'K
> RCI2 + H2O
xet: nx = - nHci = 0,1 mol -» M x = — = 32
2
0,1
-^R<32<RO->
^•
5^.^, ,
Cac phuong trinh phan ung:
RO + 2HC1
nBa(HC03)2 =
V i dii 8: Hoa tan hoan toan 2,42 gam hon hop gom hai muoi cacbonat cua kim
B.2,64.
> Mg^2
^ ^ ' D . 12,96.
nhom IIA) va oxit cua no can vua du 200ml dung dich H C l I M . K i m loai R
Dap an B.
Dap anB.
M g - 2e
"NH4NO3
'! .
- ^vftH m l •
Chatkhu:
,.!(,,
-> m = mMg(N03)2"^NH4N03 = 0,08.142 + 0,01.80 = 12,16 ->• Dap an A.
^.„,r,-..
Bao toan nguyen to Ba: nBa(0H)2 = "BaCOg + nBa(HC03)2 = 04
A. 2,20.
B. 11,36.
'^N20= 0,01 mol.
D. 0,06.
0,04
Bao toan nguyen to C: nco2 = ^BaCO^ + 2nBa(HC03)2
ling,
„ „ ^
nMg= O'O^
M :5Bi:M ;•
Lai giai:
-> a = 0,04M
,
Dap an B.
A. 12,16.
Ba(OH)2 a mol/1, thu dugc 15,76 gam ket tua. Gia tri cua a la
•^ih^A
<-9 'i
'
mgammuoi. Giatricuamla
^oe^H ^
V i d v 7: Hap thu hoan toan 2,688 lit khi CO2 (dktc) vao 2,5 l i t dung djch
Cach 2:
1
>//t«nxef: nco2= nH20 = 2"HCl =0,02.
" '••fc.J'j
ling xay ra hoan toan thu dugc 0,224 lit khi N2O (dktc) va dung dich Y chua
> BaS044'
^0,04
nBa(OH)2 "
> 2MC1 + CO2 + H2O
Vi du 9: Cho 1,92 gam M g tac dung voi dung dich HNO3 (du). Sau khi phan
.
Vay: a = 0,02.58 + 0,02.233 = 5,82 gam - > Dap an B.
Mol:
M2CO3 +.2HC1
i
Lai gidi:
0,02
A. 0,032.
> M'Ch + CO2 + H2O
m = 2,64 gam
, t
•
<
j^'COs + 2HC1
B^o toan khoi lugng: 2,42 + 0,04.36,5 = m + 0,02.44 + 0,02.18
0,02
0,02
0,02
, ,
^
'i\
Mg2* + 2 0 H "
Mol:
H2O
>
Ba^* + SO 4
Mol:
. ;,
Lai gidi:
Kit-ai ^
R < 3 2 < R + 16
^ ~* 1 6 < R < 3 2 -> R = 24(Mg) -> Dap an C.
11: Sue 0,448 lit khi CO2 (dktc) vao 100 m l dung djch X g o m Ba(OH)2
O'lM va N a O H 0,1 M . Sau khi cac phan ung hoan toan thu dugc m gam ket
tua. Gia tri cua m la
^•1,970.
B. 2,364.
'
C. 0,985.
D . 1,379.
"
215
Cty i M i l l i M I V DVVH KhangVi^t
Ca'm nang On luy^n thi dgi hge 18 chuy6n d § H6a hgc - Mytiyeii V i n HJi
Lot gidi:
2nBa(OH)2 = O'^l + 2.0,01 = 0,03 m o l
" O H - " '^NaOH+
Trong X:
" B a 2 + = 0,01
mol;
n^^^ = 0,01
mol.
^
, ^, ^^
•
J,
N h o m oxit
+
Tinh chat luang tinh:
M2O3 + 2 N a O H
^.
^ 2NaA102
J ' t - ^'"'^^ '• "
+ H2O
' '
n.
Nhan thay:
=
Phan ling tao ra 2 muoi: Cacbonat VP
= 1,5 > 1
0,02
"CO2
hidrocacbonat.
, .
^
N h a n thay: n^^2- = n^^z^ ^
^
m = 0,01.197 = 1,97 g a m D a p an A.
3. N H O M
.
.,,...,„../
+
+ 3O2 —
• ;
-
'-'I'M*--A
^
§M
>
2A1 + 6 H 2 S 0 4 ( d a c )
^
2AICI3
. ^. 1 ^ .
f_
A1(N03)3+ N O t + 2 H 2 O
Al
+
.srf! 8
-
> NaA102 + - H 2 t
2
«,",;UH:VJ
+ N a O H + H2O
Phan ung nhiet nhom
^
. •, •
•
'-'"-^^ ' ' ' " \ ^ & 8 ; , 4 ; ^ ' n . . - '
> A l ( O H ) 3 ® + 3NH4CI
Al(OH)3@ + 3 N a C l
AlCl3 + 3 N a O H
'*
Al(OH)3 + N a O H
^ NaA102+2H20
,
/
x :% q ^ , g « . . .
>• + i?(JiO)!:saf
KC
NaA102 + C O 2 + 2 H 2 O
^ Al(OH)3@ + NaHC03.^£ + i ( H O ) f i S
A l ( O H ) 3 + 3HC1
> AICI3 + 3 H 2 O
i
Viet gon la KA1(S04)2.12H20.
4 r-f*.?. •
.= f;
,X cr':v
Phen chua c6 ung dung lam trong nuoc, cam mau vai sgi,...
:
^ , ^ ,,
f.^"'
Phen nhom la mot loai muoi kep c6 cong thuc K2S04.Al2(S04)3.24H20 hay
O nhift dp cao, nhom k h u dugc nhieu oxit kim logi n h u Fe203, Cr203,.
thanh kim loai:
Dieu che
d. Phen chua
sf
'-^
:,;;.•,>•',.;-
+
NaA102 + H C l + H 2 O
> Al(OH)3© + N a C l
^
Ltfw 1/: Ke't tua A l ( O H ) 3 ^ se tan dan neu cho H C l d u :
+ 6H2O
/vne
Tdc dungvoi dung dich kiem
i
T u muoi aluminat
Luu y: N h o m bi thu dpng hoa (khong tan) trong dung dich HNO3 dac,
+
AI2O3 + 3 H 2 O
£(|i05:)r;A
Luu y: Ket tua A l ( O H ) 3 « se tan dan neu cho N a O H dvr:
UiM'~BO.
> Al2(S04)3 + 3SO21
nguQi v a H2SO4 dac, ngupi.
^
+ 2H2O
Phan ung nhiet phan
AICI3 + 3NH3 + 3 H 2 O
;WfU:
,•.„.„.—
A l + 4HN03(loang)
> NaAXOi
TumuoiAl(III)
. nuwjw*. •
> 2AICI3 + 3 H 2 t
2Al(OH)3 —
"
2Al(OH)3 — ^ — A . I 2 O 3 + 3H2O
;
2A1 + 3Cl2 —
2AI2O3
+ 6HC1
'O'iW «
iKOfUxo';
Tdc dung v&i dung dich ax it
2A1
+
+
Tdc dung voi phi kirn
4A1
+
.^.^^^
n^^co^ = " ^ ^ 2 - = a O l mol.
'
2AI2O3
Al(OH)3 + N a O H
„„,^
, „
a. N h o m
^
+ Tinh chat luang tinh:
= 0,03-0,02 = a O l mol.
= "OH--"C02
+ 3O2 —
N h o m hidroxit
. •
Ap dung cong thuc tinh nhanh so'mol C O 3" :
"co2-
4A1
*
j.; j ,,
VIDVMAU
V i dy 1: D e dieu che dupe 84 gam Fe tu Fe203 (du) bang phuong phap nhi?t
2A1
+
+ Fe203 —
Sdnxuainhom
^
AI2O3 + 2Fe
, ^.nif/jVi H,' .
'
, '-tfj^ftx
. ...^ •
.
\. r
Di$n phan nong chay AI2O3 voi criolit (3NaF.AlF3 hay Na3AlF6) trong binh
di^n phan voi hai di?n eye bang than chi, thu dugc nhom:
2AI2O3
'^""^ > 4A1 + 3O2
Chu f. K h o n g k h u dupe ion A P * trong AI2O3 bang nhung chat k h u thong
nhom v a i hi^u suat cua phan ung la 90% thi khoi lug-ng bpt nhom can dung
toithieu la: '
"
*
A. 81,0 gam.
B. 54,0 gam.
•
Lcrigidi:
Phuong trinh phan ling:
2A1
+ Fe203 —
^
C . 40,5 gam.
AI2O3 + 2Fe
D . 45,9 gam.
^..^
5,,
, ;,„.
<,..
^rW>>;|lftX'''S,0 =• , i r
'
thuong n h u C , C O , H2.
216
217
Cty TMHH MTV DVVH Khang Vigt
Cam riaiig On l u y § n thi dai hgc 18 chuygn dg Hoa hgc - Nguyin van H5i
„,
. , ,
84 ^ _
,
Theo phan ung: np^ = — = 1,5 mol
-> mAi
= 45 gam
n^l = np^
100
0,1M; sau khi cac phan u n g ke't thiic thu dugc 15,54 gam ke't tiia. G i a tri cua
•
'
Ba2- + S O 4 '
B. 0,10.
C.0,20. *
D.0,15/""
Cac phan u n g hoa hoc:
+ ;(}{';/),
> Ba(A102)2 + 4H2O
Ba(OH)2 + 2 A l ( O H ) 3
Truong hop 1: C h i xay ra phan ung (1)
(2)
30H-
Ket tua gom: BaS04 = 0,07 mol v a A l ( O H ) 3 = ^
0,04
3
Truong hop 2: Xay ra phan ling (1) va (2).
'
Gia tri c u a m l a
3Ba(OH)2
Mol:
+ Al2(S04)3
3a
3a
0,07-3a
1,1 :
0,14-6a
B. 1,56.
Ba + 2 H 2 O
(2)
,
Al
^,
Mol:
, ; v " " n ' -
a
a
'
N a + H2O
Mol:
2x
> NaOH
->
2x
Ba + 2H2O — ^
Mol:
X
- >
Theo bai: nH2 = 0,2 ^
*
.
u'Uti iii'i' / n o f i
., . . . . ,
x
2x = 0,2 ^
X
x = 0,1. '
>'
C ;
X
ij
:f-
""'^''^^ ^ ^^«0)IA
:! r
—
= 0,02 mol
'^f fii
AIO2
^1
H
Khi cho tu tir Y
H*
Mol: 0,02
AlO 2
Mol: 0,02
>
H2O
U
. ICS
0,02
+
n
!
H^ + H 2 O
> Al(OH)3l
0,02
Al(OH)3 + 3H*
0,01
< - 0,03
Ba2^ + S O ^
Mol: 0,02
0,02
^
218
X
2+ = 0,02 mol; n
Ba
.
•
0,02
> AV* + 3H2O
Mol:
Ba(0H)2 + H2t
X
ll
a
= 0,02 mol; n
OH- +
--t>5*'.^«>b'W^
1
A
+ -H2T
• *' ' ' ' " ^
Trong 100 m l Y : n , = 0,07 mol; n o _ = 0,02 mol; n„,. = 0,03 mol.
D . 45,06.
Lot gidi:
Cac phan ung hoa hoc:
D . 5,44.
AlO"+ | H 2
>
OH
thu dugc dung djch X v a 4,48 lit H2 (dktc). C h o 300ml dung dich Al2(S04)3
C . 48,18.
' '
> Ba2* + 2 0 H - + H2 _
0,2M vao X, sau khi phan ung hoan toan, thu dugc m gam ket tiia. G i a tri
c a . . , i: B. 29,54.
'
C . 3,11.
+ O H - + H2O
Trong X: n
V i d u 3: C h o mot m a u hop kim Na-Ba (ti le mol 2:1) tac dung voi nude (du),
A . 32,66.
''
Ta c6: n ^ j = 2,5a = 0,05 m o l - > a = 0,02 mol.
V = 0,20 lit -> Dap an C .
'
. '
Lcngtat:
(1)
Khoi iugng ket tua = 3a.233 + 78(8a - 0,14) = 15,54 gam
cuamla
'
vao n u a c (du), thu dugc
nH2 = 0,05 mol. Cac phan u n g hoa hgc:
Ke't tiia gom: BaS04 = 3a mol; Al(OH)3 = 2a - (0,14-6a) = (8a-0,14) mol.
- > a = 0,02 mol
o-n
:IF-'oi-Tf i
2a
> •Ba(A102)2 + 4H2O
Ba(OH)2 + 2Al(OH)3
Mol:
^''^ "
> 3BaS04l + 2 A l ( O H ) 3 i
a
4 -
4
„^
j, ^ j -
Loai.
+ 2H2O
0,2M. C h o tu tu den het 100ml dung dich Y vao X, thu dugc m gam ket tua.
mol.
*
> AIO2
dung dich X v a 1,12 lit khi H2 (dktc). D u n g dich Y gom H C l 0,3M v a H2SO4
"Sfo s;
.78 = 19,95 gam
» r l fioH ^>mi nc.
0,12
0,04
A. 0,78.
' - > Khoi lugng ket tua = 0,07.233 + ^
> Al(OH)3>L
+ OH-
<-
'. S^h ,r
.R;; . ii: <~
.
Vi du 4: C h o hon hop gom Ba v a A l (ti 1^ mol 1:1)
" f,(HO)l
Ba(OH)2 het.
0,1
uh
a = 0,1.233 + 0,08.78 = 29,54 gam ^ Dap an B.
"^i"'^"^'
> 3BaS04i + 2Al(OH)34 (1)
Mol:
->
- > 0,36
Al(OH)3
Lcngtat:
•3Ba(OH)2 + Al2(S04)3
+
Mol: 0,12
s<,4i T t ;
> BaS04i
Mol: 0,1 ' u,n M
AP*
•
A.aSO.
t
fOii tron X voi dung dich gom: AP+ = 0,12 mol; S O 4" = 0,18 mol: ^ .
Dap an D .
V i d u 2: C h o 700 m l dung dich Ba(OH)2 0,1M vao V lit dung dich Al2(S04)3
WW'
•Mi
p u n g dich X gom: Na* = 0,2 mol; Ba^* = 0,1 mol; O H - = 0,4 mol.
5
,
= — mol
>
'
oi'i / M ,«f /•
BaS04l
0,02
•.'.'1 .t/< ..; .
"hi)
, u ' r > f
^'
m = mAi(OH)3 + mBaS04 = 0,01.78 + 0,02.233 = 5,44 gam
Dap an D .
219
Ca'm nang 6n luygn thi dai hpc 18 chuySn 6i H6a hpc - Nguygn Van Hai
Cty TNHH MTV DVVH Khang Vi§t
V i d\ 5: H o a tan hoan toan m gam hon hop gom N a v a A l vao nude, thq
dugc k h i H2 v a dung dich X trong suo't. Them tu tu dung dich H C l IJvj
A l tan va sui bgt khi: A l + N a O H + H2O
vao X, khi het 100ml thi bat dau xuat hien ket tua; khi het 150ml hoap
]y[g khong phan ung.
350ml thi deu thu dugc a gam ket tua. G i a tri cua m va a Ian lugt la
A . 7,3 v a 3,9.
B. 7,3va7,8.
1
Laigidi:
Cac phan ung hoa hoc:
Na
C.5,0va3,9.
+ H2O
D . 5,0 va 7,8.
' i^i.-
.
• ij;
J
'
Vi d|i 7: Nho tu tit den d u dung dich N a O H vao dung dich AICI3. H i e n tugng
xay ra la
A. Co ket tua keo trang va c6 khi bay len. ..m
<
'
+ -H2
> NaOH
-> NaA102 + - H 2 t
B. Co ket tiia keo trang, sau do ket tua tan.
,f.jw>l-
C. C h i CO ket tua keo trang.
i H 1 ,4 *
•!"
.' M 1 > '
D. Khong CO ke't tiia, c6 khi bay len.
Al
+ N a O H + H2O
.^.(i/Ji
> NaA102 + - H 2
j.'! , ' y
NMn xet: D u n g dich X trong suo't —> A l tan het.
NaOH
Mol:
0,1 < -
+ HCl
> NaCl
, ,
>Al(OH)3i
0,05
\'-''f.H
nuoc (du). Sau khi cac phan ung xay ra hoan toan, thu dugc 8,96 lit khi H2
(dktc) v a m gam chat ran khong tan. G i a tri cua m la
>
Nhqn thdy: K h i cho 350ml dung dich H C l vao X, toan bp NaA102 se chuyen
A. 10,8.
^
B.5,4.
Mol:
X
Al(OH)3
Mol:
3
X
. .
+ 3HC1
•
/^'-"iA
* \ '
Na +H2O — >
> AICI3 + 3H2O
, ,
Led gidi:
Cac phan ung hoa hgc:
+ NaCI
X
D . 43,2.
C.7,8.
he't thanh ke't tua A l ( O H ) 3 , va sau do bi hoa tan mot phan trong H C l :
> Al(OH)3i
+ . fA
Vi dv 8: C h o hon hgp gom N a va A l c6 ti 1 | so mol tuang ung la 1 : 2 vao
.
0,05
NaA102 + H C l + H2O
• i tCtet^:
> NaA102 + 2H2O
- > D a p an B.
+ NaCl
- > a = 0,05.78 = 3,9 gam - > Loai phuang an B va D .
> A l ( O H ) 3 i {keo trang) + 3 N a C l '
Al{OH)3 + NaOH(dM')
/
0,1
NaA102 + H C l + H2O
Mol:
+ H2O
Lai gidi:
AICI3 + 3 N a O H
Mol: a
<_ 0,25-x
->
NaOH
1H2
+
a
,'
„ .
^ui.
b,5a
A l + N a O H + H2O
"J
> NaA102 + | H 2
:,s^c '
(2)
''"['
De thay: nAi(OH)3 ^ ^'^^
~^ ^
'
~
^'^^
Mol: a
^ ^'^
npjj = 2a =
T u day suy ra ban dau: N a = 0,2 mol; A l = 0,1 mol.
- > m = 0,2.23 + 0,1.27 = 7,3 gam -> Dap an A.
V i du 6: C h i dung dung dich N a O H de phan bi^t dugc cac chat rieng bif t trong
nhom nao sau day?
A. L i , Na,Be.
'
B. AI2O3, A l , Mg.
Laigidi:
<— a
1,
C. Zn, A l , Na.
i.
D . Fe, C u , Ag.
A , f
8,96
22,4
-»
= 0,4 mol
Loai D : F 3 , C u , A g deu khong tac dung voi dung djch N a O H .
- > D a p an B. H i e n tuong n h u sau:
220
a = 0,2 mol.
D a p an B.
Vi d v 9: Dot chay hoan toan 8,7 gam hon hgp Mg, A l v a Z n trong khi O2 (du)
thu dugc 15,1 gam hon hgp oxit. The tich khi O2 (dktc) da tham gia phan
UTig
la
A. 4,48 lit.
B. 8,96 lit.
C . 17,92 lit.
D . 11,20 lit.
f.:
Lai gidi:
O bai nay, cac e m ap diang bao toan khoi lugng se cho ket qua rat nhanh:
Loai C : Z n , A l , N a deu tan trong dung dich N a O H va sui bgt khi.
AhOa tan: AI2O3 + 2 N a O H
1,5a
Chat ran la A l d u = 2a - a = a mol - > m = 27.0,2 = 5,4 gam
Loai A : L i , Na, Be deu tac dung v6i H2O trong dung dich N a O H v a sui bpt
khi.
a
> 2NaA102 + H2O
nikimio?! + moxi
^
=moxit
->
m o x i = 15,1
n o 2 = — = 0,2 mol
~> Dap an A.
,
- 8,7 = 6,4
gam.
= 4,48 lit
,..
,
J/
,
•
.-v,
221
ca'm nang 6n luyQn Ihi dai hoc 18 chuySn dS H6a hpc - Nguyin Van H5i
Cty TNHH MTV DWH Khang Vigt
V i d\ 10: H 6 n h o p X g o m a m o l N a va b m o l A l . C h o X vao m o t l u o n g
p h a n l i n g xay ra hoan toan t h u d u g c 0,224 l i t k h i N2 (dktc) v a d u n g d i c h Y
t h i d u o c 1,75V l i t k h i (cac the tich d o 6 c i m g d i e u k i e n nhiet do, ap sua't^
chua m g a m m u o i . Gia t r i ciia m la
M o i quan he g i u a a va b la
A . 19,18.
A . a = 2b.
+
B. 3a = b.
C. a = b.
K h i cho X tac d u n g v o i nude ( d u ) : ^,,
N a + H2O
Mol:
> NaOH
a
—>
,
,^•
+ -H2
a
Al
Mol:
a
<-
.^.^ ^^^^^
,
. g t i k f o a i fiu) B>l
drt^'qf'
1,5a
n H = 2a m o l . *
o . lA f / • ' ,\ry ..'-»:
^2
K h i cho X tac d u n g d u n g d i c h N a O H ( d u ) :
' '
"""^
ojr;
'"
Luu y: N a tan he't d o tac d u n g v o i n u o c t r o n g d u n g d i c h , d o n g t h o i A l ciing
tan he't t r o n g k i e m d u (theo h a i phan u n g n h u viet a tren).
1
nAi=~=0'06mol;n^,=
Chatkhu:
A l - 3e
Chat oxi hoa:
2N*^ + lOe
D . 16,78.
,v • : f
3
)• A F
> N2;
+
3.0,06-10.0,01
m = mAi(iM03)3
lb
^ = 0 , 0 1 mol.
Bao toan electron: 3 n ^ i = 10 n
- > nNH4N03
> NaA102 + - H 2
2
a
C. 13,58.
Lddgidi:
0,5a
+ NaOH + H20
B. 12,78.
D . 2a = b .
D e n day cac e m l u u y, A l se tan m p t p h a n t r o n g N a O H v u a sinh ra:
+
V i dv 12: C h o 1,62 g a m A l tac d u n g v o i d u n g d i c h HNO3 (loang, d u ) . Sau k h i
n u a c t h i thoat ra V l i t k h i . M a t khac, n e u cho X vao d u n g d i c h N a O H (d^.^
^
,^ - ,
i n i ..n--.
2N*5
+ Se
> NH4NO3
8 nNH^NOi
,
=0,01 m o l .
^ ,
Ei,n'f
,
.
• •
h ' u b / w r *i4 v.:' =
"^NH4N03 ^ 0,06.213 + 0,01.80 = 13,58 gam -)• Dap an C.
4.BAITAP6NLUYEN
Bai 1 : C h o cac chat: NaHCOs, C O , A l ( O H ) 3 , Fe(OH)3, H F , CI2, N H 4 C I . So chat
tac d u n g d u g c v o i d u n g d j c h N a O H loang 6 dieu kien t h u a n g la
A . 4.
B. 5.
C. 3.
D . 6.
^
Bai 2: H a p t h y hoan toan 2,24 l i t k h i CO2 (dktc) vao 100ml d u n g d i c h X g o m
N a O H 2 M v a K O H 0 , 1 M t h u d u g c d u n g d i c h Y. Co can Y t h u d u g c a g a m
Theo bai ta c6:
^
=—
0,5a + 1,5b
=
1,75V
- > 2a = b —> D a p an D .
1,75
chat ran khan. G i a t r i cua a la
A . 17,05.
B. 14,50.
u,..,,
C. 16,20.
D . 17,90.
V i d\ 1 1 : C h o m g a m h o n h g p X g o m A l , C u vao d u n g d i c h H C l ( d u ) , sinh ra
Bai 3: C h o 3,2 g a m b g t C u tac d u n g v o i 100ml d u n g d i c h X g o m K N O 3 0 , 4 M va
3,36 l i t k h i (dktc). N e u cho m g a m X vao m o t l u g n g d u axit HNO3 (dac,
H2SO4 0 , 6 M . Sau k h i cac phan u n g xay ra hoan toan, s i n h ra V l i t k h i N O
n g u p i ) , t h u d u g c 6,72 l i t k h i NO2 (san p h a m k h u d u y nhat, a dktc). Gia trj
(san p h a m k h u d u y nhat, 6 dktc). Gia t r i cua V la
ciia m la
A . 0,746.
A . 11,5.
C. 12,3..fti |.,Ci
B. laS.
,,
D . 15,6.
Lotgtat:
K h i cho X tac d u n g v o i axit H C l ( d u ) - > C h i c6 A l p h a n u n g (do C u dung
sau H t r o n g d a y dien hoa):
Al
+ 3HC1
^
> AICI3
M o l : 0,1
+
•
^
,
y
+ fH2,,p,
M o l : 0,15
<-
Bai 4: H o a t a n hoan toan 1,794 g a m k i m loai k i e m M vao 400ml d u n g d i c h H C l
0,15
loai k i e m M la
A. Na.
B. K.
C. Rb.
D. Li.
N a O H 0,03M v a Ca(OH)2 0,01 M , t h u d u g c m g a m ket tiia. Gia t r i cua m la
A . 2,00.
'
B. 1,00.
C.1,25.
D . 0,75.
Bai 6: M o t coc n u a c c6 chiia cac i o n : N a * (0,02 m o l ) , Mg^* (0,02 m o l ) , Ca^* (0,04
m o l ) , C I - (0,02 m o l ) , S04~' (0,01 m o l ) va H C O J (x m o l ) . D u n soi coc n u a c
A l b i t h u d p n g , k h o n g tan):
+ 4HN03
/
D . 0,672.
Bai 5: H a p th\ h o a n toan 0,672 l i t k h i CO2 (dktc) vao 1 l i t d u n g d i c h X g o m
K h i cho X tac d u n g v o i axit HNO3 (dac, nguoi) - > C h i c6 C u p h a n i m g i^'^
Cu
C. 1,792.
0,1M. Co can d u n g d i c h sau phan u n g t h u d u g c 3,316 chat r a n k h a n Y. K i m
Nhan xet:
+
B. 0,448.
tren cho d e n k h i cac p h a n u n g xay ra hoan toan t h i n u o c con l a i t r o n g coc
> Cu(Na)2 + 2NO2 + 2H2O
0,3
- > m = ai.27 + ai5.64 = 12,3 gam - > Dap an C.
" '^^ '^
A. La nuoc mem.
B. Co t i n h c u n g vTnh c u u .
C. Co t i n h c u n g toan phan.
D . Co t i n h c u n g t ^ t h a i .
223
cam
nang on luy^ii thi dai lipc 18 chuyen de Hoa hgc - Nguyjn Van
Cty TNHH MTV DWH Khang Vl^t
Hki
Bai 7: Mot coc dung dich chua mot lugng nhieu cac ion: Ca^*, Mg^*, HCO3
C\-, SO l" . Nude trong coc tren thuoc loai gi?
A. Nuoc mem.
C. Nuoc
Cling
B. Nuoc cung vTnh cuu.
tam thoi.
D. Nuoc cxing toan phan.
Bai 8J Tien hanh phan ung nhiet nhom hon hop gom A l va 16 gam Fe203 de'n
khi phan ung hoan toan, thu dupe hon hop ran X. Cho X tac dung vua dii
voi V l i t dung djch NaOH I M , sinh ra 3,36 lit H2 (dktc). Gia tri cua V la
A. 0,30.
B.0,10.
C.0,20.
D. 0,15.
Bai 9: Cho khi C O (du) di vao ong s u nung nong dung hon hop X gom
AI2O3,
Fe304, CuO thu dupe chat ran Y. Cho Y vao dung dich NaOH (du),
khuay k i , tha'y eon lai phan khong tan Z. Phan khong tan Z gom
A. AI2O3, Fe, Cu.
B. Fe304,Cu.
C . Fe, Cu.
D . A l , Fe, Cu.
Bai 10: Day gom cae oxit deu bi A l khu 6 nhi^t dp cao la:
B. PbO, K2O, SnO.
C. FeO, MgO, CuO.
D . Fe304, SnO, BaO.
J'IATJ
Bai 11: Nhi^t nhom hon hpp gom a gam bpt A l va b gam Fe203, thu dupe hon
hop X. Hoa tan X trong H N O 3 du, thu dupe 2,24 lit khi N O (dktc). Gia tri
A.2,70.
^ •''a ^ / ' r r f ^ o i ff«;,n,-ii rv!+nf;! j
i'-nOi''
wniVffi-
B. 5,40.
C . 1,35.
D.8,10.iG:
Bai 12: Tien hanh phan ung nhif t nhom voi m gam hon hpp X gom A l va
Fe304 den phan ung hoan toan, thu dupe eha't ran Y. Cho Y tac dung v a i
i
dung dich NaOH (du) thu dupe dung dich Z, phan khong tan va.3,36 lit k h i
01
H2 (dktc). Sue khi C O 2 (du) vao Z, thu dupe 39 gam ket tua. Gia tri ciia m l a
A. 45,6.
B.48,3.
C . 36,7.
pai 1^*
^''^ ^^"^
400ml dung dich H C l 2M, thu
(Jupc dung djch X. Cho tung gipt den he't V lit dung djch N a O H 2 M vao X,
thu dupe 7,8 gam ket tiia. Gia trj Ian nhat cua V la
'
= ' Fe203 den phan ung hoan toan, thu dupe chat ran X. Chia X thanh hai phan
A . 29,82.
B. 32,22.
5. H I / C ! K N G D A N - L 6 I
Al(OH)3 + N a O H
Ch + 2 N a O H
NH4CI
> NaAlCh
+ NaOH
> NaF
~> Dap anB.
+ H2O
•
.
K O H (du), thu dupe a mol hon hpp khi va dung dich X. Sue k h i C O 2 (dif)
! 1 vao X, lupng ket tua thu dupe la 46,8 gam. Gia tri eua a la
B.0,60.
C.0,40.
D . 0,45.
Bai 15: Hoa tan hoan toan m gam hon hpp X gom Na20 va AI2O3 vao H 2 O thi'
dupe 200ml dung dich Y chi chua chat tan duy nhat c6 nong dp 0,5M. Tho'
-
,
'
,
.
^'•'f)
mol
•
Tie-"
-Kt; ....
n^, + = 0,2 mol; n^+ = 0,1 mol.
el: ^!^2H1 = M = 3 > 2 ~^ O H - du, phan ung chi ra tao muoi
Nhqn xel:
nroo
0,1
"CO2
o
Mol:
D . 0,55.
, f,
•
" O H - " " N a O H + n K O H = 0'2 + 0,1 = 0,3
Gia tri cua a la
Bai 14: Hoa tan hoan toan 0,3 mol hon hpp gom A l va AI4C3 vao dung dich
,., .....r
>fn l O . O - . j
-!
ro n - « *
Phuang trinh ion: C O 2 + 2 0 H -
De hoa tan het phan hai can vua du dung dich chua a mol HCl.
A. 0,55.
•
' H0^|
.{,
BAi2:
'
> NaCl + N H 3 + H 2 O
HF + N a O H
Trong X:
+ 2H2O
NaCl + NaClO + H2O
>
+
C . 0,35.
' -
^ Na2C03 + H 2 O
cacbonat.
B. 0,65.
'
D. 33,02.
^ ij/t
+ Phan mpt phan ung vira du vai 150ml dung dich N a O H I M .
>
D. 0,05.
C. 31,42.
GIAI
NaHCOs + N a O H
bangnhau.
.
0 ^ C. 0,25. J ; =
B.a35.
A. 0,45.
Bai 17: Hoa tan hoan toan 3,78 gam A l bang dung dich H N O 3 loang (du), thu
dupe dung dich X va 0,448 lit (dktc) hon hpp khi Y gom N 2 O va N2. Ti khoi
ctia Y so vai hidro la 18. Co can dung dich X, thu dupe m gam chat tin khan.
Gia tri cua m la
D . 57,0.
Bai 13: Tien hanh phan ung nhif t nhom vai 24,1 gam hon hop gom A l va
A. 0,40.
/.
dnhi i"r . •••i
D. 8,2 va 7,8.
Biil:
A. FeO, CuO, Cr203.
fr cua a la:
C O 2 (du) vao Y thu duac a gam ket tua.
Gia tri cua m va a Ian lupt la
, •
A. 8,3 va 7,2.
B. 11,3 va 7,8.
C. 13,3 va 3,9.
-r
0,1
0,2
Pfvn
^ CO3"
«Jf •
PfiO » 4 +
gfj'y r.
+ H2O
0,1
V gom: Na* = 0,2 mol; K* = 0,1 mol; O H " = 0,1 mol va CO
3"
Bao toan khoi lupng: a = m^^+ +
^n-/'
S-^t^
->DapanC.
^ai3:
B :<>
+ m ^ ^ . + m^^2-
= 0,1 mol.
= 0,2.23+ ai.39 + a i . l 7 +0,1.60 = 16,2 gam.
'
/
*
nc„= — = 0,05 mol.
64
. .nartko ^
'Mr^'l
^
.ry:,..^
225
Cty TNHH MTV DVVH Khang Vigt
C^m nang 6n luygn thi dji hqc 18 chuy6n ii H6a hqc - Nguygn VSn H&\
Nhdn xet: D u n g d i c h X c h i i a m u o i K N O 3 va H2SO4 l o a n g - > Can giai theo
phuong trinh ion.
ei t'w;
o ,
n + = 2nH2S04 = 2.0,06 = 0,12; n
= nKNOj = 0,04.
,i,Crw
+
' M o l : 0,045
^
+ 2NO3'
8H*
0,12"^
^ 3Cu2*
0,03
\::\>
V = 0,03.22,4 = 0,672 l i t ^
Bai4:
2N0'''
^ MCI + -H2
2
h> i
1
M =
' ' '
M g ^ Ca^* con du, H C O 3 he't.
Bai 7:
A U «Jj *
+
n ^ ^ . = 0,006 m o l .
'
''
' ' ^"''"^''''''
'
^
Nhqn xet: D61 chieu v o i d i n h nghia, m a u n u o c t r o n g co'c c6 chua ca t i n h
Cling tarn t h a i (chua H C O 3 ) va c6 ca fa'nh c u n g vTng c u u (chua CI", SO^").
-> ThuQC loqii n u o c c u n g toan phan - » D a p an D.
= 39 (K) ^
0,046
D a p an B.
Sir
Bai 5:
,
Bai8:
'
= 0,lmol ^
nFe203 = ^
npe = 0 , 2 m o l .
Nhqn xet: Bai nay neu cac e m viet p h u o n g t r i n h p h a n t u se gap n h i e u kho
Nhdn xet: K h i cho X + d u n g d i c h N a O H —> Sinh ra H2 nen t r o n g p h a n u n g
khan. D o vay, nen giai d u a theo cac p h u o n g t r i n h i o n .
" O H - " "NaOH + 2nca(OH)2 = 0,03 + 2.0,01 = 0,05 m o l
T r o n g X:
n 2+ = 0,01 m o l ; n + = 0,03 m o l .
nhiet n h o m t h i A l con d u va FezOa het.
_ Q H _ = 9J91 > 1
nco2
t
.
Phan l i n e tao ra 2 m u o i : Cacbonat va hidrocacbonat.
0,03
CO2
Mol: a
^
a
CO2 + 2 0 H - ^ C 0 3 " + H 2 0
Mol: b
2b
b
Ca^* + C O 3 "
^ m = aOl.lOO = 1,0 g a m
ai
Al
+ N a O H + H2O
^ CaCOa
1
-
0,2
^
0,1
'
3
> NaA102 + - H 2
dAm
> 2NaA102 + H2O
^N.OH= | n H 2 + 2nAi203 = ^
„
" ' " "' ' '
+ 2.ai
mmi^finml <::'.::
'.diui
^
=0,3mol
V = 0,31it. ^
. ,
i^f^
.^^p
Bai 9:
§nw.
Luu y: K h i C O k h o n g k h u d u g c AI2O3 nen k h i cho k h i C O d u tac d u n g v o i
X, xay ra 2 p h a n u n g :
D a p an B.
Fe304 + 4 C O — ^
T i n h nhanh: n^^2- = n ^ j ^ - - nco2 = 0,05 - 0,03 = 0,02 m o l .
D o n^^2+ <"co2- ^ ^^€3003= n^^2+ = 0,01 ^
'
> 2Fe + AI2O3
Fe203
->Dap^nA.
Ta c6: a + b = 0,03; a + 2b = 0,05 - > a = 0,01 m o l ; b = 0,02 m o l .
Phan umg tao ket tua:
+
AI2O3 + 2 N a O H
+ O H - -> HCO3
a
2A1
M o l : 0,2
Cach 1: Giai theo p h u o n g t r i n h i o n :
Bai
i.
A
-> ThuQC lo^i n u o c c u n g v i n h c u u - > D a p an D.
1 794
= 0,046
f V>ri. ?
N u o c t r o n g co'c sau k h i d u n c6 chua cac ion: Mg^^ Ca^*, C h v a SO 4 "
^ MOH + -H2
2
m ^ ^ . = 3,316 - 1,794 - 0,04.35,5 = 0,102 g a m ^
"
C a C O s i + CO2 + H2O
N h a n thay: n^^2+ + \^2+ = 0,06
'
+
''
,
A ...ib ynu,")
i-
my =
.
Ca^- + 2 H C O 3
v,
va M +H2O
Nhdn xet: Bao toan k h o i l u g n g
nM= "HCl + n
i;> i / \
> M g C O s ^ + CO2 + H2O
dun nong
0,03
,.u
>
Mg2^ + 2 H C O ;
+ 4H2O
D a p an D .
Cac p h a n u n g hoa hpc:
^
+
•
.j,
M + HCl
1.0,02 + 2.0,02 + 2.0,04 = 1.0,02 + 2.0,01 + 1.x ^ x = 0,1.
Den day, cac em luU y: Khi dun nong, muoi hidrocacbonat cua kim loai hoa
tri 2 bi phan huy:
P h u o n g t r i n h i o n r u t gon:
3Cu
So' m o l di^n tich d u o n g = So' m o l dien tich am
mcaCOa =1,00 gam.
6:
N/ifln xet: T r u d c het cac e m can ap d \ i n g d j n h luat t r u n g hoa d i ^ n de tim so
CuO
+ CO
— ^
3Fe + 4C02
" ^ •r f . g l ^ ( v , , ,
'
C u + CO2
- > Chalt An Y g o m : Fe, C u , AI2O3.
^ :
iV/
*
/
K h i cho Y tac d u n g v o i N a O H d u t h i AI2O3 b i hoa tan hoan toan:
f
AI2O3 + 2 N a O H
> 2NaA102 + H2O
Vay Z g o m : Fe, C u
D a p an C.
'' '
^
mol ion H C O J :
227
Ca'm nang On luy?n thi djii hpc 18 chuyfin dg H6a hpc - Nguygn Van Hii
Cty 1NHH MTV DVVH Kliang Vi$t
Bai 10:
Luu y: Al chi khu dug-c cac oxit kim loai dung sau Al trong day di§n hoa.
, -> Loai B, C, D vi Al khong khu dugc K2O, MgO va BaO
—> Dap an A.
., .
Cac phan ung hoa hoc:
^.^^^j^ ^ . ; e ' L ^ . . . . . , ^ ^ 1 ,
2A1 + 3?eO — ^ AI2O3 + 3Fe,^^.,^^.^;, s--*,.
^
2A1 + 3CuO
Ah03 + 3Cu
„ ^^g,,^
2A1 + Cr203
^ AI2O3 + 2Cr
'
Bai 11:
^
Cac qua trinh hoa hoc xay ra theo so do sauf'^J'''
f '
J Al, FejOj — ^ X
) Fe(N03 )3 + AKNOg )3 + NO
N/zfln xet: So oxi hoa ciia Fe khong thay doi a trang thai dau va cuoi (deu la
Fe*^), con so'oxi hoa ciia Al tang tir 0 len +3.
Bao toan electron: 3 n A i = 3 n N o —>• nAi = nwo = 0,1 mol.
a = 0,1.27 = 2,70 gam -> Dap an A.
'
Bai 12:
Nhqn xet: Bai nay ra't nhieu thi nghiem nen neu cac em giai theo phuong
trinh phan ung se mat nhieu thai gian.
j^^, .
So do duong di ciia Al trong cac qua trinh hoa hpc nhu sau:
^Fe304,t°
^j^o^
^
^
J^^^JQ^
2
r 1-
3
mol. t
ri*^ij4:'>,
^ Ban dau: n ^ r 2.0,15 = 0,3 mol -> nFe203= " ^ ^ ^ ^ ^ ^ ^ ^
'-
V$y: a = - nAi + 3nAI4C3 "0,6
,.
-> Dap an B.
Bai 15:
rO,0,8^-M..-
Nhan xet: Khi cho X vao nuoc, Na20 se tac dung ngay vai nu6c:
Na20 + H2O
af^faqfi^i
> 2NaOH
tinh luong tinh nen tan trong dung dich NaOH vvra tao ra
Al + NaOH + H2O
> NaA102 + - H 2 t
AI2O3 CO
2
Mol: 0,1
Mat khac n^i = —^H2 ^'^ ~^
tham gia phan ung nhiet
3
nhom
0,4 mol:
^ m =la m^i
+ mFe304 = 0,5.27 + 0,15.232 =£48,3 gam Dap an 3.^^ M
qeU
Bai 13: 8A1 + 3Fe304 — ^ 4AI2O3 + 9Fe 5 ,
^
Mol:
-> 0,15
"NaOH0,4
= 0'15.1
= 0,15 mol.,
N/zfln xet: Khi cho phan 1 + dung dich NaOH, tat ca Al va AI2O3 deu tac
dung va chuyen thanh NaA102.
Do vay, bao toan nguyen to Al, ta c6: nAi= nj^jaOH = O'^S mol.
99S
i '
^ ^^^^^^^
39
Bao toan nguyen to Al: n^i (x) = nAi(OH)3
Fe203 + 2A1 — ^ AI2O3 + 2Fe
Mol: 0,1 -> 0,2
0,1
0,2
-> Phan 2 gom: n^i = 0,05 mol; nFe= 0,1 mol; n^ijOa = 0/05 mol.
- > HHCI = 3 n A i + Znpe + 6nAi203 = 0,65 mol.
• '
Dap an B.
>r
r
w
Bai 14:
So do phan ung:
^
Al, A I 4 C 3
KAIO2
^ ^ ° ? ^ " 2 0 > Al(OH)3
'
Bao toan nguyen to Al: n^, + 4nAi4C3 = nAi(OH)3 ^ "Al + 4nAi4C3 = OA
Theo bai: nAi +nAi4C3 = 0,3 -> nAi =0,2; nAi4C3 = 0,l. ^
0,1
0,1
T h 6 i k h i C 0 2 : NaA102 + CO2 + 2H2O
> Al(OH)3 +NaHC03
Mol:
0,1
0,1
-> a = mAi(OH)3 = 0,1.78 = 7,8 gam;
.,
va m = m N a 2 0 + mAI2O3 = 0,05.62 + 0,05.102 = 8,2 gam.
-> Dap an D.
Bai 16:
>
nAi = 0,2 mol; nnci = 0,8 mol; nAi(OH)3 = 0,10 mol.
Dung dich X gom: AlCh = 0,2 mol; HCl = a2 mol.
^ ; 1, ,5 .
£ach 1: Giai theo phuang trinh hoa hpc:
NaOH + HCl
> NaCl + H2O
, ;v,: , ; .
AICI3 + 3 N a O H
> Al(OH)3i + 3NaCl . ^
Al(OH)3 + NaOH
> NaA102 + H2O
-> VnNaOH=
->
n,^^.
OH
= 0,45
= 0,20+
litnQ^.0,60+
= Dap
0,9moI
0,10
an A.
= 0,90 mol.