Cẩm nang ôn luyện thi đại học 18 chuyên đề hóa học 3
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Cty TNHH MTV DWH Khang Vi$t
Ca'm nang an luy^n thi d^i hgc 18 chuy6n 6i H6a hpc - Mguyjn Van Hii
X CO chxia: n^^3+ = 0,04;
= 0,08; n ^ ^ ^ , = 0,02; Na* va
D . k h u H2O v a oxi hoa i o n CI".' •* V** '"'•^
C. k h u i o n Na"^ va o x i hoa i o n CI".
SOl'
Laigidi:
3Cu
Cu
+
8H* + 2 N O ;
+ 2Fe3*
->
> 3Cu2* + 2 N O + 4 H 2 O
> Cu^^ + 2Fe2-
= 0,03 + 0,02 = 0,05 m o l
4. P H A N L T N G D I E N P H A N
*
' >'*
; £0.0 -
""P"' > 2 N a + CI2
l^»d m i uO -y.
. ... ,
"^""^ ) M g + CI2
MgCh
2NaCl
+ 2 H 2 O - ^ E ^
'
2 N a O H + Cht
' >: '
Cyc a m (catot) c6 H2O va Na* - > H2O b i d i e n phan:
2H2O + 2e
)• H2 + 2OH-
(Qudtrinhkhu)
Cue d u o n g (anot) c6 Ch va H2O
;i..r-,;r:
,
Ch b i dien phan:
^ , ,
,
,
V i d y 2: D i ^ n p h a n d u n g d i c h muo'i MSO4 ( v o i d i e n cue t r o ) v o i c u o n g d p d o n g
'
+ H2t
''^i*'-
di?n k h o n g d o i bang 1,5A. Sau 1351 giay t h i d u n g d i ^ n phan, 6 catot ehua
CO b p t k h i va kho'i l u g n g catot tang 0,672 gam. C o n g thuc muo'i la
T r u o n g h o p 2: C h i c6 i o n k i m loai b i d i ^ n phan:
CuS04 + H2O _ i E ^ Cu +
'
Nhuvay:
2C1> Ch + 2e .^j^,^^^^ {Qua trinh oxi hoa)
D a p an D .
Dien phan dung dich
T m o n g h o p 1: C h i c6 go'c axit b i dien phan:
n
D a p an A .
w.'
.,n is'ufh 6-.J /
: i ;, •
" A n h la anot n h u o n g e - E m la catot nhan e t h o i m a "
'
= 0,05.64 = 3,2 g a m
VCXQ^
- '
Van d u n g cau t h o :
. . . . ^ l i ;+.«fcO£. *^f^.tr1-^
a. L i thuyet:
. m,-t.,-n
* Dze« p/jflji nong chay
2NaCl
s '
't:'! fh>
A.CUSO4.
-O2 + H2SO4
B.ZnS04.
C.FeS04.
D . NiS04.
2
Phan u n g d i e n p h a n :
2 A g N 0 3 + H2O - ^ E ^ 2 A g + ^ 0 2 + 2HN03
T r u o n g h o p 3: Ca goc axit va i o n k i m loai d e u b i d i ^ n phan:
CuCh
*
"P"" > C u + CI2T
j^nu)
Qua trlnh oxihoa-khie tren cdcdien cue
''^
! i) - ,
+ H2SO4
™
T: A
A.I.t
Theo cong t h u c Faraday: m =
>
1 , gmu
j^nntirf
^,
Cau tho: A n h la A n o t n h u o n g e, E m la Catot n h a n e t h o i ma.
A n o t - qua t r i n h o x i hoa; Catot - qua t r i n h k h u .
Alt
*
MSO4 + H2O - J P ^ M + i o 2
Dinh luat Faraday ( t i n h l u p n g chat t h u d u o c 6 cac d i ^ n c^rc): m =
f
nF
A : kho'i l u g n g m o l n g u y e n t u ( m o l p h a n t u ) cua chat t h u duc^c b d i e n cure,
c u o n g d p d o n g d i e n (ampe)
d nfi qE<4 f -
t: t h a i g i a n d i ? n p h a n (giay)
-
B. 4,8.
C . 2,4.-
y: T i r cong t h u c d i n h luat Faraday, so mol electron trao d o i t r o n g qua
b. V i d y mau
. .;• •
CUSO4
j lii.as;? y
u,*£.0
i
V i d y 1: K h i d i e n p h a n d u n g djch N a C l ( v o i d i ? n c\rc t r a , c6 m a n g n g a n xo'p)
t h i cac qua t r i n h xay ra a eye a m va cue d u o n g Ian l u g t la
A . o x i hoa H2O va k h u i o n C h .
B. k h u i o n Cl~ va o x i hoa i o n Na*.
Cui
+|02t
a
aSa
+H2SO4
|!^: •
-
Orii*
Nhan xet: Kho'i l u g n g d u n g djch g i a m = m ^ u + n i o j
= ^
Day la cong thiec giiip gidi nhanh chong nhieu bdi todn dien phan.
- J s ^
a
64a + 32.0,5a = 8
^
. .
Laigiai:
Phan u n g d i | n p h a n :
CuS04 + H2O
Mol:
t r i n h d i e n p h a n d u g c t i n h theo cong thuc:
D . 3,2.
'
F: h l n g so Faraday (F = 96.500)
Luu
.On-- .
d u sau d i ^ n p h a n can d u n g 1,12 l i t H2S (dkte). Gia t r j cua C la:
n: so' electron trao d o i m a i o n da n h a n de tao t h a n h n g u y e n t u ( p h a n t u )
I:
.jUI BI nf.r[::
V i d y 3: D i f n p h a n 500 g a m d u n g djch CuS04 eo n o n g d g C%. Sau k h i d u n g
CuS04 eon
-
n.96500
D a p an A .
A . 5,6.
'<
_ , _ - u t uri?,,
= 0,672
- 4 A = 32n - > n = 2 va A = 64 (Cu).
m : kho'i l u ^ n g chat t h u duQfc 6 d i f n cvrc (gam)
|;,;
A.1,5.1351
d i ^ n p h a n thay kho'i l u g n g d u n g djch g i a m 8 g a m . De ke't tua he't l u g n g
nF
t u a n g u n g 6 d i | n cue.
'
'
Mol:
a = 0,1.
0,05
"-^^ ' ' ' • '
>CuSi
+H2S
<-
+H2S04^
0,05
'
^nubficricrr.."'
"
ncuso4 = 0,1 + 0,05 = 0,15 m o l ^ C % = ^'^^-^^^.100% = 4 , 8 % .
500
D a p an B . -
•'•"•"^^ - -
^
•''''''^'
•" •
Cty TNHH MTV DWH Khang Vi?t
dm nang On luy^n thi dgl hpc 18 chuySn ai H6a hpc - NguySn van H5i
V i dv 4: Hoa tan 12,5 gam tinh the muoi ket tinh CuS04.nH20 vao nuac, thu
dugc 1 lit dung dich X. Dien phan X (voi di^n cue tro) den khi bat dau c6
khi thoat ra 6 catot thi dimg dien phan, thay khoi lugng catot tang 3,2 gam
va dung dich sau dien phan c6 p H bang a. Gia t r i ciia n va a Ian lugt la
A.5;l.
B.5;2.
C. 3; 3.
D. 7; 1.
Lod gidi:
phan ling "quy doi" gia dinh:
CuS04
Phan ung dien phan: CuS04 + H2O
"P^" > Cu + ^ 0 2 + H2SO4
CuCh - ^ S ^ Cu + CI2
Mol:
bj dien phan het (bat dau den H2O bi di^n phan).
m c u = 3,2
gam
McuS04.nH20 =
61'
^
ncu
= O C U S O A = i ^ C u S 0 4 . n H 2 0 = 0/05
mol.
[H^]=M = o,l = 10"^
pH = l
Mol:
' '
^ = 250 -> 160 + 18n = 250 ^ n = 5.
Dung dich sau dien phan: n^^^ - 2nH2S04 = 2ncu = 0,1 mol.
^
^Q^^-J
,
B. 0,672.
C. a896.
D. 1,120.
<
Laigidi:
1 t,,;,
-A .ns
NMn xet: Day la phan ung dien phan hon hgp —> nen dung cong thuc h'nh
...
u u XX .
It 2,68.2.3600
,
nhanh so mol electron: n,,= — =
= 0,2 mol.
"
F
96500
X chua: n^^2+ = 0,12; n^^. = 0,04; n^^^^ = 0,04 va n^^z- = 0,12 (mol),
Tai anot:
4H2O -
Mol:
4e
0,02
> 02!
0,16
^
C u + - O 2 + H2SO4
0,04
••'
a04<-0,02
r v i x . ; ^
B. k h u H2O va oxi hoa ion Cu^*.
D . k h u C u va oxi hoa H2O.
-'
Van dung cau tho:
+
dJ'/::^':
o r / r .,,
:>
^.A..,.
" A n h la anot nhuong e - E m la catot nhan e thoi m a "
+ Cue am (catot) eo Cu^* va H2O -> H2O bi dien phan:
> Cu
Cu^* + 2e
{Qua trinh khu)
+ Cvre duong (anot) eo H2O va ion SO4'
4H2O - 4e
,_tL.
H2O bi di?n phan:
> O 2 1 + 4H*
{Qua trinh oxi hoa)
-•
Dap an A.
Vi
rO':--
.1' -fV"': •
8: Di^n phan 1,5 lit dung dich AgNOs 0,1 M voi di^n eye tro trong t gia,
100%), thu dugc chat ran X, dung dich Y va khi Z . C h o 12,6 gam Fe vao Y ,
sau khi cac phan u n g ket thiic thu dugc 14,5 gam hon hgp k i m Idai va
+4H*
0,04
V = 0,04.22,4 = 0,896 lit
+ H2O -^S^
cuong do dong dien khong doi 2,68A (hi^u suat qua trinh di^n phan la
> Cht , ^ ^
Mol: 0,04 - > 0,04 - >
CuS04
Laigidi:
r-'v-nh.
2e
0,04
C. k h u Cu2* va oxi hoa 504'.
di^n phan la 100%. Gia tri ciia m la:
2C1- -
0,04
A. oxi hoa H2O va k h u ion Cu^"^.
2,68A, thu dugc V lit khi (dktc) bay ra a anot. Biet hieu suat ciia qua trinh
'-'I'lh
->
thi cae qua trinh xay ra 6 cue am va cue duong Ian lugt la
,,. s n i j nerf
1;,:;,
0,04
' i::;/ ^
'••'^•^>
V i dy 7: K h i dien phan dung dich CuS04 (vai di^n cue tro, eo mang ngan xop)
cue tra, mang ngan xop) trong thoi gian 2 gio voi dong di$n c6 cuong dp la
A^c;
. ;
-> Dap an D .
V i du 5: Dien phan dung dich X chua 0,12 mol CuS04 va 0,04 mol NaCl (dien
..
' •J
_> Khoi lugng dung dich giam = 64.0,08 + 0,04.71 + 0,02.32 = 8,6 gam.
ny-
DapanA.
1
A. 0,560.
+ Na2S04
T h u tu dien phan n h u sau:
Theo bai, dung dien phan khi 6 catot bat dau c6 khi thoat ra —> CuS04 vua
Ta c6:
> CuCh
+ 2NaCl
Mol: 0,04
0,04
' l i i
X gom: CuCh = 0,04 mol; CuS04 = 0,08 mol va Na2S04 = 0,04 mol.
Laigidi:
.
v
Ob^^^--
khi N O (san pham k h u duy nhat eua N"*"^). G i a trj ciia t la
Dap an C.
)• >,
„ ,
A. 0,8.
,
B.1,2.
C.1,0.
D.0,3.
V i d v 6: Dien phan dung dich X gom 0,08 mol NaCl va 0,12 mol CuS04 (dien
eye tro, mang ngan xop) den khi thay thoat ra 6 anot 1,344 lit khi (dktc) thi
Phan ung di?n phan dung dich AgNOa:
/ ngung dien phan. Kho'i lugng dung dich sau dien phan giam di bao nhieu
2AgN03
gam?
A. 3,48 gam.
B. 6,04 gam.
C. 2,56 gani.
D. 8,60 gam. ^
Mol;
2a
+ H2O
;^ T
> 2Ag + i 0 2 + 2HNO3
.
2a
^.
,
':
• _ 2a
123
Cty TNHH MIV UVVH Khang Vi$t
Ca'm nang an luy$n thi djii hgc 18 chuy6n ai H6a hqc - Nguygn Van H&'\
Nhdn xet: V i cho Fe vao d u n g d i c h Y t h u d u o c h o n h o p k i m loai —> Y chiia
A g N O s —> A g N O ^ chua b i dien p h a n het.
Mol:
+
0,5a <— 2a
Fe
Mol:
b
. 0,5a
+
,
0,5a
2AgN03
<-
^M.-.
tOiO'
•
, fef)/)
> Fe(N03)2 + 2 A g
2b
2NaN03 — 2 N a N 0 2
•--vtJuD iiTiS;. X
i^**"'s'^'iq''3'?:< 01 i<:
> 3Fe(N03)2
2Fe(N03)3
0,25a
Mol:
+
—>
2KMn04 — K 2 M n 0 4
r O , y :h
> Fe(N03)3 + N O + 2H2O
4HN03
^
2b
Theo bai: 12,6 -56(0,75a + b) + 2b.l08 = 14,5.
'
2a = 0,1
-^DapanC.
''
'
'
—> D a p an A .
'fii''''
L i thuyet
O^Hf BOH ixo
*
MMOZ nitrat:
AgNOs —
A. 40%.
^
CaC03.MgC03
P
C84%.
Lin giai:
^
—
^
CaO
^'.ilf^y'
'->•'
.Av.-v;ty>
D . 92%.
•
+ M g O + 2C02t
0,2
.gnuG...
QA
(n-ys-f ! , '
gj^^j^^-j,.
D a p an D . " ''^'^
% n i c a C O 3 . M g C O 3 = ^ - 1 0 0 % = 92% ^
n.i
.
:*-;H-
mCaC03.MgC03 = 0,2.(100 + 84) = 36,8 g a m
io2
io2
A g + NO2 +
_
Phan u n g nhiet p h a n :
Mol:
C u O + 2NO2 +
^
,
B.50%.
•"'^
if•
2KNO2 + 02
Cu(N03)2 —
n
k h o i l u g n g ciia C a C 0 3 . M g C 0 3 t r o n g loai quang tren la
" ' '
2KNO3
i;.(f-
....>ttf^
tap cha't t r o sinh ra 8,96 l i t k h i CO2 (6 dktc). T h a n h p h a n p h a n t r a m ve
in
a.
+ O2
.
>
V i dv 2 (B-08): N h i e t p h a n hoan toan 40 g a m m o t loai q u a n g d o l o m i t c6 Ian
^ ^.^.^^^^.^
5. P H A N L f N G N H I E T P H A N , D O T C H A Y
: to f s i i s u y i r t .-n'u
,.3,..,
2KC1 + 3O2 ^
2KC103
••!'•! nfiifc
:f,
CaO + CO2
^
.
- = 0,1 m o l - > t = 3600 giay = 1 gio.
+ M n 0 2 + O2
C u O +2NO2 + i o 2
2
Cu(N03)2
CaC03 —
M a t k h a c : n^gNOa = 2a + 2b = 0,15 ~ ^ ^ =0,05 m o l ; b =0,025 m o l . : i C i
n^=
, 5 . ^ f i ^ t j i , pffifu:i
cac mum v6 co. Cac phuong trinh phan ling:
Cac p h a n u n g k h i cho Fe vao Y: i''"»ii'''V ^i'v....:^.,.' ' 'f:^'.-']/!; . .,, vv^'uvjr"'"'
Fe
Lai giai:
l^han xet: D a y la dang cau hoi li thuyet kiem tra cac em ve do ben nhi^t ciia
J^^e
V i d\ 3: N u n g n o n g cac h o n h o p bot ran t r o n g b i n h kin: C u + Cu(N03)2 (1); C u
*
Muoi cacbonat:
CaCOs
C a O + CO2
2FeC03 + -O2
*
,
*
+ KNO3 (2); Fe + S (3); MgCOs + M g (4). So t r u o n g h g p xay ra s u o x i hoa
k i m loai la:
A. 1.
FeaOs +2CO2
Mudihidrocacbonaf'
,0
2NaHC03
' ' f . ' i ' ^ ^ ' V 'b.i> - m . b t.t
: ) H l i q .^x* ?
V'
'
' -"f '(Ofl,* ( o i b '"•'^h r^*-' '}T('jr
Na2C03 + CO2 + H2O , >, . ,
Ca(HC03)2 —
CaC03 + CO2 + H2O j ^ ^ . ,
^
4FeS2 + I I O 2
,0
"• "''fidA v i j b 'Uff
-^—^
2Fe203 + 8SO2
.1,1 .M
.
, ,
.8,(J„A
b. V I D V M A U
KCIO3. So
muoi
t r o n g day k h i bj n h i ^ t phan tao ra so m o l k h i I o n h o n so m o l muo'i t h a m gia
p h a n l i n g la:
^
B.3.
'
'
C. 4.
D . 5.
''
C. 3.
+ O2
'" > 2 C u O
2KNO2 + O2
(2) 2 K N 0 3 — ^
2Cu
+ O2 —
+
2CuO
'" > FeS
(4)MgC03 — ^
Mg
^
CO2
'|f
D . 4.
^, j |,,^ j^^,^,
CuO + 2NO2+
(1) Cu(N03)2
(•^) Fe + S
V i du 1: Che day cac m u o i : K M n 0 4 , N a N 0 3 , Cu(N03)2, CaCOs,
A. 2.
B. 2.
Cac p h u o n g t r i n h hoa hpc:
2Cu
MMO? sunfua, disunfua:
f
, ;
io2
'
HEoH f
-
;M'.;e
•
'• -
C u b } o x i hoa
j ,
^ ,
,
.BV
/
>i
3 ^
^
Cubioxihoa
->
Febioxihoa
M g O + CO2
'" > M g O + C
->
M g b i o x i hoa
- > D a p an D .
125
Ca'm nang On luy^n thi dgi hgc 18 chuy6n dg H6a hgc - NguySn Van H5i
Cty TNHH MTV DWH Khang Vigt
Lieu y: 6 hon hg-p (3), Fe la chat khir —> bi oxi hoa; 6 hon hgp (4), M g c6 the
chay trong khi CO2.
V i d^ 4: Nhiet phan 18,8 gam Cu(N03)2 mot thoi gian, thu dugc 12,32 gam
chat ran. Hieu suat cua phan ung nhift phan la /fiv/
A. 40%.
B.60%.
C.80%. ..^^^v^,, D.50%.
Ldigidi:
f^OM:' -*0u*') <- " - -
-,00
10,0
„ ^
,
"Cu(NO3)2=-^=0,lmol.^
Nhan xet: Khoi luong chat ran giam = khoi lugng khi bay ra.
Phuong trinh hoa hoc:
CU(N03)2
• Mol:
X
m^02
^
CUO + 2 N O 2 + - O 2
nSritt finfifli
2x
0,5x
5 ,
,;t r t f i k o i t .tsrir)'qi
+ r " 0 2 = 18,8 - 12,32 = 6,48 gam -> 46.2x + 32.0,5x = 6,48
-> x = 0,06 mol ^
H = - ^ . 1 0 0 % = 60%
Dap an B.
6. B A I TAP O N L U Y E N
Bai 1: Dung djch X gom Ba^ Na" (0,01 mol) va OH" (0,05 mol). Dung dich Y gom
Na^ HCO3 (0,02 mol) va CO 3' (0,01 mol). Trpn X voi Y thay tao ra m gam ket
tua. Gia t i l cua m la
A. 5,91.
B.3,94.
C. 7,88.
D. 1,97.
Bai 2: Hoa tan hoan toan 9,2 gam hon hop MCO3 va M'C03 vao dung dich
HCl, thu duoc V lit khi (6 dktc) va dung dich chiia 10,3 gam muoi. Gia tri
cua V la
A. 1,12.
B. 1,68.
C.2,24.
D. 3,36.
Bai 3: Cho 5,76 gam hon hop X gom muoi cacbonat va hidrocacbonat cua kim
logi kiem M tac dung het voi dung dich HCl (du), sinh ra 1,12 lit khi (0
dktc). K i m loai M la
A. Na.
•
-v^-^'
B. K.
C. Rb.
D. L i .
Bai 4: Cho 14,9 gam hon hgp hot X gom Zn va Fe vao 600ml dung dich CuSO*
0,25M. Sau khi cac phan ling xay ra hoan toan, thu dug-c dung dich va 15,2
gam hon hgp k i m loai Y. Phan tram ve khoi lugng cua Z n trong hon hgp
dau la
A. 56,38%.
B. 37,58%.
C. 64,42%.
D. 43,62%.
Bai 5: Cho m gam hon hgp bgt Zn va Fe vao lugng d u dung dich CuS04. Sau khi
ket thiic cac phan ung, Igc bo phan dung djch thu dugc m gam bgt ran
Thanh phan phan tram theo khoi lugng cua Zn trong hon hgp bgt ban dau l3
A. 90,28%.
B. 85,30%.
C. 82,20%.
D. 12,67%.
gai 6: Cho 5,1 gam hon hgp X gom Mg va Fe vao 250ml dung dich CuS04 0,3M.
Sau khi cac phan ung xay ra hoan toan, Igc thu dugc 6,9 gam chat ran Y va
dung dich Z chua hai muoi. Khoi lugng cua M g trong X la:
A. 0,9 gam
B. 1,2 gam
C. 1,6 gam.
D. 2,4gam.
Bai 7 (B-07): Khi cho Cu tac dung voi dung dich chiia H 2 S O 4 loang va NaNCh,
vai tro cua NaN03 trong phan ling la
A. chat xiic tac.
B. chat oxi hoa.
C. moi truong.
D. chat khu.
Bai 8: Di?n phan dung dich X chua 0,2 mol CuS04 va 0,12 mol HCl trong thoi
gian 4 gid voi dong dien c6 cuong do la 1,34A, thu dugc m gam kim loai bam
vao catot. Biet hif u suat cua qua trinh dien phan la 100%. Gia tri cua m la:
A. 6,4 gam.
B. 3,2 gam.
C. 12,8 gam.
D. 9,6 gam.
Bai 9: Di^n phan dung dich X gom 0,1 mol KCl va 0,15 mol CuS04 (di^n cue
tro, mang ngan xop) den khi khoi lugng dung dich giam d i 10,75 gam thi
ngimg dien phan. The tich khi (dktc) thoat ra 6 anot la
A. 1,12 lit.
B. 1,68 lit.
C. 2,24 lit.
D. 2,80 lit.
Bai 10 (CD-08): Nhiet phan hoan toan 34,65 gam hon hgp X gom K N O 3 va
Cu(N03)2, thu dugc hon hgp khi Y. Ti khoi cua Y so v o i hidro bang 18,8.
Khoi lugng ciia Cu(N03)2 trong hon hgp ban dau la
A. 20,50 gam.
B. 11,28 gam.
C. 9,40 gam.
D. 8,60 gam.
Bai 11 (A-09): Nung 6,58 gam Cu(N03)2 trong binh kin khong chua khong khi,
sau mgt thoi gian thu dugc 4,96 gam chat rSn va hon hg-p khi X. Hap thu
hoan toan X vao nuoc de dugc 300ml dung dich Y. Dung djch Y c6 p H bang
A. 4.
B. 2.
C. 1.
D. 3.
Bai 12 (B-08): Nung mgt hon hgp ran gom a mol FeC03 va b mol FeS2 trong
binh kin chua khong khi (du). Sau khi cac phan ung ket thiic, dua binh ve
nhi|t dg ban dau, thu dugc chat ran duy nhat la Fe203 va hon hgp khi. Bie't
ap suat khi trong binh truoc va sau phan ung bang nhau, moi lien h$ giiia a
vabla
io;,
A . a = 0,5b.
B.a = b.
C.a = 4b.
D . a = 2b.
7. H l / O N G D A N - L 6 I G I A I
Bail:
"
Dung dich X trung hoa di?n: 2 n 2+ + 0,01 = 0,05 -> n ^ 2+ = 0,02 mol.
HCOi
Mol: 0,02
+
OH"
> CO3" + H 2 O
0,02
,iv0,02
^
=r:/
n^^2-= 0,02 + aOl = a03 m o l .
CO3
Ba2*
Mol: 0,02
+ CO 3"
0,02
> BaC03i
0,02
mBaC03
= 0,02.197 = 3,94 gam.
Dap an B.
197
Ca'm nang 6n luygn thi dgi hgc 18 chuy6n dg H6a hpc - Nguygn Van Hii
Cty TIMHH MTV DWH Khang Vi$t
Bai 2:
Nhati xet: T r o n g p h a n u n g , m u o i cacbonat bien t h a n h m u o i clorua, nen:
> 2 m o l CI-
i s 1 '"ol
Theobai: a m o l
1,1
0,1 m o l
-> a =
- » khoi Iirong tang 71 - 60 = 11 gam.
,
,
,
tang 1 0 , 3 - 9 , 2 =1,1 gam.
M2CO3 + 2HC1
a mol M g
{t>i>
IMK. imfi
> 2MC1 + CO2 + H2O
i'l/,,
-^oifo ...
MHCO3 + H C l
> M C I + CO2 + H 2 0 . - . t - a
.roBSli,.^ :
N h a n thay : nx = n c o j = 0/05 m o l .
< ' i ^ b S""^' f ^
—
5 76
-> M x = ^ — = 115,2 -> M H C O 3 < 115,2 < iVbCOs.
0,05
M + 6 K 115,2 < 2 M + 60 -> 27,6 < M < 5 4 , 2 ^ M l a K - ^
i:;is::S c.;j,K, f i & o l ns
g^j^,
Dap anB.
;\«l,t
^ -lnr,i
^
> a mol Cu
b m o l Fe
Theo bai:
> h mol Cu
mtang
r ; -.^'^ :
kho'i l u g n g tang (64-24)a = 40a.
->
tang (64 - 56)b = 8b
= 6,9-5,1 = 1,8 gam ^ 8b + 40a = 1,8
_> a = 0,0375 m o l ; b = 0,0375 m o l .
- > m M g = 0,0375.24 = 0,9 (gam)
Bai 7:
'
'^
^ ^. ^ ^ ,
D a p an A .
=V *
—
"
Qua t r i n h d i ? n l i :
^
+ NOJ
> 2H* +
H2SO4
SO
I"
''
D u n g d i c h di§n l i c6 m a t H * va N O 3 —» hoa tan C u theo p h a n u n g :
+ 2NO:
> 3Cu^* + 2 N O t
Fe = 56 da p h a n u n g m o t phan tao thanh C u = 64 —> Y g o m Fe d u va Cu.
3Cu
G p i so m o l b a n d a u : Z n = x; Fe = y
- > C u la chat k h u , N O J la chat o x i hoa
65x + 56y = 14,9.
b + 5a = 0,225.
Bao toan electron: 2 n 2+ = 2nMg + 2npe(p.) -> 0,075 = a + b.
NaNOs
Z n phan u n g truac r o i den Fe. D o mx < m Y —>
N^fln xet: T i n h k h u Z n > Fe
'
- > Fe = 56 d a p h a n u n g m g t p h a n de tao t h a n h C u = 64 —> Z g o m
Gpi so m o l M g ban d a u = a m o l ; npg( p.) = b. N h a n thay:
22,4
V
D a p an A .
^
I\lhan xet: T i n h k h u M g > Fe —> M g p h a n l i n g t r u o c r o i m o i d e n Fe. D o mx <
MgS04vaFeS04.
1 12
B a i 3: n(-02 = — = 0,05 m o l . Cac p h i r o n g t r i n h p h a n u n g :
^
=— ^
.100% =
.100% = 90,28%.
65x + 56a
65.8a + 56a
pai 6:
jnv
n c o 2 = n ^ ^ 2 - = 0/1 m o l -> V c o j = 2,24 l i t .
->DapanC.
->
"
o/„^
•^^^^
+ 8H*
+ 4H2O
^
, ''
, i i J g n o i< > >
D a p an B.
Dat: nFe(p.)=a.
Nhan xet: Day la p h a n u n g d i ^ n p h a n h o n h g p —> nen d u n g cong t h i i c t i n h
Bao toan electron: 2 n ^ 2+ = 2n2n + 2npg(py^ -> 0,15 = x + a.
g
M a t khac: mv = 15,2 - » 56(y - a) + 64(x + a) = 15,2
^ % m z „ = .5:1:^.100% = 43,62%
x = 0,1; y = 0,15; a = 0,05.
D a p an D.
2"
14,9
Cach 2: A p d u n g tang - g i a m k h o i l u g n g :
.,rt:
^
x mol Zn
> X mol Cu
-> k h o i l u g n g g i a m (65 - 64)x = x
a m o l Fe
> a mol Cu
^ k h o i l u g n g tang (64 - 56)a = 8a. V
Theo bai:
mwng
^
' •^ ^ *
'^^zn +
^^Ve{pu)
~^
Cu
Catot:
C u - . 2e ' — ^
Mol:
0,1 < - 0,2 - >
B a i 5: G o i so'mol ban d a u : Z n = x; Fe = y. N h a n thay:
X mol Zn
> X mol Cu
a m o l Fe
> a mol Cu
Theo bai: mtang = 0 -> x = 8a.
128
,
0
y
f
- » k h o i l u g n g g i a m (65-64)x = x. I
->
Cu
' ^-'^^'^^^
SO^
4r^-~m^^<^
ai
;
^^^^ *
•
^
„ j^H < <
f*,0 •
Phan u n g " q u y d o i " gia d i n h :
^,15 = x + a.
a = 0,05; x = 0,1 ^ %my„ = ^ ^ ^ . 1 0 0 % = 43,62% .
^"
14,9
H
-> m = ai.64 = 6,4 gam -> Dap an A. ^ „ ^
CuS04
^
CI
Bai 9:
= mv - mx = 0,3 g a m -> 8a - x = 0,3.
Bao toan electron: 2 n ^ 2+ =
« --^
u
u I . .
I t 1,34.4.3600 „ ^
f
n h a n h so m o l electron: np= — =
==0,2 m o l .
^
F
96500
X chua: n ^ 2+ = 0,2 m o l ; n ^ , . = 0,12 m o l ; n^+ = 0,12 m o l v a n^^2- = 0,2 m o l .
tang (64-56)a = 8a. '
Mol:
+
0,05 < -
2KCI
> '
> CuCh
0,10
0,05
+
K2SO4
. + ,n')<,'^r
0,05
- > X g o m : C u C h = 0,05 m o l ; CuS04 = 0,10 m o l va K 2 S O 4 = 0,05 m o l .
T h u t u d i ^ n p h a n n h u sau:
CuCh
Mol:
0,05
'^P"' > C u + C k
->
0,05
0,05
'
mgiim
'-^"^ '
'
= 0,05.(64 + 71) = 6,75 gam.
K h o i l u g n g d u n g d i c h con g i a m tiep 10,75 - 6,75 = 4 gam.
!^ '
129
elm nang On luyQn thi dgi hpc 18 chuySn dg H6a hoc - Nguyin VSn HSi
Mol:
dpdd ^ „
1 „
> C u + - O 2 + H2SO4
+ H2O
CuS04
a
aSa
^
a
64a + 32.0,5a = 4 ^ a = 0,05 mol.
' ^
- > Vci2 +
= (0,05 + 0,025).22,4 = 1,68 lit
Bai 10: Phuang trinh hoa hoc:
Cu(N03)2
Cty TNHH MTV DWH Khang Vi$t
l i X H CHAT CIJA CAC OXIT
;
Dap an B.
,
.,
iiCL- '
CuO + 2 N O 2 + I 0 2
, ion" Cir, r ;
a*
...
* phan ung trao aoi
Mol:
Mol:
=° ^'^S-^vd " i(.*m rifid o'X:
y
^^^'^-^ =
'^^"S '^""g * u c cua so do duong cheo: "'^^^'^^ ° '
n NO2
32-37,6
5,6 2
2x
2
46-37,6
8,4
3 ' 0,5(x + y ) ~ 3 ^ ^ ~ ^ ' ' '
, , ,. '
K Mat khac: I88x + l O l y = 34,65
x = 0,05 mol; y = 0,25 mol.
CU(N03)2
- > m^joj +
,>:••...'.•
+ O2 + 2 H 2 O
,SMV',
A. 80.
^
a
' '
=
02Mol:
D . 120.
it o r
„.
p-g,a^
^O'^"'
~* " o 2 -
.
0,1
2H^
.H20
0,2
' *''
•
'
'
''
•.MAm^Oi'A:.
"'
= 0,1 mol -> V^d H2SO4 =
Vay: n H 2 S 0 4 =
:; '
= 0/1 lit = 100ml.
Dap a n C.
V i du 2: H o a tan hoan toan 5,44 gam hon hop X gom Fe203, FeO, CuO can 80
m l axit H2SO4 I M (loang). K h u hoan toan 5,44 gam X bang khi H2 (nung
nong) thu dugc m gam kim loai. Gia tri cua m la
a
0,25a + 2,75b = a + 2b ^
' '
C. 100.
B.60.
trong axit tao thanh H2O theo phuong trinh:
, j i t . .*,tj '••,,,r „ j : , ;»»u*iD}:
^ I'^'-^^'^'P^tm&lig^^fim.d'^bmy^^
'
Khi cho oxit kim loai tac dung v a i axit, ion O^" trong oxit se ket hg-p voi
''^
A. 4,32.
4FeS2 + 1 1 0 2 — ! ^ -> 2Fe203 + 8SO2
b
2,75b - >
2b ^
Dap an B.
> Fe2(S04)3 + SO2 + 4H2O
•• i
Bao toan kho'i l u o n g t a c6: m o 2 = 5,82 - 4,22 = 1,6 gam
-> Fe203 + 2 C O 2
N;zfl«^e^ V i a p s u a t k h o n g t h a y d o i
;
Way.
' . f ^ ^ ' ^ n f M l g
0,25a
v. '
Lcngidi:
IH1=^
= l O - ' - > p H = l->Dapand:^
0,3
+ - 0 2 ^
2
> 3Fe(N03)3 + N O 2 + 5H2O
. i j,,,
V i dv 1: Dot nong 4,22 gam hon hop X gom Mg, A l va C u trong khi oxi d u , thu
V ^ " " N 0 3 = n N 0 2 - > n j ^ , = 0,03 m o l .
2FeC03
„
b. V i du mau
-> 4 H N 0 3
Bai 12: Cac phan ung hoa hoc:
.->gfti^O??^P'-fv'iv *n'Mvni,» 'J .4 if,,
> 3Fe(N03)3 + N O + 5H2O
2FeO+ 4H2S04(^flc)
0,5a
V^JJ-M
> 2Fe(N03)3 + 3H2O
H 2 S O 4 I M . Gia trj cua V la
= 6,58 - 4,96 = 1,62 gam - > 46.2a + 32.0,5a = 1,62.
- > a = 0,015 mol. ••••
130
> CuS04 + H2O
Fe304 + IOHNO3
CuO + 2 N O 2 + -O2
2a
Mol:
C u O + H2SO4
,'. } i i W i t. ' . • •
1
L
:'.i.ih 'itn.ih tfrti ' •
dugc 5,82 gam hon hop Y . De hoa tan het Y can to'i thieu V m l dung dich
mart xet: K h o i lugng chat rSn giam = khol l u g n g k h i bay ra.
Phirong trinh hoa hpc:
'' '^ymii..fmf'Mttml.
t « MiirM
Mol:
> 2 A I C I 3 + 3H2O
3FeO+ IOHNO3
m c u ( N 0 3 ) 2 =0,05.188 = 9,4gam ^ D a p a n C .
Bai 11:
4N02
AI2O3 + 6HC1
» Phan ung oxi hoa-khu
••'-'-^ , . . « r . ; n o i t o t e a « m c .
' t' /I
i^ififel I
Fe203 + 6HNO3
KNO3 — ^ KNO2 +-O2
Ml
OXIT B A Z O + A X I T
a = b. 1
t
.
B.4,16.
C. 4,64.
Laigidi:
B. 4,48.
- a " ! ^ y.,,.
• '>>''"'
^ H 2 S 0 4 = 0,08.1 =0,08 m o l - ^ n ^ + = 0,16 mol.
'Nhan xet: K h i cho oxit kim loai tac dyng voi axit, ion O^- trong oxit se ket
hgp voi H * trong axit t^o thanh H2O theo phuong trinh:
Ccim nang On luygn thi dgi hpc 18 chuySn ai H6a hpc - Nguygn VSn Hii
02-
+
2H*
ai6
M o l : 0,08
>
.
K
H2O
^
^
J
•
d u n g d i c h H2SO4 I M (loang, v u a du). Sau k h i cac p h a n u n g xay ra hoan
t * * i t i,
toan, t h u d u g c d u n g d i c h X. C o can X t h u d u g c m g a m muo'i k h a n . Gia t r j
. 'riXA + 0 . i f t * l i : '
^,
, „
Dap an B.
^ m = 5,44 - 1,28 = 4^16 gam
5: H o a tan hoan toan 10,56 gam h o n h g p b g t Fe304 v a C u t r o n g 160 m l
yi
^
Vgy: m o - =0,08.16 = 1,18 gam.
Vi
Cty TMHH MTV DWH Khang ViQt
3: H o a tan het 20,88 gam m o t oxit sat bang d u n g d i c h H2SO4 dSc, nong
cua m la
«. > •
A. 15,36.
,
C. 17,28.
. 1 B. 23,36.
dktc). Co can d u n g dich X, t h u dugc m gam m u o i sunfat khan. Gia t r i o i a m la
Gpi so m o l : Fe304 = a; C u = b. Theo bai: 232a + 64b = 10,56.
A. 36.
Nhqn xet: C h i c6 Fe304 p h a n u n g true tiep v o i axit. Cac p h a n u n g hoa hgc:
B.24.
C. 72.
D . 54.
'^'1-
Lcngidi:
- i'-OlAm
Nhan xet: O x i t sat FexOy tac d u n g v o i H2SO4 dac, n o n g
i K, .
Fe304 +
a = 0,04
Cac ban l u u y rang 1 m o l FeO hoac Fe304 deu chiia 1 m o l Fe^^ nen deu c6
Cu
kha na ng n h u o n g 1 m o l electron, d o do:
= npe^Oy =
2nso2 ^
^ F ^ x O y = ^
i"Fe2(S04)3=
mm tfh •
Vi
= 232 (Fe304).
B. Fe304 va 4,64.
C. Fe203 va 3,20.
D . Fe304 va 2,32.
0,02
''' "
> xFe2(S04)3
x.0,01
m = 232.0,02 = 4,64 gam - > Dap an B.
0,04
n-BO,'.
C. 75ml.
D. 90ml.
,n
'
^'"'^^
Oey, OK>1 BR.': •
+ moxi = m o x »
= 4,46 - 3,5 = 0,96 gam
->
.
H20.
~
nH2S04 =
a045 m o l
'
''
* ^fTfeisb ,i;:ilb
D a p an A .
2. OXIT BAZO + C H A T KHU" (NHIETLUY5N)
a. L i t h u y e t :
*
n h u o n g 1 m o l electron, d o d o :
X = 3 ^ O x i t sat la Fe304.
moxi
- ^ VHCI = 0,045 l i t = 45ml ^
« »OPrH'i : y « '
= 2-0,01= 0,02 m o l
mkimid
Suy ra: n^+ = a 0 9 m o l ^
T r o n g 1 m o l FeO hoac Fe304 deu chua 1 m o l Fe^* n e n d e u c6 k h a nang
|§. -> x.0,01 = 0,03 ^
0,02
, >
0 96
= - r r - = 0,045 m o l
32
vjiAii p a^rtfJ Ywi> 'sr-i.'
-> n o = a 0 9 m o l - > nQ2-(„^i,) = 0,09 m o l .
g
. , . u . j,^ £,
r.O
Nhqn xet: O x i t sat FexOy tac d u n g v o i H2SO4 dac, n o n g - > SO2 t h i oxit la
2FexOy
+ 2FeS04
B.60ml.
o ^ + 2H^ — >
FeO hoac Fe304.
BaotoannguyentoFe:
> CuS04
, • •
K h i cho oxit bazo tac d u n g v o i axit tao ra nuoc:
Lmgtat:
12
0 224
n F e 2 ( S 0 4 ) 3 = 7 ^ = ^'^^ m o l ; nso2 = - r r — = 0,01 m o l .
"Fe^Oy
'
6: C h o 3,5 g a m h o n h g p X g o m A l , Fe v a C u 6 d a n g b g t tac d u n g hoan
-»
, ,
22,4
0,04
Lcngidi:
d i c h chiia 12 g a m m g t loai m u o i sat d u y nhat v a 0,224 l i t SO2 (dktc). Cong
= npe^Oy = 2nso2 ^
+ Fe2(S04)3
Bao toan k h o i l u g n g :
A . FeO v a 1,44.
-
- > b = 0,02 m o l .
A. 45ml.
yjj
thijrc cua oxit sat va gia t r i m Ian l u g t la
+ 2H2O
tich d u n g d i c h H 2 S O 4 I M vira d u de p h a n u n g het v o i Y la
0 , 0 9 . | .400 = 54 g a m - > Dap an D .
400
0,04
Fe2(S04)3
toan v o i O2 t h u d u g c h o n h g p Y g o m cac oxit c6 k h o i l u g n g 4,46 gam. The
^^.o..
> 3Fe2(S04)3
+
- > m = 0,02.160 + 0,02.400 + 0,08.152 = 23,36 g a m - > D a p an B.
V i d u 4: H o a tan hoan toan m gam Fe^Oy bang H 2 S O 4 dac nong, t h u d u g c dung
.i
0,16
M o l : 0,02 - > 0,02
1 008
" F e x O y = 2 - ^ J J = 0,09 m o l
B a o t o a n n g u y e n t o F e : 2Fe304
> FeS04
4H2SO4
M o l : 0,04
SO2 t h i oxit la
FeO hoac Fe304.
'
j i a!
Laigidi:
(du), t h u d u g c d u n g d i c h X v a 1,008 h't k h i SQz (san p h a m k h u d u y nhat, 6
MI,
D . 13,28.
^h- lorn
Tac dung voi CO, Hz
,?t B M fj-j'iS
Fe203 + 3 C O —
^
2Fe + 3C02
Fe304 + 4 C O —
^
3Fe + 4C02
CuO
+ H2 —
^
C u + H2O
lim y: K h i CO, H2 chi k h u d u g c cac oxit ciia k i m loai tir Z n ve sau.
Tac dung oai Al (nhiet nhom)
2A1
+ Fe203
—
^
AI2O3
+ 2Fe
.
.
'
.1/;) .J'
"
-
133
Cty Ti\ : '/ITV DVVH Khang Vi?t
Cim nang 6n luygn thi dgi hgc 18 chuygn dg H6a hgc - Nguygn Van HSi
+ 3Fe304
'** 8A1
3CuO
14
2A1
+ Cr203
->
-
Laigidi:
+ 9Fe
— 4 A I 2 O 3
+ 3Cu
X, xay r a 2 p h a n u n g :
AI2O3
+
Fe203 + 3 C O
2Cr
CuO
V i d\ 1 (A-09): C h o l u o n g k h i C O ( d u ) d i q u a %1 g a m h o n h g p g o m C u O va
,
-!,.•.,
B. 8,3 g a m .
.'. V' -
C. 2,0 gam.
.'•
+ CO
Cu
i f • I'lKG
n o = ~ = 0 , 0 5 m o l - > n^uo = 0,05 mol.
> 2NaA102 +
Vay Z g o m : Fe, C u
D a p an D .
SO.O :!o
^
A. Fe304 va 0,224.
B. Fe304 v a 0,448.
^^'l '
C. FeO v a 0,224.
D . Fe203 v a 0,448.
;n ;
Lai
nung
d u , t h u d u g c 7,5 g a m k e t t u a . Chat
k h u d u y nhat, 0 dktc). Gia t r i ciia V la
C. 1,12.
FexOy
Mol:
a075.4
0,84
0,075 m o l
traodoi
->
VNO
3 n jvjo
'^NO
>
Fe, C u .
B. Fe203,Cu.
C. A l , Fe, C u .
- =
0,015 3
- 5 ^ = 4
V c o = 0,02.22,4 = 0,448 l i t - > D a p an B. iD
:k.;
"CO2
= 0,02 - »
Vco2 = 0,448
B.4,48.
C. 2,24.
1 mol CO
L o a i A v a C.
, „,
g i a m 16 gam.
->
lmol(COvaH2)
Theobai:
r
(C02vaH20)
G i a m 16 gam.
giam 4gam.
0,25 m o l
> V = 0,25.22,4 = 5,60 l i t
D . Fe, C u .
D . 5,60.
!-2-> C O 2 - > K h o i l u g n g chat ran g i a m 16 gam.
lmolH2
f - - -
lai p h a n k h o n g t a n Z . Phan k h o n g tan Z g o m
AI2O3,
ax = 0,015; ay = 0,020 ^
^
Laigidi:
V i d v 2: C h o k h i C O ( d u ) tac d y n g v o i h o n h g p X g o m A I 2 O 3 , FezOa, C u O
t h u d u g c chat r a n Y . C h o Y v a o d u n g d j c h N a O H ( d u ) , k h u a y k i , tha'y con
A.
X
= 0,015 m o l
A . 6,72.
= 0,15 m o l .
~ 0,05 mol.
= 1,12 l i t - > D a p an C.
ay
••'••;,'.(1J:'
hoan toan, k h o i l u g n g h o n h g p r a n g i a m 4,0 gam. G i a t r i ciia V !a
Bao toan electron:
ttg trao doi —
ax
...8
dv 4: C h o V l i t h o n h g p k h i (6 dktc) g o m C O va H2 tac d y n g v o i m g t l u g n g
d u h o n h g p r a n g o m FeO v a Fe304 n u n g n o n g . Sau k h i cac p h a n u n g xay ra"
> CaCOa + H2O
I\QQ = TXQQ^ =
yC02
D a p an B.
.
2nQQ. M a
56
xFe +
M a t k h a c : n p e i n o = 0,015:0,02 = 3 : 4 - > Fe304
Nhqn xet: K h i cho h o n h g p oxit tac d u n g v o i k h i C O t h i :
H e traodcS =
^
a
Nhdn xet: n o (oxit) = "co=
_^
O day, cac e m can s u d u n g so d o phan l i n g :
j
^C0
CuO, F e 2 a
•> X
) Muoinitrat + N O
CO2 + Ca(OH)2
yCO —
Cach 2:
Laigidi:
;
Mol:
+
-> O x i t sat la Fe304
D . 3,36. ,
ncaco3 = 0,075 m o l
gidi:
Giai theo p h u o n g t r i n h hoa hgc
r a n X p h a n l i n g v o i d u n g d j c h H N O 3 d u t h u d u g c V l i t k h i N O (san p h a m
B.4,48.
•'
3: K h u h o a n toan m g t oxit sat FexOy 6 n h i f t d g cao can v i r a d u V l i t k h i
n o n g , sau m o t t h o i g i a n t h u d u g c chat r a n X va k h i Y. C h o Y h a p t h u
A . 2,24.
"
Cach 1 :
' ^
V i d u 3: D a n l u o n g k h i C O d i qua h o n h g p g o m C u O v a Fe203
'!
H2O
C O (dktc), sau p h a n u n g t h u d u g c 0,84 g a m Fe v a 0,02 m o l k h i CO2. Cong
m e g e,fl o t O * -
- » n c u o =0/05.80 = 4 , O g a m - > D a p an D .
h o a n toan v a o d u n g d i c h Ca(OH)2
^ C u + CO2
thiic ciia oxit sat va gia t r i ciia V la
+ CO2
T a n g - g i a m k h o i l u g n g : m o (trong CuO) = 9 , 1 - 8 , 3 = 0,8 g a m
y
t, <
' 1
f I f + fit i ! n
AI2O3 + 2 N a O H
Vi
Nhan xet: K h i C O chi k h u d u g c C u O theo phan u n g :
CuO
D . 4,0 g a m .
Lbigiai:
n e n k h i cho k h i C O d u tac d u n g v o i
'° > 2Fe + 3CO2
+ CO —
,
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 h o a n toan:
Kho'i l u g n g C u O c6 t r o n g h o n h g p ban d a u la:
A . 0,8 g a m .
AI2O3
,
Chat ran Y g o m : Fe, C u , AI2O3.
n u n g n o n g d e n k h i p h a n u n g hoan toan, t h u d u g c 8,3 g a m chat ran.
AI2O3
y: K h i C O k h o n g k h u d u g c
im
AI2O3
„
'
.ri
Dap an D .
135j
dm
Cty TNHH MTV DVVH Khang Vi^t
nang On luy$n thi djii h(?c 18 chuy6n d6 H6a hqc - Nguygn Van Hil
K i m loai sau phan ung gom: A l = 0,02 mol; Fe = 0,09 mol.
V i dv 5: C h o 6,72 lit khi C O (dktc) di tu tir qua ong su nung nong dung 10,44
3,
jChi tac dung voi axit: A l -> - H 2
gam mpt oxit sat den khi phan ung xay ra hoan toan, thu duoc hon hgp khi
Y CO ti khoi so v a i hidro bang 18,8. Cong thiic cua oxit sat va phan tram the
tich cua khi CO2 trong Y la
- t u.j'...
>
A.FeO;28%.
B. Fe203; 72%.
C . Fe203; 28%.
Laigidi:
D . Fe304; 72%.
_> nH2 = 0,12 m o l - > V H 2 = 2,688 lit
^CO
44 - 37,6
2 _ 0,12
S'
riC02
28 - 37,6
3 ~ 0,18
Ta c6: no
(oxit)=
Fe304
dii dung dich chua a mol H C l . Gia tri cua a la
A. 0,9.
•i :E u
0,18.16 ^ ^^^^^
56
A. 81,0.
^
Ban dau: nAi = 2.0,3 = 0,6 mol
"Fez03 =
48,2-0,6.27
160
,
B.54,0.
C.40,5.
D . 45,0.
Mol:
0,2
AI2O3
2A1
Fe203
2Fe
0'2
0,4
•
+
,
= 0,2 mol.
1.0
0,4
Cac chat trong phan 2: nAi = 0,1 mol; nFe= 0,2 mol; nAi203 = 0,1 mol.
n H c i =3nAi +2nFe+6nAi203 = l , 3 m o l .
.Cr203
+
2A1
->
AI2O3
+
2Cr
- » Dap an B.
1,5
m^i = 1,5.27.
100
90
2.
= 45 gam
Dap an D
OXIT L I / O N G
Sau khi phan ung hoan toan, thu du(?c 9,66 gam hon hgp
tin
X. C h o X
phan ling voi axit H2SO4 (loang, du) thoat ra V lit khi H2 (dktc). G i a trj cua
...J
ah qsQ'
Cac oxit luong tinh thuong gap: Z n O , AI2O3
'< Tac dung voi axit
^Al203 + 6HC1 IznO
^ Via
B. 2,688.
T I N H
a. L i thuyet:
V i dv 7: N u n g hon hgp bpt gom 6,96 gam Fe304 va m gam A l a nhi^t d^ cao.
A. 1,344.
= 0,3 mol.
Do vay, bao toan nguyen to'Al, ta c6: n A i = nfsjaOH = 0,3 mol.
0^
Mol:
D.1,5.
d y n g va chuyen thanh NaA102.
Dap an D .
78 , ,
;
ncr= — = 1,5 mol
'
C.0,5.
Nhan xet: K h i cho phan 1 + dung dich N a O H , tat ca A l v a AI2O3 deu tac
phap nhi^t nhom voi hi^u suat cua phan ung la 90% thi can toi thieu m gam
,
B. 1,3.
n N a O H = 0,3.1
.
V i dv 6 (CD-09): De dieu che dug-c 78 gam C r tu Cr203 (du) bang phuong
bpt A l . G^a trj cua m la
gom A l va Fe203 (trong dieu ki?n
vira dii voi 150 ml dung dich N a O H 2M. De hoa tan het phan hai can v u a
%Vco2 =72% ^ C h Q n B h o a c D .
- > npe : n o = 0,135 : 0,18 = 3:4 ^
'
thu dugc sau phan ung thanh hai phan bang nhau. Phan mot phan ung
=0,3mol.
n c o , = 0-18 mol ^ nj^,= 10,44
'
khong CO khong khi) den khi phan ung xay ra hoan toan. C h i a hon hgp
22,4
A p dung cong thuc cua phuong phap duong cheo, ta c6:
in
^-DapanB.
V i dv 8: N u n g nong 48,2 gam h6n hgp
,
Theobai: M y =18,8.2 = 37,6 va n Y = n c o =
va Fe -> H2
C . 1,792.
D . 2,016.
Lai gidi:
Nhan xet: Bai nay cac em se gap kho khan neu khong tinh dugc so mol A l ban
+
H2SO4
-> 2A1C13
-> ZnS04
+
3H2O
+ H2O
\Tdcdung vai baza
lAhOs
+ 2NaOH
—>
ZnO
+ 2NaOH -
->
2NaA102
+
2H2O
Na2Zn02 + H2O
dau.
b-Vid\imau
L u u y rang, trong phan ling nhi|t nhom, thuong ap dyng djnh lu|t bao toan
V i d^ 1: C h i diing dung dich K O H de phan bi^t dugc cac chat rieng bi?t trong
khoi lugng, cu the:
i"Fe203 + m A i = m x - » m^i= 9,66 - 6,96 = 2,7gam ^
Phuong trinh phan ung: 3Fe304 + 8A1
Mol:
0,03
0,08
n A i = 0,1 mol.
> 9Fe + 4Ah03.
0,09
0,04
nhom nao sau day?
A.Al203,AL
B.K,Na.
C. Zn, Al.
D . Fe, Mg.
Laigidi:
Loai B: K , N a deu tac dung v a i H2O trong dung djch K O H va sui bgt khi.
^r^7
Cty TNHH MTV DVVH Khang Vijt
Ca'm nang 6n luy^n thi d ^ i hgc 18 chuy§n dg H6a hgc - Nguyln Van Hai
yi
Loai C : Z n , A l deu tan trong dung dich K O H v a siii bpt khi.
.
4: C h o 6,63 gam hon hop gom BaO v a AI2O3 (ti 1§ mol 3:2) vao nude (du),
Loai D : Fe, M g deu khong tac dung voi dung dich K O H .
thu duoc dung djch X. D u n g dich Y gom H C l 0,3M va H2SO4 0,2M. C h o t u
—>• D a p ? n A. Hien tugng n h u sau:
tu den het 100ml dung dich Y vao X, thu dugc m gam ket tiia. G i a tri ciia m
Jv
, AhOstan:
AI2O3+
2KOH
l^^^^M tan v a sui bpt k h i : , •
^»rr A l + K O H + H2O
> KAIO2 + | H 2 t
<
A. 5,44.
m.^abn'iiM -'ii
+ H2O
> 2KAIO2
b r a S 1,0-
^
.,,5,:.,,,,
,
^J®^'
" •
khi CO2 (du) vao Y thu dugc a gam ket tiia. Gia tri cua m v a a Ian lugt la
C . 13,3 v a 3,9.
• ,:
Na20 + H2O
i
> 2NaOH
A l + N a O H + H2O
Mol: 0,1
Thoi khi CO2: NaA102
Mol:
^
> NaA102 + - H 2 t
2
0,1
'
0,1
+ CO2 + 2H2O
0,1
V i d\ 3: C h o khi H2 (du) di vao 6'ng s u nung nong dvrng hon hgp X gom
AI2O3, FezOs, C u O thu dugc chat ran Y . Cho Y vao dung dich K O H (du),
C . Fe, C u .
^
D . A l , Fe, (fu.
Ldigiai:
'
Lieu y: K h i H2 khong k h u dugc AI2O3 nen xay ra 2 phan ung:
Fe203 + 3H2 — ^
CuO
+ H2
2Fe + 3H2O
C u + H2O
0,02
AIO2
Mol:
^
'
'
K h i cho y tac dung voi K O H d u thi AI2O3 bi hoa tan hoan toan do c6 tinh
AI2O3 + 2 K O H
0,01
Mol: a02
V a y Z gom: Fe, C u —> D a p an C .
" '
:'
,
.'HOu-
> 2AIO2 + 2H2O
• '
+ 3H*
<-
^ t i,
, <, i c^,v
1
- u. <
y
TIXAl.
> Al(OH)3>l.
0,02
'
\^i\.
> A P - + 3H20'^'''^
0,03
" ' '
^
H*-^^''' •+ '
^ BaSOii
<-0,02
a02
.
•
' H '
- > m = mAi(OH)3 + mBaso4 = 0,01.78 + 0,02.233 = 5,44 gam.
- > D a p an A .
V i d\ 5: H o a tan hoan toan m gam hon hop gom N a v a AI2O3
vao nuoc,
thu dugc k h i H 2 v a dung dich X trong suo't. T h e m tu t u dung dich H C l
I M vao X, khi het 100ml thi bat dau xuat hien ket tua; khi het 200ml hoac
400ml thi deu thu dugc a gam ket tua. G i a trj cua m v a a Ian lugt la
A. 13,40 va 3,90.
.
C 21,05 va 3,90.
B. 13,4 va 7,80.
;
D . 21,05 va 7,80.
^^^->^'>''-^^'*' Lmgidi:
v-ac phan ung hoa hgc:
Na
> 2KAIO2 + H2O
0,06
0,02
+ H * + H2O
- > Chat ran Y gom: Fe, C u , AI2O3.
luong tinh:
,., ^
, pj^,
> H2O . • i'iii^ '^h ••••
H*
Ba2^ + S O)24"
' s*^"^^
' *
—> Dap an D.
B. Fe203,Cu.
0,03
Al(OH)3
«
khuay ki, thay con lai phan khong tan Z . Phan khong tan Z gom
^
,
Ba> + 2 0 H -
Mol: 0,02 - > 0,02
0,1
'
D . 0,78.
Mol: 0,02
0,04
0,04
Trong X: n ^ ^ _ = 0,02 mol; n „ 2+ = 0,03 mol; n .
= 0,02 mol
OH
Ba
AIO2
- » Trong 100ml Y : n , = 0,07 mol; n^^2- = 0,02 mol; n^,. = 0,03 mol.
H
SO^
CI
Mol.
'
va m = m N a o + mAbO-, = 0,05.62 + 0,05.102 = 8,2 gam
'
Mol: 0,03
OH- +
* ' '>'^
>
Khi cho tu tu Y vao X:
> Al(OH)3 + N a H C 0 3 " '
a = mAi(OH)3 = 0,1.78 = 7,8 gam;
A . AI2O3, Fe, C u .
B a O + H2O
AI2O3 + 2 0 H -
AI2O3 CO tinh luong tinh nen tan trong dung dich N a O H vtra tao ra:
I
•
So mol moi oxit ban dau: B a O = 0,03 mol; AI2O3 = 0,02 mol.
D . 8,2 v a 7,8.
NMn xet: K h i cho X vao nuac, Na20 se tac dung ngay voi nuac: ^ ^, ^
C.3,11.
Cac phan ung hoa hoc:
dugc 200ml dung dich Y chi chua chat tan duy nhat c6 nong do 0,5M. Thoi
B. 11,3 va 7,8.
B. 1,56.
Laigidi:
V i d\ 2: Hoa tan hoan toan m gam hon hop X gom Na20 va AI2O3 vao H2O thu
A . 8,3 v a 7,2.
,
+H2O
AI2O3
> NaOH + i H 2
+ 2NaOH
'
> 2NaA102
'
•
.^^^^
+ H2O
.HcJi/ri/''''. "^r. ,
HORK;:
.
.< '
139
dm
Cty TNHH MTV DWH
nang On luygn thi dgi hqc 18 chuy6n 66 H6a hpc - Nguyjn Van Hi\
SO2
sc + 2 N a O H
Nhan xet: Dung dich X trong suo't
AbOa tan het.
N a O H + HCl
> NaCl + H2O
Mol:
0,1
<-
ai
rt--.-v;f;Vijt!^-•
NaA102 + HCl + H2O
Mol:
0,1
> Al(OH)3>l' + NaCl
0,1
'
.14-.c .
'
Nhan thay: Khi cho 400ml dung dich HCl vao X, toan bp NaA102 se chuye'n
het thanh ket tiia Al(OH)3, va sau do bi hoa tan mot phan trong HCl:
NaAlCh + H C l + H2O
Mol:
X
X
>
P2O5 + 6 N a O H
CI2O7 + 2NaOH
> 2Na3P04 + 3H2O
> 2NaC104 + H2O
Al(OH)3 + 3HC1
MZI
Mol:
X
nAi(OH)3 "
x -
^
x = 0,15
mol.
+
D. Nuacbrom.
—>
> CaCOsI
2CO2 + Ca(OH)2
Phan ieng oxi hoa-khu
SO2
b. V i
+
Br2
+ Br2 + 2H2O
Dap
,
gtat:
>
s
H 2 S O 4 + 2HBr
]
gnoi
ciia V la
B. 224.
n B a S 0 4 = - ^ =0,01
SO2
C. 560.
D. 336.
mol.
+ 2Br2 + 2H2O
H2SO4
+ BaCb
> H 2 S O 4 + 4HBr
> BaSOd
+
2HCI
• -
.
i
,
-> Dap an B.
+ 2HBr
Vi dy 4: Hap thu hoan toan 336ml khi CO2 (dktc) vao 1 lit dung dich Ca(0H)2
0,01M, thu dugc dung dich X va m gam ket tiia. Gia tri cua m la
A. 1,0.
loang?
B.1,5.
li
C. 6.
D.5.
nco2 = ^
Laigidi:
Cac oxit Si02, CuO khong tac dyng voi dung djch NaOH loang.
Cac phan ung hoa hpc aia cac oxit voi dung dich NaOH loang (du):
>
—->
D.0,5.
= 0,015 mol; n ^ ^ . = 2nc,(OH)2 = 2.1.0,01 = 0,02 mol. ^ ,
a = - S H I = -2l£2. ^
nco,
0,015
Nhan xet: Day la cau hoi kha tong hop ve tat ca cac loai oxit. L u u y them:
C.2,0.
Laigidi:
,«.^.,;v..:v
B. 8.
Cr203 + 2NaOH
' *
-> nso, = ngaSOA = O'Ol mol -> V = 0,01.22,4 = 0,224 lit = 224ml.
CuO. Co bao nhieu oxit trong day tac dyng dup-c voi dung djch NaOH
+ 2NaOH
*
dich X. Them dung dich BaCh d u vao X, thu dugc 2,33 gam ket tiia. Gia tri
.
V i d^ 1: Cho day cac oxit: NO2, Cr203, SO2, CrOs, C O 2 , P2O5, CI2O7, Si02,
NO2
"''' '
Laigidi:
mlu
A. 7.
it'z-w
2 33
+ H2O
> H2SO4
a • ikrlt
,r
an D.
A. 112.
> Ca(HC03)2
+ H2O
^
Vi dy 3: Hap thu het V m l khi SO2 (dktc) vao nuoc brom d u , thu dugc dung
HO
> NaHC03
CO2 + Ca(OH)2
*
>•
> Na2C03 + H2O
CO2 + NaOH
•ri<
C Dung dich NaOH.
Tdc dutiQ vai baza
CO2 + 2 N a O H
<--
B. CaO.
SO2
= 0,1
.
f^,/>- ^tini.>>;^
- > CO the lam mat mau nuoc brom:
X
—
^
> , ! iO^'
A. Dung djch Ba(OH)2.
^'"-^
T u day suy ra ban dau: Na = 0,25 mol; AI2O3 = 0,075 mol.
^ m = 0,25.23 + 0,075.102 = 13,4 gam
Dap an A.
;
4. OXIT AXIT
a. L i thuyet
*
n ^ . - w n n
^
f :
Nhan xet: Khi C O 2 va SO2 deu c6 tinh axit, tuy nhien khi SO2 con c6 tinh khu
>
mol ->
an A.
Lot
> AICI3 + 3H2O
a3-x
Dap
-
Na2C03+H20
"
0 3—
Ta c6:
—>
> Na2Cr04 + H2O
Vi du 2 (CD-09): De phan biet CO2 va SO2 c6 the dung thuoc t h u la
> Al(OH)3^ + NaCl
•
Na2S03+ H2O
>
Cr03
+ 2NaOH
Cr
r 0 2 + 2NaOH
- > a = 0,1.78 = 7,8 gam —> Loai phuang an A ua C.
Khang Vijt
33 _^ tao 2 loai muoi: Cacbonat va hidrocacbonat.
Cach 1: Giai theo phuang trinh phan ung.
CO2
NaN02 + NaNOs + H2O
+
Ca(OH)2 — — > CaCOsi
+
H2O
2NaCr02 + H2O
J ^ ^ ^
141
C^m
n a n g O n l u y g n t h i d j i i UQC 1 8 c h u y § n 6i
2CO2
Mol:
2y
H6ahpc-
+ Ca(OH)2
-i:
N g u y § n van H S i
, Cty TNHH M T V DVVH
> Ca(HC03)2
vy ,
. .nvw
y
nc02 =0,12 mol; nggcog = 0,04 mol. Phuong trinh hoa hgc:
X = 0,005; y = 0,005.
Ba(OH)2
—> m c a c o a = 0,005.100= 0,5 g a m - > Dap an D .
C a c h 2: T i n h n h a n h : n ^ o = n ^ ^ . - n r o o = 0,005 m o l .
CO-^
Mol:
..y.
OH
Nhan t h a y : n^^2- > \^2^-
Mol:
3
-> m = 0,005.100 = 0,5 g a m
:r'-
•!
Dap an B.
V i d u 5: Hap t h u h o a n t o a n 0,672 l i t k h i CO2 ( d k t c ) vao 200ml d u n g d i c h X
g o m N a O H 0,1M va K O H 0,1M t h u dugc d u n g d i c h Y . C o c a n c a n t h a n Y
t h u d u g c a g a m c h a t r a n k h a n . G i a t r i Ion n h a t c i i a a la
A. 2,26.
B.3,06.
C . 2,54.
D . 2,88.
^
+ = 0,02
mol;
n^+ = 0,02
n^^2- = n^^^. - nco2 = 0,01
nhanh:
mol;
0,04
0,04
.
• ,. ' , ,
,.', '
| . ,r . ;
,v
> Ba(HC03)2
0,08
" '
'.
'
-> riBa(OH)2 ~ 0,08 mol -> a = 0,1 M -> Dap an D .
5, B A I T A P O N L U Y E N
Bai 1: E)6't 4,2 gam Fe trong khi oxi, thu dugc hon hgp X gom FeO, Fe203,
V'
(san pham k h u duy nhat) va dung dich chua m gam muoi. G i a tri ciia m la
B. 18,15.
C.24,2.
D . 20,57.
Bai 2: Hon hgp X gom C u O va Fe203. Hoa tan hoan toan 16 gam X bang dung
,
dich H2SO4, thu dugc 36 gam muo'i. Mat khac, neu k h u het 16 gam X bang
,
C O (du), dan hon hgp khi thu dugc vao dung dich Ca(OH)2 (du), tao thanh
m gam ket tiia. G i a tri cua m la
N h a n t h a y : ^ =^
= i-.Taora2mu6i.
n(_02
0,03 3
Tinh
+ 2CO2
A. 12,10.
mol.
^
'
Fe304. Cho X tac dung voi dung dich HNO3 loang (du), thu dugc khi N O
n ^ ^ . = n N a O H + " K O H = 0'02 + 0,02 = 0,04 m o l
n
0,04
> BaC03 + H2O ...
' '
.j,
Lai gidi:
Trong X:
+ CO2
0,04
Ba(OH)2
( I
l l '
riQ^co^ = ^^o^- = ^''005 m o l .
3
Vi$t
Lad gidi:
:
Ta c6: X + 2 y = 0,015 ; X + y = 0,01 ^
Khang
A. 10.
'
n^^^^ = n c o 2 " "^^2- = 0,02
mol.
B. 15.
C.25.
D.20.
Bai 3: Hon hgp X gom MgO, Fe203 va C u O . Hoa tan het 9,6 gam X bang dung
djch H2SO4 (du), thu dugc dung dich chua 22,4 gam muoi. Mat khac, neu
Y
gom:
Na^ = 0,02
= 0,02
mol;
Bao t o a n k h o i l u g n g : a = m
mol;
CO
++m ++ m
Na
K
= 0,01
mol;
HCO
gam
-> Dap an B.
V i d^ 6: Cho 0,1 mol P2O5
vao
dung dich
,
A. 16.
, ,
,,
K O H 2M. Sau
khi
phan
A. K3PO4 v a K2HPO4.
Gia tri ciia V la
D . H3PO4 va KH2PO4.
->
+ 3H2O
P2O3
"H3PO4
=
0,2
> 2H3PO4
= 0,2
mol.
Vi
dv 7: H a p t h u h o a n
thanh
ran X phan ling voi dung dich HNO3 d u thu dugc V lit khi N O (san pham
khu duy nhat, 0 dktc). Gia tri ciia V la
H3PO4 va
xet
ti 1$ m o l
n h u tren.
t o a n 2,688 l i t k h i CO2 ( d k t c ) vao 800 m l d u n g d i c h
Ba(OH)2 a mol/1, t h u d u g c 7,88 gam k e t hia. G i a tri ciia a la
A. 0,03.
142
B.0,04.
D . 4,48.
hoan toan vao dung dich Ba(0H)2 du, thu dugc 29,55 gam ket tiia. Chat
A. 2,24.
P2O5
C . 3,36.
nong, sau mot thoi gian thu dugc chat ran X va khi Y . Cho Y hap thu
nH3P04 = 2np205
= 4 > 3 ^ K O H d u va t a o ra muoi K3PO4.
can chuyen
B.2,24.
Bai 5 (B-12): D a n luong khi C O di qua hon hgp gom C u O v a Fe203 nung
- > Dap an C .
Luu y: Cac em
D.24.
HNO3 du, thu dugc 1,12 lit NO (san pham k h u duy nhat ciia N*^ a dktc).
A. 1,68.
Lcngidi:
Nhanxet:
C . 12.
dugc hon hgp X gom FeO, Fe203 va Fe304. Cho X tac d y n g voi dung dich
ling
B. K2HPO4, KH2PO4.
:\. K3PO4 va K O H .
B.20.
Bai 4: Cho V lit khi C O (dktc) tac dung het voi 10 gam Fe203 nung nong, thu
x a y ra h o a n t o a n , d u n g d i c h t h u d u g c c6 cac c h a t la
f
khu hoan toan 9,6 gam X bang CO (du), cho hon hgp khi thu dugc Igi tu tu
qua dung dich Ca(OH)2 (du) thi thu dugc m gam ket tua. G i a tri ciia m la :
HCO3
^
400ml
mol.
2- + rn
3
= 0,02.23 + 0,02.39 + 0,01.60 + 0,02.61
->• m = 3,06
= 0,02
i
C0,05.
D . 0,10.
B.4,48.
C . 6,72.
D . 3,36.
Sai 6 (B-10): K h u hoan toan m gam oxit MxOy can vira dii 17,92 lit khi C O
(
, ,,, ,
B. C r O .
C . Cr203.
D . Fe304.
143
Cty TIMHH MTV DWH Khang Vijt
calm nang On luyjn thi d^i hgc 18 chuy6n 6i H6a hgc - Nguygn Van HSi
Bai 7: Syc 4,48 lit khi CO2 (dktc) vao 1 h't dung dich X chiia hon hp-p
Ba(OH)2 0,05M va NaOH 0,2M. Sau khi cac phan ung xay ra hoan toan thu
dug-c m gam ket tiia. Gia tri ciia m la
A. 19,70.
B.9,85.
C. 11,82.
D. 15,76.
Bai 8 (CD-12): Cho 1,42 gam P2O5 tac dung hoan toan vol 50nJ dung dich
K O H I M , thu duQ-c dung dich X. Co can dung dich X thu dug-c chat ran khan
gom
<'
A. K3PO4 va KOH.
'
C. KH2PO4 va K2HPO4.
no (x) = 0,2 mol.
Khi cho 16 gam X tac dyng vai C O thi: no (x) = n c o 2 = " C a C O s = 0,2 mol.
m =0,2.100 = 20 gam ^ D a p a n D .
,„Kn:.
D. H3PO4 va KH2PO4.
Bai 9: Dan luong khi C O di qua hon hgp gom CuO va Fe203 nung nong,
sau mot thoi gian thu dugc chat ran X va khi Y. Cho Y hap thu hoan toan
Bai3:
Nhan xet: Khi cho X + H2SO4 thi oxit chuyen thanh muoi sunfat va mgt
a mol
tang 22,4-9,6 = 12,8 gam.
12 8
.
-> a = — ^ = 0 , 1 6 mol -> no(X) = 0,16 mol.
:
vao dung djch Ba(OH)2 du, thu dugc 11,82 gam ket tua. Chat ran X phan
Ltfu y: Khi cho X tac dung vai CO thi: no (X)
ling vai dung dich HNO3 du thu dugc V lit khi NO (san pham khu day
->
nhat, 6 dktc). Gia tri ciia V la
A. 1,120.
•
B. 2,016.
C. 0,896.
ri«o .:,„f
nguyen tii 0^~ trong oxit dugc thay the bang mgt ion SO l~ .
1 mol 0 2 > 1 mol SO 4 ' - > Khoi lugng tang 96 - 1 6 = 80 gam.
^ B. K2HPO4 va K3PO4.
'P'-^
20
_^ a= — =0,25 mol
80
= nco2 = " C a C 0 3 '( ''''^
'
' '
n c a C 0 3 = 0,16 mol.
- » m = 0,16.100 = 16 gam -)> Dap an C. c O * - - ^ - • - m , . : Y S A
D. 1,344.
Bai 10: Hoa tan hoan toan 8,16 gam hSn hgp gom Fe304 va FeS2 trong dung
Bai 4:
,
dich axit HNO3 (dac, du), thu dugc 4,032 lit khi NO2 (dktc) va dung dich X.
•
'I r . O - S . 0 • €.0 -,,f0(1 - f r - . 5 n :finfirin j ' .:,
" 22^4 "
Cho X tac dung voi dung djch Ba(OH)2 du, Igc ket tua va nung trong khong
+3
+3
khi den khoi lugng khong doi thu dugc m gam chat ran. Gia tri ciia m la
So do phan ling: FezOg
A. 16,18.
Nhan xet: So oxi hoa ciia Fe 6 san pham cuoi cimg hoan toan giohg vai ban
B. 15,86.
C. 12,66.
D. 30,26.
>X
dau.
6. HI/6NG DAN-L61 G I A I
V'
B a i l : nFe = 0,075 mol.
^^
^
, ,,
"i^foi
'
nhieu thai gian.
2nco = 3r»NO
Mol: 0,075
> Fe(N03)3
(£l-lf) v
ff,tt^-r'iih
'^.P . ^ I K ' '
0,075
ricr
-> m = 0,075.242 = 18,15 gam -> Dap an B.
Bai 2:
Nhan xet: Khi cho X + H2SO4 thi oxit chuyen thanh muoi sunfat va mnguyen tu O^" trong oxit dugc thay the bang mgt ion SO 4 .
Tang giam khoi lugng:
1 mol 0 2 a mol
> 1 mol SO 4"
<-
" c o = 0'075 mol -> V = 1,68 lit
Dap an A.
2g gg : 91 fl im 6'/ »OHfJi Atmh aUsH rrayuitj nK> ma 3 E 3 ;\ U . .
'""^"^ > Fe(Na)3
vaap dung bao toan nguyen to Fe:
Fe
"'^"i.meH"
Bai 5:
O day, cac em can su dung so do phan ung:
«'P
> X, sau do X nhuong so
electron vua nhan dugc cho N"^^ de tra thanh Fe(in).
Bao toan electron:
> X (FeO, Fe203, Fe304)
^' -"^^ '
lAc dau C O nhuong electron cho Fe203
^ .^.^
Nhan xet: Bai nay neu dua theo phuong trinh phan ung se ra't ciai dong va ton
Fe
"""^"^ > Fe(N03)3 + NO
~ * " C O = " C O 2 = " B a C 0 3 = O'^S moi
" 0'^^
So do phan ling:
'
Khoi lugng tang 96 - 16 = 80 gam. '^^
'
*
'
*
:.
cOD .•:
+3
+2
+3
+2
Pe2 0 3 , C u O
>X
^"^"^3 ) Fe(N03)3 + Cu(N03)2 ^ NO
N/ian xet; So' oxi hoa ciia Fe va Cu a san pham cuoi ciing hoan toan giong
voi ban dau.
. ^_ f
, . ,,,, ,
Luc dau C O nhuang electron cho Fe203, CuO
'
tang 36 - 1 6 = 20 gam.
"BaC03 "
> X, sau do X nhuong so'
electron vua nhan dugc cho N'^ de tra thanh Fe(III), Cu(II).
Bdo toan electron:
2nco =3nNo
^
Dap an C.
- > nNO
r
I'.i7f-t.
= 0,lmol
V = 2,24lit
md<.>SvO- , . ,
'
^
145
Cty TNHH MTV DVVH Khang Vigt
Ca'm nang On luy^n thi dgi hpc 18 chuyen 66 H6a hqc - Nguyjn Van Hi)
Bai6:
1 iw.
nco =0/8 mol; nso2 =0/9 mol.
Gpi so mol: Fe304 (a mol) va FeSz (b mol). Ta c6:232a + 120b = 8,16.
Nhan xet: Vi oxit MvOy bj khu boi khi C O -> Logi B va C vi C O khong khir
dugc oxit ciia crom.
O cac phuong an con lai, M deu la Fe —> M la Fe.
Cac phan ung khu:
Ta co: no (oxit) = nco = 0,8 mol.
»
Bao toan electron: 3npe = 2nso2
= 0,6 mol.
-> i}F^ = M = l ^ OxitsatlaFe304
no
0,8
4
'"
'"' 'v^
Fe304-le
n„
Nh^n thay:
Mol: a03
Na
°"
FeS2
= — = 1,5 > 1 -> Phan ling t?o ra 2 muoi cacbonat va
a2
nc02
^,
>• -Fe203
2
2nB3(OH)2 = 0,2 + 2.0,05 = 0,3 mol
: .^^^
hidrocacbonat.
!>
,
, 1 ,
an C
,
1/0", u i i ; i i j i i <
sfUw'EK*
,1 ..ti
!i
•
> ) 11
'
f J'fl
'
'>
fj||.ii[|
' .;
'4..,,,,,
' wsV;--Vvh f M i s - . P
* f fi.jt.'i.
IM
•
Hi:
*- -
J
a045
> -Fe203 + 2BaSa
Mol: 0,01
0,005
0,02
m = 160.0,05 + 0,02.233 = 12,66
-> Dap
Tinh nhanh: n^^2- = "QJ^- - " 0 0 2 = 0,3 - 0,2 = 0,1 mol!
j„
':;•.;..>. .
a + 15b = 0,18
a =0,03; b = 0,01.
Bao toan nguyen to'Fe va S:
DapanD.
= 0,05 mol; n^, + = 0,2 mol.
2+
Ba
^
nN02 = "Fe304 + 15nFeS2
Fe304
Trong Z:
>3Fe^3
FeS2-15e
> ¥e*^ + 2S^
Bao toan lectron:
Bai7:
" O H - = ""NaOH +
OH'
•*
...
,
>U
• • ifh i/1
„ .jjfg,^^ -^^^
;
3
-> Nhan thay: n^^2^ < n^^2-
nBaCOa = "332+
3
= 0,05 mol.
->m = a05.197 = 9 , 8 5 g a m ^ DapanB.
Bai 8:
Nhan xet:
>
P2O5 + 3H2O
2H3PO4
-> Tao2mu6i:K2P04vaK3P04
Lim y: Cac em can chuyen
...........
DapanB.
P2O5
thanh
H3PO4
,
u ',(h
va xet ti 1? mol nhir tren.'
Bai 9:
npaCOa = 0,06 mol. 6 day, cac em can sir dung so do phan ung:
CuO,Fe2a
> X
C O 2 + Ba(OH)2
) Muoinitrat + NO
> B a C a + H2O
% Mol: a06
,.
a06
'
'
.
,, Nhan xet: Khi cho hon hgp oxit tac dyng voi khi C O thi:
ng
trao doi
= 2nco • Ma n^o = " 0 0 2 ~ ^'^^ ^'^^ ~^
Bao toan electron:
^VNo=a8961it
ngtraod6i
=3
n]sjo-> nj^o
= 0,12 mol.
0,04 mol.
I
-» Dap an C.
U7
elm nang 6n luy^n thi dgi hgc 18 chuySn dg H6a hpc - Nguyjn Van H5i
Cty TNHH M T V D V V H Khang Vigt
Chuyen de 6
phat b i e u diing la:
CAC mmm
TO P H I KIM mm
umn
1, H A L O G E N
a. L i thuyet
*
+ 3Cl2
F2 + H2
*
s.i^w
2FeCl3
Cu+ C k
Tdc dung vai hidro
> 2HF
c ^
> K C l + KCIO + H2O
3Cl2 + 6 K O H
*
.Dn!5qM<
>• 2FeCl3
* ' '-IdSwiv.},
- ;.
2 K M n 0 4 +16HC1
^
MnCh + C h t
.tOnlV'. f ,
Laigiai:
Loai C vi F2 chi c6 tinh oxi hoa, khong c6 tinh khu.
> 2KC1 + 2 M n C l 2 + S C h t + 8 H 2 O
'
Mol:
-> a =
•,
V i dv mau
V i di^ 1: Phat bleu nao sau day la sai?
t'.
< , t^^ n .
»>
A. Halogen la nhung chat oxi hoa manh.
B. K h a nang oxi hoa cua halogen giam tir flo den iot.
^
.
'~rA'^
^•.f^n^r
, , 4 -
B.0,48.
C.0,40.
D . 0,2C.
Laigiai:
•^M^
3Cl2 + 6 K O H
""'^ ) 5KC1 + KCIO3 + 3H2O
0,03
<-
0,25
0,06
= 0,24 mol/1
,
'^'''^^
0,05
,
Dap an A.
d\ 5: C o cac thi nghi^m sau:
. ,
.
(1) Cho dinh sat vao dung djch H2SO4 {loang, nguoi);
Nhan xet: D o flo la nguyen to'co dp am di^n manh nhat -> flo chi the hi§n so
"
'
-C>C
^ nV,t >tsit:'.
So' thi nghi?m xay ra phan ung oxi hoa-khu la
Q , , | j r ^. '!
A-2-
•\:,p,4..
B.I.
c3.
LMgidi:
Phuong an C sai.
,
,^
(4) Cho la A l vao dung dich H2SO4 (dac, ngupi).
LMgidi:
,
;.
(3) Cho nuoc brom vao dung dich N a l ;
D. Cac halogen c6 tinh chat hoa hpc tuong t\ nhau.
,
H;>lt!JH UlJJ >X
(2) Sue khi SO2 vao nuoc brom;
:
C . Cac halogen deu c6 the c6 so oxi hoa : -1, +1, +3, +5, +7.
oxi hoa -1 trong cac hop chat
^
ffj (jjyC'j -
Loai D v i CI2, Br2,12 deu it tac dung voi nuoc.
+2H2O
mn
, (|/;
...
Cac phan u n g hoa hpc a cac thi nghi^m:
-> Dap a n C .
V i d\ 2: C h o biet cac phan ung xay ra sau:
2FeBr2 + Br2
>• FeBra
2NaBr + C h
, V|SOnM .S
A.a24.
2NaCl + 2 H 2 O - ^ S ^ 2NaOH + C I 2 T + H 2 t
b.
D. Tac dung manh voi nuoc.
!
sec. Sau khi phan ung hoan toan, thu dupe 3,725 gam K C l . G i a tri cua a la
>2HBr+H2S04
—
iCfiq
n wMrt d 9 IdA gm-ml t-n uh'
Vi dv 4: Cho 1,344 lit khi CI2 (dktc) d i qua 250ml dung dich K O H a mol/1 0
Dieu che'clo
M n 0 2 + 4HC1
^^nub wb .gnv/ui i o v g m i
C. V u a CO tinh oxi hoa, vira c6 tinh khtr.
-^DapanB.
Tdc dung vai hap chat
Br2 + SO2 + 2 H 2 O
*
V i dv 3: C h p n nhan dinh dung. Cac halogen deu c6 tinh cha't chung la
I
'! + 5.(X0.Ooi a i ,
+ I2
, ,'
('» r i i / l :
Lo^i A v i F2, CI2 la cha't khi; Br2 la chat long va I2 la chat ran.
> 2NaBr
CI2 + 2FeCl2
tiHf
{Clorua voi)
> 2 N a C l + Br2
Br2 + 2NaI
•—^DapanD
B. Co tinh oxi hoa.
"Day" halogen ditngsau
CI2 + 2NaBr
•() li / : ; \
Tinh oxi hoa CI2 > Br2; tinh k h u Br- > CI",
A. Chat khi 6 dieu kien thuong.
^.^ ^
{Nu&c Gia-ven)
>CaOCh + H2O
t^han xet: (1) —>• Tinh oxi hoa Br2 > Fe'^; tinh k h u Fe^* > Br".
(2)
\is ri\
i ijof*
tinh oxi hoa CI2 > Br2 > Fe^*; tinh k h u Fe^* > B r > Ch.
.
> 5KC1 + KCIO3 + 3 H 2 O
CI2 + Ca(OH)2 {sua voi)
D . CI2 c6 tinh oxi hoa hon Fe^*.
01,
> 2HC1 t o :
CI2 + H2
Tdc dung v&i dung dich kiem
CI2 + 2 K O H
*
CuCh
j'nrv„>^.ff)fy^-. ., , « r o... r't;» r +
,
B. Br2 c6 tinh oxi hoa manh hon CI2.
Q B r CO tinh k h u manh han Fe^*.
Laigiai:
,,,,;.|. ,
Tdc dung vai kim loai:
2Fe
*
t-g^i ^
^.^ . ,
A. CI" CO tinh k h u manh han B r .
> 2 N a C l + Br2
0 ) Fe + H2SO4 (loang, ngupi)
' 'ii
•
^
(1)
(2)
,
(2) S02 + Br2 + 2H2O
(3) Br2 + 2NaI
> FeS04 + H 2 t ^
> H2SO4 + 2HBr
> 2NaBr + I2
'• '
*
' '
- * ~ ''
'
'
"
149
Cty TNHH MTV DWH Khang Vigt
Ca'm nang On luy$n thi dgi hgc 18 chuySn dg H6a hpc - Nguygn Van HSi
—> Dap an C.
S + HNO3 (dac)
Lim y: 1- Fe tan trong H2SO4 loang, nguoi; 2- A l , Fe khong tac di^ng voi
5 + 2H2SO4 (dac)
H2S04dac, nguQi.
Vi
^ :
*
6: Hon hgp khi nao sau day khong ton tai 6 nhi|t dp thuong?
A. C 0 v a 0 2 .
B. Cl2va02.
C. H2SvaN2.
Nhan xet: H2 va F2 c6 the phan ling manh liet ngay ca trong bong to'i va 6
nhi^t do rat thap, c6 the gay no va toa nhieu nhi^t, tao thanh HF. 5
^DapanD.
2Ag + O3
c, Hidrosunfua
*
'
D. K2Cr207. >
MnCk + C h t
KMn04 + 8HC1
+ 2HCI
*
+2H2O
2KC1 + 2CrCl3 + SCh
-—>
+
VHiO
K2Cr207 nhan electron nhieu nhat -> tao khf Ch nhieu nhat
Dap an D.
*
O2 + S — ^
*
Fe + S — ^
S + H2 — ^
*
FeS
*
*
4Fe(OH)2 + O2 + 2H2O
2H2S + O2
> NaHSOs
Dieiiche
'° > 2SO2 + 2H2O
,
,
,
(
-f,
> 2S03
A:
> 2HBr + H2SO4
— ^
,
' •
,j, ^, ^
^. | ,
> 3 S + 2H2O
: l;;,M>c.ri
4FeS2 + I I O 2
> 4Fe(OH)3
*
> Na2S03 + H2O
Tinhoxihoa
s + 02
CO2
> H2SO4 + 8HC1
Tinhkhie
SO2 + 2H2S
Tac dung v&i hap chat
f
>• FeCh + H 2 S t
SO2 + Br2 + 2H2O
H2S
>
2S + 2H2O
2SO2 + O2 <
> HgS
SO2
CO + O2 — ^
'
2SO2 + 2H2O
2NaOH + SO2
''
2H2O
,
" *'
* ' 1
NaOH + SO2
O2 + 2H2 — ^
'
e. Luu huynh dioxit
a. Oxi - luu huynh
Hg + S
1
Dieii che
FeS + 2KC1
2. OXI - L l / U H U Y N H
3Fe + 2O2
'° > Fe304
Tac dung vai phi kim
.» 1
+ 2HN03
+ H2SO4
Voi clo: H2S + 4Ch + 4H2O
Mn02 CO the nhan vao Ian lugt la 1, 5, 6, 4.
MgO
i-
Tinhkhiemanh
2H2S + O2 — ^
Cach 2: Nhm xet: So mol electron ma 1 mol moi chat: CaOCh, KMn04, K2Cr207,
*
> FhSi
> CuSi
j
6 dieu k i f n thuong, dung dich H2S bi oxi hoa dan thanh S:
Dap an D.
M g + O2 — ^
'•:(')
2H2S + 3O2 — ^
,
CaCh + Ch + H2O
Tac dung vai kim loai
M f^ , „
Voi oxi: H2S chay trong khong khi voi ngpn lua mau xanh nhat:
> KCl +MnCl2 + - C I 2 T + 4H2O
2
K2Cr207 + 14HC1
CaOCh
Tinhaxityeu
H2S + CuS04
Cach 1: Dua theo cac phuang trinh phan ung:
,
> O2 + I2 + 2 K O H
H2S + Pb(N03)2
Laigidi:
Mn02 + 4HC1
'° > Ag20 + O2
O3 + 2KI + H2O
phan ling vai lugng d u dung dich HCl dac, toi phan ling hoan toan thi
'"
*
> 2KC1 + 3O2
chat tao ra lug-ng khi CI2 nhieu nhat la:
C. CaOCh.
K2Mn04 + Mn02 + O2
b. Ozon:
.„;,",
'
1 ri
Ozon la mpt chat oxi hoa manh, manh han ca oxi.
V i d^ 7: Neu cho 1 mol moi chat: CaOCh, KMn04, K2Cr207, Mn02 Ian luot
B.Mn02.
3S02 + 2H2O
2KC103
£ »,;
A.KMn04.
> SO2 + NO2 + H2O
Dieu cheoxi
2KMn04 — ^
D. H2vaF2. 4 '
Laigidi:
'°
•
SO2
J.I
-
•
^'
'° > 2Fe203 + 8SO2
Na2S03{rin) + 2 H 2 S 0 4 C d j c , d u ) — ^
2NaHS04 + SO2 + H2O
151
Cty TNHH MTV D W H Khang Vigt
C^m nang On luygn thi d j i hpc 18 chuyfin 66 H6a hpe - NguySn v a n H i i
f.
L i r a h u y n h trioxit - O l e u m
SO3
+ H2O
Lai gidi:
> H2SO4
nSOa + H2SO4 ^
: .2.
D a p an C. Cac p h a n u n g hoa hoc:
>• H2S04.nS03
(Oleum)
---H',
VIDVMAU
2SO2 + 0 2
I'--' •'
Vi dv 1: K h i n h i ^ t p h a n h o a n toan 100 g a m m o i chat sau: KClOa (xiic tac
Mn02), K M n 0 4 , KNO3 v a A g N O s . Chat tao ra l u g n g O2 I o n nhat la
A.KCIO3.
C.KNO3.
B.KMn04.
Lai
2KMn04 —
^
D.AgNOa.
KNO3 — ^ KNO2 +
io2
i^'
rttt!
^^,,^^3 ,
D a p an A
'
.
,
nuh ^tvfmili
B.Zn.
C. A l .
tt^-$Bip'
NfcOn
2M
n02 — ^
+
+
2nHCl
^
2M20n
-obio
'
+
n
n
nH2t
" •
.. j ^ ;
- r - ^^^^^y
•
M
n
"e = 4no2 + 1 " H + " ^ " 0 2 + 2nH2 = 0,28 m o l .
A . 3.
J o m € J .0 - '
B.l.
'
C.2.
Tat ca cac phat bieu d e u d u n g . D a p an D .
~
«> i .
D.4f-'"
T'
'
X-Offfj
"
JrW ;6x
.
»^ • "^^^
V i d u 5: C h o 9,2 g a m h o n h g p X g o m Cu2S, CuS, FeS2 v a FeS tac d y n g het
v o i H N O 3 (dac nong, d u ) t h u d u g c V l i t k h i c h i c6 N O 2 (dktc, san p h a m
dich BaCl2, t h u d u g c 23,3 g a m ket tua; con k h i cho toan b g Y tac d u n g v o i
A . 19,04.
B. 12,32.
A . D u n g d i c h BaCl2, CaO, n u o c b r o m .
>,
^
D . 5,60.
••A. rfi niXi -i-
K h i cho d u n g dich Y + d u n g dich BaCh:
> BaSa^^
«"soJoe;/
va N O i . i ;V .
»
• ' ^ , ^^*^^'?*^*'
^
'
"
^nol.
K h i cho Y + d u n g dich N H s d u :
Fe3^ + 3NH3 + 3H2O
> Fe(OH)3>l'
> Cu(OH)2>l
Lm y: Cu(OH)2 tan trong NH3 d u tao thanh phuc chat:
Vi dv 3: D a y chat nao sau d a y deu the hien t i n h oxi hoa k h i p h a n iVng v o i SO2?
B. D u n g d i c h N a O H , O2, d u n g d i c h K M n 0 4 .
C. 8,40.
Nhan xet: D u n g d i c h Y chua cac i o n : F e ^ Cu^^ SO|",
Cu2^ + 2NH3 + 2H2O
M = 12n ^ n = 2; M = 24 ( M g ) - > D a p an D .
C. O2, n u o c b r o m , d u n g d i c h K M n 0 4 .
v.' i H
So phat bieu d i i n g la
Bao toan nguyen to S: ng (X)= r»BaS04
Cach 2: N h a n thay, M Ian lugt tac d u n g v o i 2 chat oxi hoa la O2 va H2SO4 (H*).
D . H2S, O2, nuoc b r o m .
KK;.
Led gidi:
• VK.
Vay: n = 2 va M = 24 ( M g ) ^ D a p an D .
^ • n = 0,28 ^
iM nrij : , o l f -.if i '
ih) $H Jif S f « , S x/ub u d t !
(d) M o o c p h i n va cocain la cac chat ma t u y .
Ba2- + sol
n
> K2SO4 + 2MnS04 + 2H2SO4
,
•
> M2(S04)n
- ^
d u n g d i c h N H 3 d u t h u d u g c 5,35 gam ke't tua. Gia t r i ciia V la
>• 2MCln + n H 2 0 '
+ nH2S04
4<^;«
k h u d u y nhat) v a d u n g d i c h Y. Cho toan b g Y vao m g t l u g n g d u d u n g
Nhqn xet: K h i cho Y tac d y n g v o i axit H C l giai p h o n g k h i —> M c o n d u .
4M
5SO2 + 2 K M n 0 4 + 2H2O
Lai gidi- '^'''^ ^'^^^ *
Mg. ^
Lai gidi:
Cach 1 : Giai theo p h u o n g t r i n h hoa hoc.
<
> 2HBr + H2SO4
"
t h u d u o c chat r a n Y. H o a tan het Y t r o n g d u n g d}ch H2SO4 loang, t h u dugc
0,896 l i t k h i H2 (dktc). K i m l o ^ i M la
,
(c) K h i d u g c thai ra k h i quyen, freon (chii y e u la CFCI3 v a CF2CI2) p h a h u y
Vi dv 2: C h o 3,36 g a m k i m loai M (hoa t r i k h o n g doi) tac d u n g v o i 0,05 m o l O2,
A.Ca.
SO2 + Br2 + 2H2O
tangozon.
N/ifln xef: KCIO3 cho lu(?ng O2 nhieu nhat
.
^ '.^ i -
(b) K h i SO2 gay ra hien t u g n g m u a axit.
' * '^'^^^^'^ *
Ag . NO2 + - O2
AgN03
'
(a) K h i CO2 gay ra h i ^ n t u o n g hieu u n g n h a k i n h .
> 2KC1 +3O2
2KC103
<
^^
V i d v 4 : C h o cac p h a t b i e u sau:
gidi:
K 2 M n 0 4 + Mn02 + O2
,
+ ?
^
Cu(OH)2 + 4NH3
> [Cu(NH3)4](OH)2
5^35
^''^^ -./CiSS.i
Bao toan nguyen to Fe: npg (X) = ripe(OH)3
^ ^'^^ m o l . - .-^
Bao toan k h o i l u g n g : m x = m c u
"^Fe "^s
~» mcu = 9,2 - 0,05.56 - 0,1.32 = 3,2 gam
'
'
, , ',
dm
nang On luygn thi djii hpc 18 chuyfin ii H6a hgc - Nguyin van H5i
32
ricu(x)==0'05mol
Cty TNIHH MTV DWH Khang Vi$t
yi
>. i
Qui doi X ve hon hop gom cac don chat:
Fe = a05 mol; Cu = 0,05 mol vaS = ai mol.
^ n e ( x ) =2ncu +3nFe +6ns =0,85mol^^ ^
i|?
Q^j.^r;
nN02 " " e ( X ) =0,85 mol
-> V = 0,85.22,4 = 19,04 lit
'
^.n >, ,,f).;
,0;.^
^Dap an A.
V i dy 6: Hoa tan het 3,76 gam hon hop X gom Al, Mg va Zn trong dung djch
8: Hon hgp khi X gom O2 va O3, c6 ti khoi hoi so voi H2 la 19,2. Dan X
syc tu tu vao dung dich KI du, thay c6 75ml khi di ra. Cac the tich do 6
ciing dieu kien nhi^t dg, ap suat. Phan tram the tich ciia O2 trong X la
^.30%.
B.45%.
C.60%.
D.40%.
Laigidi:
Theo bai: M = 19,2.2 = 38,4
Ap dung cong thuc cua phuong phap duong cheo, ta c6:
HCl, thu dugc 2,912 lit Ha (dktc). Mat khac, dot chay hoan toan 3,76 gam
M
no3
A. 5,84.
Phuong trinh hoa hgc:
C. 4,28.
D. 4,96.,^,:
.....qiiOoM ( h :
Laigidi:
2 912
nH2==0,13mol.
' M-gniAu%Mlsfki'cy:.
Nhan xet: Khi cho X + HCl, ion H* da nhgn electron de tro thanh khi Hz:
:;,•{„.?„,,';;. 2H^ + 2e
,
48 - 38,4
hon hg-p X trong oxi, thu dugc m gam oxit. Gia tri cua m la tniih u& (a;
B.4,80.
Mol:
v.bn ^
> Hat
0,26 <-
ai3
.-|'|sf',^ itp^^,)(.if^gfi.,.(i^,
>
O3
6,4
x = l,5y.
Oit
+ 2KI + H2O
Mol: y
-» Thetichkhithoatra:x + y = 75 -> x =45;y = 30. gnsorW OB-> gb
%Vo, = - . 1 0 0 % = 60%. ^ Dap an C.
Mol: a065 <- 0,26
*
V i d\ 7: Nhi^t phan hoan toan 4,03 gam hon hop X gom KCIO3 va KMn04, thu
dugc 0,784 lit khi O 2 (dktc). Phan tram khoi lugng cua KMn04 trong X la
€.39,2%,^.:^^ ,
B.19,6%.
Fe voi S la
A. 40%.
B.60%.
2KMn04 — ^
Fe + S
Mol:
) 2Ka
y
'
+ 302
Fe + H2SO4 •
FeS + H2SO4
•
= 0,02; y = 0,01.
s'J =--(X, ..i<
2
.r^nn^a'AM
-> %mKMn04 = °^"^Q3^^-100% = 39,2%.
, r, - ,,fr
,1.8
-^DapanC.
Sf:! ,0.-;^.
.
'
'" > FeS
X
-> FeS04 + H2t
> FeS04 + H2St
- > H2 = (0,l-x)mol.
^
H2S = xmol.
Ap dung cong thiic cua phuong phap duong cheo, ta c6:
l,5y
Theo bai: 122,5x + 158y = 4,03 va l,5x + 0,5y = 0,035
-> X
X
Cho Y tac dung voi HCl:
K2Mn04 + MnOa + O2
0,5x
2KC103
X
- > Y gom: Fe = (0,1 - x) mol; FeS = x mol; S = (0,15 - x) mol.
!
X
D. 80%.
Cach 1: Giai theo phuong trinh hoa hgc:
D. 58,8%.
; Vj
Mol:
C. 70%.
np = M = 0,1 mol; nc= ^ = 0,15 mol; My = 10,6.2 = 21,2.
56
32
•
Mol:
Cac phan ung dieu che O2:
' '
r-.
Laigidi:
'
„
•'^'^
............ ,,. •
Bao toan khoi lugng: m = mx + moj = 3,76 + 0,065,32 = 5,84 gam.
A. 78,4%.
,
75
"--2
I^I'-'M
hgp khi Y. Ti khoi cua Y doi voi hidro la 10,6. Hi^u suat cua phan ung giua
2(y-
- > Dap an A.
+ 2KOH
+ hi
CO oxi), thu dugc chat ran X. Cho X vao dung dich H C l du, thu dugc hon
Mat khac, khi cho X + O 2 , 0 2 da nhan electron de tro thanh 02-:
+ 4e
32-38,4
3
2
Vi dv 9: Nung 5,6 gam Fe voi 4,8 gam S 6 nhiet dg cao (trong dieu kien khong
= "HCl = 2nH2 = 0,26.
O2
ilr«.;'t?v,. v
Goithetichcua:02 = xml;03 = yml.
^H2S
_ 34-21,2
2-21,2
x = 0,06 mol
3.(0,l-x) = 2x.
19,2
3
H = 0,06 .100% = 60%.
0,1
• Dap an B.
IKK
Cty TNHH MTV DVVH Khang Vigt
dm nang 6n luygn thi d^ii hpc 18 chuySn dg H6a hpc - Nguyin Van H^i
Cach 2: NMn xet: Do Fe
11^2 + ^H2S
Muoiamoni
> H2S nen
> FeS
"^Fe O'l 'Tiol. Ap dung cong thuc duong cheo, ta c6:
>»•
+
2
0,04
2-21,2
3
0,06
.{fei^,8-
• H = 60%
Dap an B.
^
2NH4CI
+
^^.^.^ .j,, _
^ .^.^^ „
*
6Li + N2
1
> 2Li3N
N2 + 3H2
*
_
2NH3
'
, •"
,
:
r
3 M g + N2 —
4
,^
^
MgsNa
2P+
*
„ J^|,; ,
6 nhi^t do cao khoang SOOO'^C (hoac nhiet do cua 16 ho quang dien), nito
\ NH4NO2
.it.!
,)
>
,
,
. J l
(> .(< ^ ' i
^
,
.
,
,
,(;,-!
,f i ,
^ ^ ^,
.
Tinh khu
.
2P + 5Cl2
.-.ti-f-i ff>|f( 5r,!/f, (-pfK^ v^no.*' i i i i i i -i.
'° > 2PC15
n-ii ,ni>.
.ir,u>-\
^ H3PO4 + 5NO2 + H2O
> N2 + 2H2O
> N2 + N a C l + 2H2O
,
V i du 1: C h o can bang hoa hpc: N2 {k) + SHi (k) <
. ;i
Tflc dung vai axit: NHa + H C l
C . Thay doi nhiet dp.
'
D . Them chat xiic tac.
Nhan xet: Theo bai, phan ung toa nhif t (AH < 0) - > can bang se chuyen dich
khi thay doi nhi^t dp —> Loai C .
> Al(OH)3 i + 3NH4C1
''rloralx- HI''
> [Cu(NH3)4](OH)2 '
^,.
^
> [Zn(NH3)4](OH)2
> [Ag(NH3)2]Cl„,
,
,
Thay doi nong dp N2 ciing se lam can bang chuyen djch - > Lo^i B.
.
- > D a p an C .
'
Luu y: Chat xiic tac chi lam anh huong deh to'c dp phan ling, khong lam
^ chuyen dich can bang.
4NH3 + 3O2 ——> 2N2 + 6H2O
> N2 + 6HC1
Tac dung vai mot sooxit kim loai:
3 C u O + 2NH3 —
^ 3Cu + N 2 +3H2O
,,
,
^
^ i d\ 2: C h o cac thi nghiem sau:
(a)DotkhiH2Strong02
> 4 N O + 6H2O
Tac dung vai do: 2NH3 + 3CI2
can bang chiu tac dpng cua ap sua't
Loai A .
Tfn/i khu
4NH3 + 5O2
; gn< *
'•' V nnrf nhrl - m » '
Lot gidi:
> NH4CI
> 2NH3 (fc); phan ling
B. Thay doi nong dp N2.
Tong so mol khi 6 hai ben khac nhau
•••
Zn(OH)2 + 4NH3
Tac dung vai oxi:
'
thuan la phan ling toa nhiet. C a n bang hoa hoc khong bi chuyen dich khi:
A. Thay doi ap sua't ciia hf.
;•'
Tflc dun^ vai muo'v. A I C I 3 + 3NH3 + 3H2O
A g C l + 2NH3
,{>m ./„
^
NH3 + H 2 0 < = ± N H ; + O H -
Cu(OH)2 + 4NH3
) 6CaSi03 + P 4 + l O C O
VIDUMAU
Tinh baza yen:
Tao phuc chat:
+ IOC
(J fi6
b. Amoniac
+
<- '
'
^^^^^^^
3C. ^ ^ , C a 3 P 2
2Ca3(P04)2 + 6Si02
NH4CI + N a N 0 2
+
risds
2NO
Dieii che
*
^ N2O + 2H2O
P + 5HNCh (dac) —
ket hgp trvc tiep voi oxi, tao thanh khi nito monoxit:
N2 + O2
^ NH3 +HC1
4P + 5O2 — 2 P 2 O 5
Ti'n/i khie
+ 2H2O
1 ^, , , 1 1
d. Photpho
* T«cd«ngt;a7b-m/o,/
„
.l.k
v
+ Na2S04 + 2H2O
> CaCh + N H s t
+ Ca(OH)2
NH4NO3 —
Tac dung vai kirn loai:
> 2NH3t
Phdn itng nhiet phan
NH4CI —
3.NITO-PHOTPHO
^5 -
Tac dung v&i dung dich kiem
(NH4)2S04 + 2 N a O H
•J(W
34-21,2
n H2S
^
> Hi; Fe
•
du;
( c ) D a n k h i F 2 vaonuocnong;
4-v
> t-
(b) Nhi^t phan K C I O 3 (xiic tac M n 0 2 ) ;
(d) Do't P trong O2 d u ;
(e) K h i N H 3 chay trong O2;
( g ) D a n k h i C 0 2 d u vao dung djch Na2Si03.
157
Ca'm nang 6n luy^n thi dai hgc 18 chuy§n dg H6a hgc - Nguyin van Hit
Cty TNHH MTV DWH Khang Vi§t
So thi nghi^m tao ra cha't khi la
A. 5.
y i dv 4: N u n g n o n g h o n hg-p g o m 1,6 gam Ca va 0,62 g a m P t r o n g b i n h k i n
B.4.
C.2.
D.3.
k h o n g CO k h o n g k h i t a i phan l i n g hoan toan, t h u d u g c h o n h g p cha't ran Y .
Laigiai:
;
Cac phan ung hoa hoc:
(a) 2H2S + 3O2
, •
, ,
i .
'° > 2SO2 + 2H2O
> 2KC1 + 3O2
(*) » '
(c) 2F2 + 2H2O
> 4HF + O2
(*)
Cho Y tac d u n g v o i n u o c d u , t h u d u g c V m l h o n h g p k h i (dktc). Gia t r j ciia
, 1
•
(b) 2KC103
(d) 4 P +
,
Via
A . 672.
'
(e) 4NH3 + 3O2 — ^
Mol:
(g) 2CO2 + Na2Si03 + H2O
(*)
> Si02 + 2 N a H C 0 3
2P
0,03
C. 336.
D. 448.
0,01
0,1
!.,
tong h g p N H s la
B. 25%.
ft
C. 30%.
Lai
D. 10%.
.1
giai:
-i tS. ,
'
M
r,
B.69,0%.
A
; • f i t i,
C.73,1%.
D . 65,9%.
Ta c6: mx = m^j + m^j = 28.1 + 4.2 = 36 gam.
,
Nhan xet:
1 m o l canxi dihidrophotphat Ca(H2P04)2 hay 1 m o l P2O5 deu chua 2 m o l P.
Ca(H2Pa)2 <
Khoi l u g n g mol:
% khoi lugng:
> P2O5
234 gam
x%
!
142 g a m
^
v H u .
40%
234 40
- > X = - —
= 65,9-> Dap an D . ,;(„ ,
>;fl
,
^Vv;-'
V i d\ 6: H o n h g p k h i X g o m N2 va H2 c6 t i k h o i so v o i h i d r o bang 3,6. N u n g
Gia thiet trong X: n^j = l m o l ; nH2 =4 m o l .
n o n g X t r o n g b i n h k i n c6 bgt Fe xuc tac. Sau p h a n u n g t h u dugc h o n h g p k h i
• *
X CO so m o l g i a m 8% so v o i ban dau. H i § u suat ciia p h a n l i n g t o n g h g p N H 3
36
Nhan xet: Bao toan k h o i lup-ng: my = mx = 36 gam -> ny = — = 4,5 m o l .
la
.
,
A . 25%.
B. 20%.
g
,r
Phan ling hoa hpc:
3x
-
"
Bao toan n g u y e n to'P theo so do:
dugc hon hop Y . Ti khoi ciia Y so voi hidro bang 4. H i | u suat cua phan ung
X
' '
Phan supephotphat kep CO t h a n h phan c h i n h la Ca(H2P04)2.
/ "
trong binh kin 6 nhi?t dp khoang 4500C c6 bot Fe xuc tac. Sau phan ung thu
Mol:
"'
i
P2O5 ve k h o i l u g n g . H a m l u g n g p h a n t r a m ve k h o i l u g n g cua
V i dy 3: Hon h^p X gom N2 va H2 c6 ti 1$ m o l tuong ung la 1:4. Nung nong X
3H2
' >
, i i if , ' i
> CaCk + H 2 t
A . 71,3%.
N a C l + N2t + 2H2O
0,1
N2 +
"
'
(PHs k h o n g tac d u n g v o i H C l )
Ca(H2P04)2 t r o n g loai p h a n b o n nay la
- > nN2 = nNH4CI = 0,1 m o l - > V = 0,1.22,4 = 2,24 l i t - > Dap an A .
,j
nu •
'
Laigiai:
N H 4 C I + NaN02 — ^
A . 20%.
'
> 3 C a C l 2 + 2PH3t
Phan ling hoa hoc:
fj
• '
- > V = (0,02 + 0,01).22,4 = 0,672 l i t - > D a p an A .
40%
= 0 ' 0 1 r n o l ; nNaN02 = 0 ' 2 m o l .
0,1
M
V i d\ 5: Phan supephotphat kep t h y c te san xuat d u g c t h u o n g chi u n g v o i
Laigiai:
Mol:
D . 448.
Ca3P2
0,02
Ca + 2HC1
ung hoan toan, thu dug-c V lit khi (dktc). Gia trj cua V la
"NH4a
+
Ca3P2 + 6HC1
V i d^ 2: Dun nong 100ml dung dich chua N H 4 C I I M va NaN02 2 M d e n phan
B. 560.
C. 336.
Y gom: Ca3P2 = 0,01 m o l ; Ca = 0,01 m o l .
-> Dap an D.
A . 224.
B. 560.
Laigiai:
3Ca
2N2 + 6H2O
.,„,, ^„
UO t
nca = 0/04 m o l ; n p = 0,02 m o l .
'° > 2P2O5
502
,
C.16%.
D . 23%. ' ' ,
Lcngidi:
Theo bai: M x =3,6.2 = 7,2.
<—>
->
2NH3
A p d u n g cong thiic a i a p h u o n g phap ducmg cheo, ta c6:
2x
N h | n thay sau phan ung so m o l giam: 4x - 2x = 2x.
28 - 7,2
"N2
= — - » Dat so m o l trong X: n ^ j = 1;
~ ^-
2-7,2
-> n x - n Y = 2 x - ^ ( l + 4 ) - 4 , 5 = 2 x - > x = 0 , 2 5 m o l - > H = 2 5 % - ) - D a p a n A .
IRQ
159
dm
Cty
nang 6n luygn thi dgi hgc 18 chuy§n dg H6a hpc - Nguyin Van HSi
Khi cho Fe(OH)3 va Cu(OH)2 vao dung dich N H s du thi Cu(OH)2 tan do tao
Phan ling hoa hpc:
N2
•
Mol: X
+
<—^
3H2
2NH3
[ I'M I d j ; ..J!
.t^b) H ^ j i ^ j r ! rre>d (M>;r .;>vjtj|i){arlj
» i j n ,('dv
J
8
N h a n thay sau phan u n g so' m o l g i a m : 4x - 2x = 2x =
a
.5
+
V i dyi 7: N u n g n o n g 12 g a m M g trong b i n h k i n chua 2,8 g a m k h i N2 den phan
ling hoan toan, t h u dupe h o n h g p chat ran X. H o a t a n h o a n toan X bang
t r o n g d u n g d i c h H C l d u , t h u d u a c V l i t k h i (dktc). Gia t r i ciia V la
C. 6,72.
.^„„, ,
i^f,:-
D . 8,96.
•
•
Mol:
0,3
Mg3N2
+ N2
0,1
K . « a
jg
A^:-
+
'In;'•>•'-••
>
3MgCl2 +
Mg+2Hci—>Mgci2+H2t-^
,
-
Cacbon
• /
, A:
' :
„ ,;
,
nHi\f r r a i f ' '
pjnvwto
,^^^.^^
2NH3
^ '^^'H
.•••..M
•
' '^xy
.
u:)
•
Tac d u n g v o i o x i : C + O2
> CO2
CJ nhiet d p cao, cacbon lai k h u d u p e CO2 theo p h a n l i n g :
> 2CO
« P>
•sfK
'h
H
-mmM;.::,'
:
^
•
+
"
^rixfcii.
J ;
CO2 + 4NO2 +
2H2O
•iw'Bin'lYujin
CO + Z n
D . T i n h chat hoa hpc.
•
vff(m
,
rsfto.
• **yr4^*'
i i ! * ! - " •f"'-^'
J^TJS
'•<*•:
•* •'•
.
2CO2
Fe203 + 3CO
2Fe + 3C02
^ u O + CO
Silic
lififh
\Bf\'oi'-f "DB hiA
C u + CO2
Tac dung voi phi kim
Nhan xet: D o nhiet d p soi ciia N2 (-196''C) thap h o n n h i ^ t d p soi o i a O2
tach r i e n g N2 (o t r a n g t h a i k h i ) va O2 (a t r a n g t h a i long) - > D a p an B.
V i d v 9: C h o d u n g d i c h X chiia cac m u o i : C u C h , FeCla, A l C b , Z n C k . Cho X tac
^^•^.^ .
| l S i + 2 N a O H + H2O
Silicdioxit
(du), t h u d u p e ket tiia Z . K e t tua Z la:
Si02 + 4 H F
C.Fe(OH)3.
> SiF4Si + O2 — S i 0 2
Tac dung vai dung dkh kiem
d y n g v a i d u n g d i c h K O H ( d u ) t h u d u p e ket tiia Y. C h o Y tac d y n g v a i N H s
A . Cu(OH)2, A l ( O H ) 3 .
,
IdK <
•• 1
|.
> Na2Si03 + 2H2
,,•
Ml
msrfi
> SiF4 + 2H2O
.,
'
'" :
ftfo/no-'
'
B. Cu(OH)2, Fe(OH)3.
,
D.Al(OH)3.
IM giai:
Nhan xet: K h i cho X tac d u n g v o i d u n g d i c h K O H d u , chi t h u d u p e hai ket
tua la Fe(OH)3 va Cu(OH)2. Cac h i d r o x i t : A l ( O H ) 3 va Zn(OH)2 c6 t i n h
l u o n g t i n h nen t a n t r o n g k i e m d u .
160
Si + 2F2
*
'
qirG
Silic tac dung vai flo 6 nhi^t dp thuong; voi elo, oxi khi dun nong.
(-183''C) nen n g u a i ta d u a vao s u khac n h a u nay de c h u n g cat phan doan
'
Tac d u n g v a i o x i : K h i do't n o n g , k h i C O chay t r o n g o x i hoac t r o n g k h o n g
2CO + O2
k h i l o n g d u a tren s u khac n h a u giira n i t o va o x i ve:
C. H o a t r j .
(
Hr'^-o:/'
+ 6 n h i ^ t d p cao, k h i C O k h u d u p e n h i e u o x i t k i m lo?i:
V i d \ 8: T r o n g cong nghiep, n g u a i ta c6 the tach r i e n g n i t o v a o x i tir k h o n g
B. N h i ^ t d p soi.
(
k h i , cho n g p n l u a m a u l a m nhat va toa n h i e u nhiet:
= 0,2 m o l - > V = 4,48 l i t - > D a p an B.
A. Dp am dien.
j (it ft'C' '
'
. ' ~ '
,C)
Tinh khu
,i^MM)e^-'
•
i;
Tac d u n g v a i h p p chat
C + ZnO —
• .--W
imA'^ni
f'^^'
^
b. Cacbon o x i t
> NH4CI
V a y k h i thoat ra chi la H2:
B.
,
oxi hoa khac n h u H N O 3 dac, H2SO4 dac, KClOa,...
Lim y: K h o n g c6 k h i N H 3 bay ra v i k h i nay c6 phan u n g v o l H C l :
NH3 + H C l
an
C + 4HN03(tMc) — ^
X g o m : Mg3N2 = 0,1 m o l ; M g = 0,2 m o l
Mg3N2 + 6HC1
..(^fj,,,^
Dap
> ^^5;.^
CJ nhiet d p cao, cacbon c6 the k h u d u p e n h i e u oxit, p h a n u n g v o i n h i e u chat
'-^v•^^
0,1
—>
C + CO2
nMg = 0,5 m o l ; n^^^ = 0,1 m o l . Phan u n g hoa hoc:
3Mg
Ket tiia Z la Fe(OH)3.
Tinh khu:
vu,'..
2x = 0,4 m o l -> x = 0,2 m o l -> H = 20% - » D a p an B.
B.4,48.
phuc chat tan [Cu(NH3)4](OH)2
4. C A C B O N - S I L I C
100
A . 3,36.
MTV DVVH Khang Vi$t
VIDVMAU
Vi
di;i 1 : Cho khi C O du di vao ong su nung nong dung hon hpp A I 2 O 3 , Fe203,
CuO. Sau khi phan ung hpan toan, chat ran thu dupe trong ong su gom
A. A l ; F e ; Cu.
B. A I 2 O 3 ; Fe203; Cu.
C. A I 2 O 3 ; F e ;
Cu.
D. A l ; Fe;
CuO.
161
dm
Cty Tr\iHM MTV n v v H Khang Vijt
nang 6n luy§n thi d?i hgc 18 chuyeii Jg Hua hoc - Nguyen VSn H4i
Lot gidi:
NMn
xet: K h i C O chi k h u dugc cac oxit ciia cac kim lo^i dung sau Z n trong
M
. - 3»; H'"\ VXM ' p.
day di?n hoa nen AI2O3 khong bi khu.
Fe203 + S C O — ^
+ 3CO2
2Fe
••
CuO
+
CO — ^
Cu +
•
CO2
,
;
->• D a p an C .
-.rrod-K
. . V , a,
»
Mol:
SO2
+
Br2 + H2O
0,02
< - a02
> H2SO4 + 2HBr
>fi •-
-» ms = 0,02.32 = 0,64 gam -> % m c = 95,1% - > D a p an A.
j
ifib,,- '
each 2:
.•'im.:,:
> SO2
Bao toan nguyen to S: S
-> ns= nso2
= H B I J
> H2SO4
= 0/02 mol ^ m s = 0,64 gam -> % m c = 95,1%.
- > D a p an A.
V i d v 2: D a n tu tu V h't khf C O (dktc) di qua mot ong s u dung luong d u hon
V i dy 5: N u n g nong trong 16 ho quang di^n hon hg-p chua 11,2 gam C a O va 5,4
hgp ran gom C u O va Fe203 (6 nhi^t dp cao). Sue toan bp lugng khf thu dupe
gam C (than coc) den phan ung hoan toan thu dugc chat ran X va khi Y
()
sau phan vao dung dich C a ( O H ) 2 du, tao thanh 4 gam ket hia. G i a tri ciia V
theo phuong trinh phan ung: C a O + 3 C
1^
A. 0,224.
'gniir
B. 0,448.
C . 0,896.
D . 1,120.
LaigiAi:
Baotoannguyentocacbon:CO
> CO2
> CaCOs
-> nco=nco2 =ncaC03 = 0/04 m o l V c o = 0,896 1ft.
-»
-
The tfch dung djch H C l 2M can thie't de tac dung het voi chat ran X la
A. 50ml.
fOVIHI* +
^
V i d v 3: C h o cac chat bot m a u trang: N a C l , Na2C03, BaCOs v a BaS04. C h i c6
A. 1.
NMn
- < , ^ ' B .
3.
^to^J/
C . 4.
J'
D . 2.
B. 100ml.
112
54
56
12
ncaO= -TT "
D a p an C .
H2O v a CO2 C O the nhan biet dugc toi da bao nhieu chat?
>• CaC2 + C O
CaO
+
3C
0,15
< - 0,45
C . 150ml.
Lot gidi:
D . 200ml.
= 0,45 mol.
— i - > CaC2 + C O
S^Hfb
Mol:
•Ofb.J/i
^ X gom: CaC2 = 0,15mol; C a O = 0,2 - 0,15 = 0,05 mol.
xet: D u n g nude de thu tfnh tan: Hai chat tan la N a C l va Na2C03; hai
CaC2 + 2HC1
CaO
+ 2HC1
0,15
> C a C h + C2H2t
> C a C h + H2O
1
ntyicf^fttf,
,
04
chat khong tan la BaCOs va BaS04.
Sue khf CO2 vao hai cha't khong tan (ngam trong nuac), chat nao tan ra la
BaCOa: BaCOa + CO2 + H2O
> Ba(HC03)2
Lay dung dich Ba(HC03)2 v u a tao thanh cho vao dung dich d y n g hai chat
-> VHCI = ^
=0,21ft = 2 0 0 m l ^ D a p a n D .
,
,
V i dy 6: H e n hop X gom C u O va Fe203. Hoa tan hoan toan 44 gam X bSng
dung dich H C l , thu dugc 85,25 gam muoi. Mat khae, neu k h u het 22 gam X
tan, neu xua't hi^n ket tua la dung dich Na2C03:
bang C O (du), dan hon hgp khi thu dugc vao dung djch Ca(OH)2 (du), tao
Ba(HC03)2 + Na2C03
thanh m gam ket tua. G i a tri cua m la
> BaC03l
+ 2NaHC03
—> N h a n dugc ea 4 chat —> D a p an C .
A. 45,5.
B. 50,0.
C . 40,5.
Lot gidi:
V i d u 4: Dot chay hoan toan 1,3 gam than (chua tap chat luu huynh) trong oxi
du. Dan hon hgp khf tao thanh qua dung dich brom (du), thay c6 0,32 gam
NMn
brom da tham gia phan ung. H a m lugng (%) ciia cacbon trong m a u than
tu O trong oxit dugc thay the bang hai nguyen tir CI.
tren la
1 mol O^-
•> [h,-
A. 95,1%
B.91,2%.
iJkHS. + t-ifc
C.93,4%.
Lai gidi:
Cach 1: C i a i theo phuong trinh hoa hpc.
C + O2 — ^
S
+ O2
CO2
'" > SO2
Khf SO2 tac d y n g voi dung dich brom:
...
xet: K h i cho X + H C l thi oxit chuyen thanh muoi clorua va mot nguyen
a mol
D . 97,8%.
Aljr
D . 37,5.
a=
* ,
> 2 mol C h - > khoi lugng tang 71 - 1 6 = 55 gam. v "
tang 85,25 - 44 = 41,25 gam.
=0,75mol -> no(X) =0,75mol.
oavlj
as&fU
Khi cho 22 gam X tac dung vai C O thi: n o (X) = nc02 = ^C^CO^ = 0,375 mol.
- > m = 0,375.100 = 37,5 gam.
—> D a p an D .
163
Ca'm nang 6n luygn thi dji hqc 18 chuyen 6i H6a hqc -
Nrjiiyeii
van Hai
5. BAI TAP O N L U Y E N
Cty TNIHH MTV DWH Khang Vijt
n^f
Bai 1: Cho hpn hg-p khi X gom O2 va CI2 tac dung vira du voi hon hop bgt gom
1,2 gam M g va 1,3 gam Zn, thu dugc 5,27 gam hon hgp chat ran Y. Phan
tram the tich cua khi O2 trong X la
A. 80%.
B.40%.
C.50%.
'
D. 60%.
Bai 2: Di#n phan 1 h't dung dich hon hgp gom NaCl 0,5 M va KCl 0,1M (dien
cvc tro, CO mang ngan xop) den khi 0 anot thu* dugc 1,12 h't khi (dktc) thi
|i;
dung di§n phan. Gia tri p H cua dung dich sau di^n phan la
A. 12.
B. 13.
C. 10.
j
j^^l^,
D.7.
Bai 3: Hon hgp M gom hai muoi NaX va NaY (X, Y la hai nguyen to'co trong tu
nhien, a hai chu ki lien tiep thugc nhom VIIA, Zx < ZY). Hoa tan het 6,03 gam
M vao nuoc roi cho tac dung voi dung dich AgNOs (du), thu dugc 8,61 gam
ke't tua. Hai nguyen to X, Y Ian lugt la
A. F va CI.
B. CI va Br.
C. Br va I .
D. CI va
Bai 4: Cho nuoc CI2 (loang, du) vao dung dich chua a mol NaBr
I.
b mol NaT
den phan ling hoan toan, thu dugc dung dich chua 1,17 gam NaCl. Gia trj
cua (a+ b) la
,
A. 0,01.
B. 0,02.
C. 0,03.
va O2. Sau phan ung thu dugc 11,5 gam chat ran va tong so mol k h i da
C.Ca.
D. M g .
H^;^;
Bai 6: Dot chay hoan toan 6,8 gam mgt chat v6 co X chi thu dugc 4,48 lit khi
,
SO2 (dktc) va 3,6 gam nuac. The tich khi O2 (dktc) da dung de dot chay X la
A. 7,4 lit.
B. 6,72 lit.
C. 4,48 lit.
D. 8,96 lit.
Bai 7: Hoa tan het hon hgp gom 0,02 mol FeS2 va 0,04 mol FeS vao dung djch
H2SO4
(dac, nong, du), thu dugc dung dich X va V lit khi
SO2
(dktc). Gia tri
cua V la
lis A. 7,392.
CuS04 0,2M. Sau khi phan ung hoan toan, Igc ket tua va dem nung trong
khong khi den khoi lugng khong doi, thu dugc m gam cha't ran. Gia tri cua m
la
A. 2,62.
C. 1,792.
C. 3,64.
D. 2,016.
J f i o J p .
i^02.
dung djch N H 3 den d u vao dung dich Y, ban dau xuat hi?n ket tua xanh,
sau do ket tua tan, thu dugc dung djch mau xanh tham. Chat X la
A.CuO.
B.Cu.
C.Fe.
D. FeO.
Bai 13: Hoa tan het 1,56 gam hon hgp gom A l va AI2O3 bang dung dich HCl
(du), thu dugc V lit khi H2 (dktc) va dung djch X. Nho dung djch N H 3 d u
vao X, Igc ket tua va dem nung den khoi lugng khong doi thu dugc 2,04
A. 0,448.
(2) NH3 + O2
,
B. 0,224.
BM14: Cho cac phan ung sau:
A. (2); (3).
t
~p
C. 1,344.
ai- ' j
>
D.a672.p^^
,8
(3)NH4N02
)
- i l ^ ;>fo o r D ; f •
(4) NH3 + CuO
Cac phan ung deu tao khi N2 la
j
B. (2); (3); (4).
— ^
*
C. (3); (4).
sCJi-
D.(l); (3); (4).
Bai 15: Cho 0,448 l i t khi N H 3 (dktc) d i qua ong su d\mg 16 gam CuO nung
nong, thu dugc chat ran X (gia su phan ung xay ra hoan toan). Phan tram
khoi lugng Cu trong X la
A. 85,88%.
B. 5,600.
B_ 2,04.**-''>
Bai 12: Chat ran X phan ung voi dung dich HCl dugc dung djch Y. Cho tir t u
(1) N H 3 + Ch
tham gia phan ung la 0,125 mol. Kim loai R la
v , * . , ^ B.C.
Bai 11: Cho dung djch N H a d u vao 100ml dung dich chua Al2(S04)3 0,1M va
gam chat ran. Gia trj ciia V i a
D. 0,05.
Bai 5: Dot chay hoan toan 3,6 gam kim loai R (hoa trj 2) trong hon hgp khi CI2
A. Be.
Bai 10: Co 4 dung djch muoi rieng bi?t: CuCh, ZnCh, FeCb, A I C I 3 . Neu them
dung dich NHs (du) roi them tiep K O H (du) vao bon dung dich thi so cha't
ket tiia thu dugc la:
.(aliifo) •sf{-jri>) i ' i ' X D v h mri ,;>/b V')l J
i>
A.3.
B.2. J A :>
c.4."j-ri.s
D. 1.
B. 14,12%.
C. 87,63%.
D. 12,37%.
Bai 16: Nen hon hgp gom 4 lit khi nito va 14 lit khi hidro trong binh phan ling
Bai 8: Dot chay hoan toan 18 gam FeS2 trong O2 d u roi hap thu toan bg lugng
6 nhi|t dg khoang 450''C va co bgt Fe xiic tac. Sau phan ung thu dugc 16,4
SO2 thu dugc vao 2,5 lit dung dich Ba(OH)2 0,1 M , tao thanh m gam ket tiia.
lit hon hgp khi (a ciing dieu ki^n nhi^t dg va ap suat). Hi|u suat ciia phan
Gia tri cua m la:
ung tong hgp amoniac la
A. 32,55.
B. 21,70.
C. 54,25.
D. 43,40.
Bai 9: Cho 8,45 gam mgt loai oleum vao nuoc du, thu dugc dung dich X. De
A. 16%.
,
B. 20%. "' •
C. 23%.
D. 25%.
Bai 17: Hon hgp khi X gom N2 va H2 co t i khoi so voi heli bang 1,8. Dun nong
trung hoa 1/10 dung dich X can 20 ml dung djch NaOH I M . Oleum do la
X mgt thoi gian trong binh kin (co Fe lam xiic tac), thu dugc hon hgp khi Y
A. H2SO4.2SO3.
CO t i khoi so voi heli bang 2. H i f u suat cua phan ung tong hgp N H s la
B. H2SO4.3SO3.
C. H2SO4.4SO3.
D. H 2 S O 4 . S O 3 .
A. 25%.
B.50%.
C.36%.
*' D. 40%.
Cty TNHH MTV DWH Khang Vi?t
dm
nang On luygn thi d^i hgc 18 chuy6n 66 H6a hpc - Nguygn Van H5i
^
Bai 18: K h u hoan toan 10,44 mpt oxit kim loai thanh kim loai 6 nhi^t dp cao can
x = a 0 2 m o l ; y = a 0 3 m o l - > %Vo^= ^ . 1 0 0 % = 40%.,^
0,18 mol C O . Toan bp luong k i m loai sinh ra durp-c hoa tan het vao dung
djch H C l d u , thu dupe 3,024 lit khi Hi (dktc). Cong thuc ciia oxit kim l o ^ la:
A.FeO.
i .C!
B. Fe203.
C . AI2O3.
D . Fe304.
M
Bai 19: N h o timg gipt dung dich X chiia Na2C03 0,2M v a N a H C 0 3 0,1M vao
\ 100ml dung dich H C l 0,5M deh khi khi ngimg»thoat ra, thu dug-c V m l khi
^^•.....-'.--e:,,,f,
-^^P^^„
B.224.
D . 336.
= 0,6 mol; n , , + = 0,5 mol; n , = 0,1 mol.
jSlhan xet: Ion C I " v a H 2 O tham gia di^n phan. Phuong trinh di^n phan dang
2C1-
D . 448.
Bai 20: H a p thu hoan toan 448ml khi C O 2 (dktc) vao 500ml dung djch gom
,..,,,„„•.'
c'
ion riit ggn:
>, C O 2 (dktc). G i a trj cua V la
A. 672.
•
Bai 2:
Mol:
[trnAslroM
+ 2H2O
> 20H-
ai
ai
+CI2
ntiorouli
+ Hz^Q.
ao5
^
'.
^ N a O H 0,02M va Ca(OH)2 0,01M, thu diegc dung dich Y . C o can Y thu dugc
m gam cha't ran khan. G i a tri cua m la
A. 1,34.
B. 1,65.
C.0,84.
D . 1,06.
p H = 13 ^
Bai 21: C h o 2 gam hon hop silic va than tac dung vdi lugng dung dich N a O H
loang, d u , tao ra 1,344 lit khi hidro (dktc). Phan tram kho'i lugng ciia silic
trong hon hgp ban dau la
A. 42,15%.
B. 84,30%.
C . 28,10%.
D . 56,20%.
gom 2,4 gam magie va 2,7 gam nhom tao ra 17,35 gam cha't ran. Than tram ve
the tich cua khi clo trong X la
-
B.55%.
C.75%.
Bai 23: C h o cac phan ling sau:
(a) H 2 S + SO2
>
(c) S i 0 2 + M g —
^
; i;,
f
(b)Na2S203 + H2SO4 (loang)
'
(d) AI2O3 +
B.2.
C.3.
D.4.
a
Bai 1:
nMg= — = 0,05 mol; n z „ = ^ = 0,02 mol. ^'
'•^^
24
65
). (I
, „
^
;
,
Nhan xet: D a y la bai toan hai chat k h u (2 kim loai) v a hai chat oxi hoa (2 phi
kim) nen can ap dung djnh luat bao toan electron.
Mol:
2e
)-Mg^2
0,05-^0,1
Chat oxi hoa:
Mol:
> 20-2
4x
CI2 + 2e
> Zn*^
- > 32x + 71y +1,2 +1,3 = 5,27 ^
32x + 71y = 2,77.
M N 3 X = — . 8 5 = 198,67 -> X = 175,67 ^
n M = ^
Y =C1.
Mgtnguyentoco
,
,
.
. •
Ltm y: K h i X = F thi N a F khong tac dyng v a i A g N 0 3 do A g F la muoi tan.
"
"Nacr
~
,
0/02 mol. Cac phan ung hoa hgc:
2NaI
+
CI2
> 2NaCl +
2NaBr
+
CI2
> 2 N a C l + Br2
-
I2
arttq^Q
i^NaBr+ " N a l = 0,02 mol -> a + b = 0,02 -> D a p an B.
iJd^H'
Bai 5:
Nhqn xet: Bai nay cac em c6 the vie't phuang trinh de giai. T u y nhien, nen ap
Chat oxi hoa:
> 2C1-
Mol:
2e
a
X
£,f5
> R*^
+ 4e —>
- > 4x
Ch + 2e
ICt^
Mol:
'
,o?fi
> 20r
y " > 2y
Bao toan khoi lugng: m o j + mciz = " ' 5 - 3,6 = 7,9 -> 71x + 32y = 7,9.
Bao toan electron: 4x+2y = 0,1+0,04 = 0,14 mol.
.
tang 8,61-6,03 = 2,58 gam.
nvi
Chat khu: R -
Mol: y —>• 2y
Bao toan khoi lugng: m x + m j ^ g + m z n " m y
khoi lugng tang 108-23 = 85 gam.
dving bao toan khoi lugng v a bao toan electron de giai nhanh hon.
2e
Mol: 0 , 0 2 ^ 0 , 0 4
O2 +4e
X —>
Zn -
> 1 mol A g X ^
B^4:
6. H i r d N G D A N - L 6 1 G I A I
Chat khu: M g -
• f,;.n •-v
:
-> Dap an A .
>
>
So phan ung t^o thanh don cha't la
A.I.
Theo bai:
+ VayX = F
^HMfi
'L-i'A(i'i
NaOH
_
nguyen tu khoi > 175,67 -> Loai v i khong c6 trong tu nhien.
D.65%.
4
• • .-fsiijinji^,,..!, • ^r,oan''«,g
¥ : G g i cong thuc chung cua hai muoi la N a X.
Nhan xet: 1 mol N a X
"
Bai 22: H o n hgp khi X gom clo va oxi. C h o X phan u n g vira het v 6 i hon hop
A.85%.
Bai 3:
+ Neu
D a p a n A.
—
Theo bai: n c i j + " 0 2 = 0,125 ^ x + y = 0,125 -> x = 0,025; y = 0,1.
.
167
Cty TNHH MTV DWH Khang ViQt
elm nang On luy?n thi dgi hpc 18 chuy6n di H6a hpc - Nguygn Van HSi
Bao toan electron: 2nR = 4no2 + 2nci2
->
MR=
=24
R
0^5 + 2.0,1 ^ ,3^5
- » nR =
GQi cong thiic oleum la H2S04.nS03.
Mol:
Bai6:
Nhanxet-.X
+ O2
SO2 + H2O
'»
'
Bao toan khoi lugng:
mx + m 02 = m
-> no2 = ^
a
' i^'' '
+n^H20 - > m O2 = ^2,8 + 3,6 - 6,8 = 9,6 gam.
Mol:
« ', v
a(n+l)
a(n+l)
-
"
> Na2S04 + 2H2O
H2SO4 + 2NaOH
r,/
I • fj
> (n+l)H2S04
H2S04.nS03 + nH20
la Mg -> Dap an D.
. / «
,^,
2a(n+l)
2a(n+l) = 10.0,02.1 ^
a(n+l) = 0,1.
^
[o/;
Theo bai: (98 + 80n)a = 8,45 - > a = 0,025; n = 3.
Cong thuc oleum la H2SO4.3SO3
= 0,3 mol -> V02 = 0,3.22,4 = 6,72 lit ^ Dap an B.
" ^
- > Dap an B.
Bai 10:
Trong X: ns= nso2 = 0> mol; n H = 2 n H 2 0 = 0/4 mol.
-> n H : n s = 2 :1 ^
6 bai nay, cac em can viet va can bang dung 2 phucmg trinh sau:
-iiK^yij,^
0,04
-> nso2 = 0,33 mol ^ Vgoj = 0,33.22,4 = 7,392 lit.
'
^ - X '(f.
tao ra, nen ne'u viet phuong trinh bao toan electron:
^ "^^ FeS
=
> Cu(OH)24 + 2KC1
ZnCh +4KOH
> YULnOi
^^sOz
,
,
^
Mol:
0,15
'_),, ; ( , > i ; V .
t '
- '
- - Q,
m^y^
y: A1(0H)3 va Zn(OH)2 c6 tinh luong tinh; Cu(OH)2 tan du(?c trong
j^Wii.) ;f<
'^x'y.^'^P''^
„. rnc'
Ah(S04)3
) 2Al(OH)3l — ^
0,01
0,02
CuS04
!'>rff S9,C
> Cu(OH)2i
AI2O3
0,01
ICu(NH3)4](OH)2rt S « « «'
m = mA,203 =0,01.102 = 1,02 gam
0,3
(Ltm y: Co the bao toan nguyen to S: ngoj = 2nFeS2 " ^'^ "^°^)^ ' . " O H - 0,5
Ta co:
=
>^
nso2
0,3
'
dung djch N H 3 .
,
2Fe203 + 8SO2.
A- "'
Dap an D.
Mol:
4FeS2 +1102
+ 2KC1 + 2H2O.
> [Cu(NH3)4](OH)2
Bai 11:
So do phan ung
"soa = 0'25 mol ^ Vso2 = 5,600 lit.
-> Dap an B - > Ket qua sai.
i
:J?iL™ j 0 5 + iHl-'U
> Fe(OH)3i + 3KC1
Cu(OH)2 + 4NH3
luu
Luu y: Do SO2 do ca chat k h u (FeS2, FeS) va chat oxi hoa (H2SO4 dac) cung
llnFeS2
€M * f H
Vay ket tua con lai la Fe(OH)3
.i^^oy.-
. •/;:,->_ Dap an A.
*pKt..rfsn..mJ
AICI3 + 4KOH
> KAIO2 + 3KC1 + 2H2O
-'
Them tiep N H 3 du, Cu(OH)2 tan hoan toan do tao phuc chat:
> Fe2(S04)3 + 9SO2 + I O H 2 O
2FeS + IOH2SO4
Mol:
'
C u C l 2 + 2KOH
FeCb + 3KOH
> Fe2(S04)3 + I 5 S O 2 + I4H2O
0,15
2FeS2 + I4H2SO4
0,02
'
Khi cho tac dung v6i K O H du, thu dugc 2 ke't tua:
Hgp chat X la H2S.
Bai7:
*'
Mol:
'
- > Dap an D.
b f i f f ? ' 7 • ; ^ r f s ^ M ^
Tao thanh 2 muoi: Sunfit va hidrosunfit. ' „ .
•
•
•
^
Bai 12:
.YS.-^*S.'
m&n xet: Loai B v i Cu khong tan trong HCl.
'
it
•
>i
i* * +*) * • ^ '
jv ,
Loai C, D v i ket tua Fe(OH)2 khong tan dugc trong NHs d u .
Tinh nhanh: n^^-^ = n ^ ^ . - ngoj = 0,5 - 0,3 = 0,2 mol.
- > Dap an A. Cac phan ung hoa hpc:
,
Nhan thay: n^.
< n^^2. -> n^^so, = ^o^- = ^'^
m = 0,2.217 = 43,4 gam
Dap an D.
'
,
.,
of*^
t*('
> CuCh + H 2 O
CuO + 2HC1
CuCh +
2NH3 +
2H2O
Cu(OH)2 + 4 N H 3
> Cu(OH)2i (xanh) + 2NH4CI
> [Cu(NH3)4](OH)2 (xanhtham)
^
' ' ''
•M'*"''''''
