DE
SO 1
DE THI HOC S1I\IH GiOl HOA HOC LOfP 9, TP. HQ CHI MINH NAM HOC 1998
Cdu
I.
Cdu
I I Viet phuang
Cdu
Viet 3 phuang
trlnli
khdc
trinh phdn
> Fe(0H)2
FeCh
> Fe(0H)3
mang
(NHJ.SO,;
dugc
NaoCOs;
de dicu
die muoi
ling de bieu dlen
FeCls
III. Co 6 ong nghiem
NaOH;
nhau
chiiSi
ZnCl2.
bien
^ FeSO,
PbiNO^)^;
so ndo dUng chat ndo? Viet phdn
CaCl..
icng minh
Fe(N0s)2
> Fe.
so tii 1 den 6 chica
Ba(N03)2;
hoa sau:
>
> FesOs
ddnh
- 1999
cdc dung
dicli:
Hay cho bict
6'ng
hga. Biet
rdng:
a) Dung
dich
(2) cho ket tila trdng
vai cdc dung
dich
(1), (3), (4). ''•
b) Dung
dich
(5) cho ket tua trdng
vai cdc dung
dich
(1), (3), (4).
c) Dung
dich
(2) khong
tgo ket tila vai dung
d) Dung
dich
(1) khong
tgo ket tua vai cdc dung
dich
e) Dung
dich
(G) khong
phdn
(5).
f) Dung
dich
(5) bi trung
i(ng vai dung
hoa bdi dung
dich
(5).
dich
(3), (4).
dich HCl.
g) Dung dich (3) tgo ket tua trdng vai HCl, khi dun nong ket tua nay se tan.
»
CduIV.
a) Nong
Cau
•
do dung
Tinh
dich
bdo hoa KCl d 40°C la
do tan cua dung
) Xdc dinh
AgNOs
lugng
dich KCl a ciing
AgNOs
tdch
28,57%.
nhiet
do.
ra khi Idm Ignh
bdo hoa a 60°C xuong
10°C. Cho biet
2500
gam dung
dich
do tan cua AgNOs
o
60°C Id 525 gam, a 10°C Id 170 gam.
V. (A) la dung
Trgn
Lay
them
dich H2SO4;
(B) la dung
0,3 lit (B) vai 0,2 lit (A) dugc
20 nd (C), them
tic tit dung
dich
dich
NaOH.
0,5 lit (C).
mot it quy tim vdo thdy
HCl 0,05M
c6 mdu xanli.
tai khi quy tim ddi thdnh
Sau do
mdu tim
thdy het 40 ??il axit.
Trgn
0,2 lit (B) vai 0,3 lit (A) dugc
Lay
20 ml dung
Sau
do them
dich
(D), them
tii tii: dung
dich
Tim
dich
do mol/l dung
I r(i n i A i n c T U I u n r C I M M n i f i i H H A H f i n Q
mot it quy tim vdo thdy c6 mdu do.
dich NaOH
tim thdy het 80 ml dung
nong
0,5 lit (D).
0,1M tai khi quy ddi thdnh
NaOH.
(A) vd (B).
mdu
Cdu
VI. Xdc dinh
cong
thiJtc ciia hai oxit sdt A vd B, biet
•
23,2 gam (A) tan vica dil trong
•
32 gam (B) khi k/iii bdng Ho tqo thdnh
r&iig:
' ^ ) ^ 60"C, t r o n g 525 + 100 b a n g 625 g a m d u n g d i c h c6 525 g a m AgNOg
v a 100 g a m H2O. T r o n g 2500 g a m d u n g d i c h c6 x g a m AgNOa v a
0,8 lit HCl IM.
sdt vd 10,8 gam IhO.
y g a m nxidc:
L6\I
Cdu I.
Cdu 11.
* Zn + CI2
ZnCla
>
.
Z n O + 2HC1
•
Z n S 0 4 + BaCl2
X
Fe + 2IIC1
+ BaS04i
> FeCl2 + H 2 t
FeCl2 + 2 N a O H
> Fe(0H)2^ + 2 N a C l
(trdng
Fe(0H)2 + H2SO4
2Fe + 3CI2
> B a S 0 4 ^ + Fe(N03)2
FeCla + 3NaOH
> FeCOIDai + 3NaCl
(ndu do) •
= 2100
(gam)
400 g a m ni/dfc.
CJ 10°C, CLf 100 g a m ntfcJc h 6 a t a n 170 g a m AgNOa
400 g a m niJcJc h o a t a n z g a m AgNOa
z =
Do do khoi
400x170
100
liTgng AgNOa
= 680 (gam) AgNOa.
ket t i n h k h i lam lanh:
2100 - 680 = 1420 (gam)
Cdu V.
Phan ufng:
— > Fe203 + 3H2O
Fe203 + 3 C 0
625
& 60°C t r o n g 2500 g a m d u n g d i c h c6 2 1 0 0 g a m A g N O s v a
xanh)
> 2FeCl3
2Fe(OH)3
Vay
> FeS04 + 2 H 2 O
FeS04 + Ba(N03)2
2500 X 525
y = 2500 - 2100 = 4 0 0 g (nxidc)
> ZnCl2 + H2O
> ZnCh
=
H2SO4 + 2 N a O H
> Na2S04 + 2 H 2 O
L a n thuf n h a t q u y t i m c6 m a u x a n h chufng t o diT N a O H . T h e m a x i t
— > 2 F e + 3CO2T
Cdu I I I . Theo cac duf k i e n de b a i neu r a , cac lo d i i n g cac h o a c h a t sau:
H C l de t r u n g h o a N a O H dif:
HCl + NaOH
f ) => L o so 5 d i i n g : Na2C03
> N a C l + H2O
(2)
g)
L o so 3 d i i n g : Pb(N03)2
L a n 2 l a m quy t i m c6 m a u do chufng t o H2SO4 dir.
d)
L o so 1 chufa: Ba(N03)2, lo 4 chuTa CaCl2.
T h e m N a O H de t r u n g h 5 a H2SO4 d i / .
2 N a O H + H2SO4
a ) => L o 2 chijfa: (NH4)2S04.
(NH4)2S04 + Ba(N03)2
> B a S 0 4 i + 2NH4NO3
(NH4)2S04 + Pb(N03)2
> P b S 0 4 i + 2NH4NO3
(NH4)2S04 + CaCl2
(3)
Theo cac p h a n ufng (1), (2), (3) t a c6 phiiofng t r i n h :
0,3y - 2
> C a S 0 4 i + 2NH4CI '
0,3x -
Cac p h a n i l n g c o n l a i hoc s i n h tii vie't.
Cdu rv.
X
0,2x = 0,05
0,2y
0,1
x
0,08
X
2
0,04
2
x
500
20
500
20
= 0,05
= 0,1
(a)
(b)
G i a i h e p h u a n g t r i n h (a), (b) => x = 0 , 7 M v a y = 1,1M.
a ) G o i S l a do t a n cua K C l d 40°C.
Cdu VI.
K h o i lUcfng d u n g d i c h t h u dUgfc: (S + 100) g a m
N o n g do p h a n t r a m cua c h a t t a n t r o n g d u n g d i c h b a o h o a :
S + 100
S + 100 - 3,5S »
> Na2S04 + 2H2O
G o i X , y l a n l u g t l a n o n g do cua H2SO4 v a N a O H .
e) =^ L o 6 chijfa: N a O H .
P h a n irng:
(1)
S + 100
G o i c o n g thufc o x i t s a t l a FexOy.
Fe^Oy + 2 y H C l
(mol)
0,4
0,8
S = 40 (gam).
I fi\l n £ T u i u n r s l u u Rini Hni Hnr. a
^^IfilAlflg THI HOC SINH GIOI
GIO HOA HOC 9
> x FeClayx + y H 2 0
(1)
Ta c6: I I H C I = 0,8 X 1 = 0,8 (mol)
PHAN B. Bai toan
Theo de:
Bai
23,2 = — (56x + 16y)
y
<=>
56x = 42y
<=>
X _
3
fx = 3
y
4
[y = 4
FexOy + yH2
vd K2SO4
dugc trgn Idn theo tl Ic
Hoa tan hon hgp vdo
102 gam nUac thi thu dugc dung dich A. Cho 1664 gam dung dich
BaCL
10% vdo dung dich A. Lgc ket tua, them H2SO4 du vdo tiudc vUa Igc tin
thdy tgo ra 46,6 gam ket tua. Xdc dinh nong do phdn tram cua NosSOj
> xFe + y H 2 0
0,6
(mol)
Co mot hon hgp gom Na2S04
I : 2 ve so mol (1 mol Na2S04 yd.2 mol K2SO4).
Vay oxit sSt (A): Fe304 (sdt tii oxit)
•
1-
(2)
vd K2SO4 trong dung dich ddu. •
Bai 2.
0,6
Tinh
CM cua dung dich H2SO4
vd NaOH,
biet rdng 10 nd dung
dich H2SO4 tdc dung vUa du vdi 30 ml dung dich NaOH. Neu lay 20 ud
10,8
Ta c6:
18
dung dich H2SO4
= 0,6 (mol)
lugng du ndy tdc dung vita du vdi 10 nd NaOH.
'
TheodI:
^ ( 5 6 x + 16y] = 3 2 «
y ^
x = 2
LdlGIAI
- = -=>•
y 3
[y = 3
PHAN
=> Cong thufc oxit (B): FeaOs-
Cdu
DE SO 2
NaOH + N H 4 C I
2 N a O H + FeCl2
PHAN A. Li thuyet
Chi dugc diing thuoc thii de nhgn biet cdc muoi sau: NH4CI,
MgCl2,
NaCl, AICI3
(Gidi thich vd viet phuang
trinh phdn
ling
2Fe(OH)2 + - O2 + H2O
> Fe2(S04)3
b) MgCOs
> Fe(0H)3
> MgS04
> MgCOs
> FcsOa
> FeCls
> MgCl2
> CO2
> Mg(0H)2
> Ca(HC03)2
III. Chi dugc dung guy tini vd dung dich AgN03
IV. Sdt nguyen chdt trong khong khi thi khong
CO tap chdt de Idu ngdy trong khong
hien tugng
a
3 N a O H + FeCls
>Fe
>FeCl2
phdn biet cdc dung dich: NaOH, NaCl, HCl, H2S,
Cdu
> 2Fe(OH)3
Mau CO ket tija nau do la FeCls:
trinh phdn ling bieu dien cdc bien hoa sau:
> FeClz
Cdu
,
2
II. Viet phuang
a) Fe
> Fe(0H)2i + 2NaCl
trang xanh
FeClo,
neu c6).
Cdu
— > NaCl + N H g t + H^O
Mau tao ket tiia trSng xanh, hoa nau do trong khong k h i la FeC]2:
TP. HO CHJ MINH NAM HOC 1998 - 1S99
FeCls,
Cho dung dich NaOH dii Ian liicft vao cac mau thif t r e n va dun nhe;
Mau thuf nao c6 k h i m i i i khai bay ra la N H 4 C I :
DE THI CHOIV flOl TUYEN KQC SINH GIOIHOA HOC 9, qUAAl TAN BINH
I.
A. Li thuyet
I. - T r i c h m6i lo mot i t lam mt\ thuf.
-
Cdu
cho tdc dung vdi 2,5 gam CaCOs thi axit con du vd
^
> Fe(0H)3i + 3NaCl
nau do
> AgCl.
Mau tao ket tua keo trSng sau d6 tan trong NaOH dif la AICI3:
> MgO
3 N a O H + AICI3
> CaCOs
> A l ( 0 H ) 3 t + 3NaCl
keo trdng
c6 sdn, neu each
A 1 ( 0 H ) 3 + NaOH
H2SO4.
> NaAlOa + 2H2O
Mau tao ket tua trSng la MgCl2:
bi han gl, nhUng sdl
khi Igi bi han gl. Hay gidi
'
thich
2 N a O H + MgCl2
nay.
^ Mg(0H)2 + 2NaCl
Mau khong c6 hien tJong gi la NaCl.
Lfll GIAI Oi THI HOC SINH GIOI HOA HOC 9
Lfll GIAI
T H I unr emu ni/\i un» unr a
^
Cdu
Cciu
II.
a)
2Fe + 6H2SO4 damdac
2Fe(OH)3
-
K h i t r o n g s a t b i I a n t a p c h a t , de l a u n g a y b i b a n g i do x a y r a s u
a n m o n k i m l o a i tufc l a bie'n sat t h a n h h o p c h a t cua sat.
— > FeaOs + 3H2O
Gidi
— > 2Fe + SCOat
l u o n g o x i n e n c h u y e n Fe -> Fe^\
F e + 2HC1
> FeCla + H a t
V a o x i hoa t a n t r o n g niidc theo qua t r i n h : O2 + 2H2O
2FeCl2 + CI2
> 2FeCl3
Sau do Fe^' k e t h o p vdri O H "
2FeCl3 + Fe
> SFeC^
M o t p h a n F e ( 0 H ) 2 b i o x i hoa t a o F e ( 0 H ) 3 (nau
MgCOg + H2SO4 loang
Mg(0H)2
> MgCl2 + B a S 0 4 i
> MgCOIDai + 2 N a C l
—>
M g O + CO2 —
MgCOa
+ CO^^t + HgO
> MgSOi
^
PHAN
Bai
1.
B. Bdi
rr.
-
T a co:
> C a C O a i + NaaCOg + 2H2O
hoac Ca(HC03)2 + C a ( 0 H ) 2
2 C a C 0 3 i + 2H2O
HI.
M a u k h o n g c6 h i e n t i i o n g l a N a C l .
•
Nhufng m a u l ^ m quy t i m h d a do: I I C l , H2S, H2SO4
Sau do cho dung dich AgNOa Ian li/gt vao cac m a u l a m quy t i m hoa do.
M a u nao tao k e t t u a t r S n g la H C l :
> A g C l i + HNO3
•
10
<-
X
1664
n
o
/
1^
= 0,8 ( m o l )
> BaS04i + 2NaCl
X
(mol)
2x
->
(1)
X
> B a S 0 4 i + 2KC1
2x
^
(2)
2x
V i k h i t h e m d u n g d i c h H2SO4 vao n L f d c loc l a i t a o k e t t u a n e n
t r o n g niidc loc con dtf BaCla.
(mol)
> B a S 0 4 i + 2HC1
0,2
«o
<= ^
»
X
(3)
0,2
= 0,2 ( m o l )
233
T h e o de: ng^ci, = ^-^ ^n^^l)
ng^^i./d) + "BaCi,/(2) + ^Baa.,im
= ^'^
+ 2x + 0,2 = 0,8 => X = 0,2
«
mdung dich = 0,2
C % N a so
n a o CO ke't t u a den l a H2S:
2 A g N 0 3 + H2S
10
=
K h o i l a g n g d u n g d i c h A l a : nidungdich = m N a . s o ^ + " ^ K , S O , + "^H^O
irdng
MSU
m,,
=1:2
B^so,/,3,
•
•
X
M a u thuf n a o l ^ m quy t i m h o a x a n h l a N a O H .
AgNOs + H C l
X
BaCl2 + H2SO4
C h o quy t i m t a m ni/dc I a n \\iat vao cac mau thijf t r e n :
•
C%
BaCl2 + K2SO4
T r i c h m 6 i lo m o t i t l a m m a u thtf.
-
n3,,,^ =
(mol)
C a C 0 3 i + H2O + C 0 2 t
hoac Ca(HC03)2 + 2 N a O I I
•
toun
BaCl2 + Na2S04
> Ca(HC03)2
—>
4Fe(OH)3
Cac p h a n ufng x a y r a :
> M g O + COat
Ca(HC03)2
do)
G o i X l a so m o l c i i a Na2S04 => Hj^^g^^ = 2x ( m o l )
MgCOg
2CO2 + C a ( 0 H ) 2
->
xanh)
Do do s a t b i h a n g i c6 m a u n a u do.
"Na^so, • " K ^ S O ,
M g O + H2O
>• 4 0 1 1 "
> F e ( 0 H ) 2 (trdng
4 F e ( 0 H ) 2 + O2 + 2H2O
> Fe(N03)2 + 2 A g C U
MgCl2 + 2 N a O H
-
T r e n be m a t k i m l o a i c6 I d p nUdrc a m , d a h o a t a n m o t
thidi:
FeaOg + SCO
M g S 0 4 + BaCla
Cau
b e n t r o n g k h o n g k h i a n h i e t do t h u 6 n g .
> 2 F e ( O I I ) 3 i + 3Na2S04
FeCla + 2 A g N 0 3
b)
1^02(804)3 + 3SO2T + 6II2O
>
Fe2(S04)3 + 6 N a O H
IV.
- s a t n g u y e n c h a t k h o n g b i b a n g i v i n6 dugc bao ve bori I d p Fe203
^ ^ 2 ^ ° "
> kgiSi
^
^
^
1 74
+ 2HNO3
M d u k h o n g c6 h i e n t u o n g g i l a H2SO4.
i f i l R I 4 I f i F TH] H f i r SlUH f;inr u n A u n r n
=
X 142
Ml GIAI DE THI HOC SINH GIOI HOA HOC 9
^ ^
200
X
+ 0,4
X
x 174
+ 102
= 200
(gam)
100 = 14,2%
04
11
Bai 2.
T a c6:
2.5
= 0,025 (mol)
100
nC a C O ,
4. Tafco
them
P h i i n ufng:
H2SO4 + CaCOa (mol)
0,025
(mol)
a
2NaOH
->
Ong 3: dung dich bac
Na2S04 + 2H2O
20
5. Mot hon hap NaCl vd MgCl2 them nuac vao hon hap ta c6 dung dich A.
= 0,0125 (mol)
Them dung dich AgNOs vao dung dich A, khi phdn iing ket thuc
30 m l d u n g d i c h N a O H chufa y m o l N a O H
2a
30
X
10
Chia B lam 2 phdn: a vd b.
Phdn a: Sau khi c6 can tiep tuc dun nong thl diigc mot hon Jigp khi
= 6a (mol)
C. Cho hon hap khi nay qua binh dUng KOH.
H2SO4 + 2 N a O H
(mol)
Na2S04 + 2H2O
Phdn b: Cho vdo lugng dU dung dich HCl thi thu dugc ket tua trdng D.
0,0125 - > 0,025
Viet cdc phuang
V i p h a n ufng t r u n g hoa n e n 6a = ^'
"2SO4
"
0,01
logi
bo chat ket tua trdng, phdn dung dich con Igi la dung dich D.
T r o n g 10 m l d u n g d i c h N a O H chijfa 2a m o l N a O H
y =
Em hay cho biet
dng ndo con chat ket tua. Gidi thich vd viet phiXang trinh phdn icng.
10 m l d u n g d i c h H2SO4 chtfa x m o l H2SO4
=
nitrat
Sau do cho them axit nitric vao ca ba ong nghiem.
T r o n g 20 m l d u n g d i c h H2SO4 chufa 0,025 m o l H0SO4
X
vao:
Ong 2: dung dicli natri cacbonat
2a
10 X 0,025
ngUdi ta cho
diig l:~dung dich kali cacbonat
-> CaS04 + C O a t + H2O
< - 0,025
H2SO4 +
3 6'ng nghi&m deu dung dung dich bari clorua,
(mol)
6
trinh phdn
dng.
6. Lam the ndo de phdn biet cdc Ig hoa chat diidi day md khong
dung theni hoa chat ndo khdc: MgCl2, H2SO4, NaCl, CuSO^,
= 1,25M
dugc
NaOH.
7. Mot hon hap gom: CuO, FeO, AI2O3. Lam cdch ndo de tdch chiing ra
0,025
^6
2,5
0,03
khoi
M
nhau.
Cdu II. Bai
todn
Cho 26,Ig MnOo tdc dung vdi dung dich HCl c6 20 gam HCl. Cho hct
OE SO 3
khi do qua mot lit dung dich NaOH lodng dU.
DE THI HOC SINH GiOl HOA HOC 9 QUAN 3 TP. HQ CHJ MINH NAM HOC 1998 - 1999
a) Lugng HCl nay c6 du de phdn dng het vdi Mn02
Cdu 1. Li thuyet
b) Tinh nong do mol 11 cua muoi thu dUgc trong phdn I'ing giOa do vd NaOH.
1. Tit 7 Ig hoa chat sau, em c6 the dieu chc nhUng chat khi nuo?
Axit sunfuric;
natri hidroxit; amoni
sunfit; sdt sunfua vd kim logi keni.
2. Ta H2SO4 CO may cdch de dieu che
nitrat;
canxi
cacbonat;
c) Nung qugng pyrit sdt 'de tgo ra SO2. Cho khi SO2 sue vdo dung
natri
chda 2 muoi
khoi
lugng pyrit
dung dich
dich
tren.
Sau do them vdo mot lugng du Ba(N03)2.
CaSOJ
3. Viet cong thUc vd ten ggi 3 muoi diing trong nong nghiep (phdn dam,
phdn Idn vd plidn kali). Hay gidi thich tgi sao ngUdi ta khong trgn tro
bep vdi phdn dam de ban rugng?
19
khong?
can dung.
Tim khoi lugng ket tua va
Biet rdng lugng SO2 tdc dung
vUa du
muoi.
(Cho Mr, = 55; O = 16; H = 1; CI = 35,5; S = 32; Fe = 56; Ba = 137)
^^F:.
,
13
L d l GIAI
Cdu I. Li
dng
thuyet
1000°C
Zn
+
-> CaO + C O a t
FeS + H2SO4
—>
—
NH4NO3
> Na2S04 + S O s t + H2O
MgCl2 + 2AgN03
Phdn
2AgN03
—>
> CaS04 + 2H2O
Phdn
2 M g O + 4 N 0 2 t + O-^t
—>
b: K h i cho H C l d u vao d u n g d i c h B t h u diJOc k e t t u a . D i e u
H C l + AgNOs
> A g C l i + HNO3
(D)
> CaS04 + C O a t + H2O
Chu y: Con nliieu cdch klidc, xin nJiuang ban doc!
3. C o n g thufc ba m u o i d u n g t r o n g n o n g n g h i e p :
6, T a n h a n t h a y : t r o n g t a t ca cac d u n g d i c h t r e n t h i c h i c6 m o t d u n g
d i c h CO m a u x a n h l a : CUSO4, cac d u n g d i c h con l a i l a k h o n g m a u .
•
CO(NH2)2 ( 4 6 % n i t a ) : U r e ( p h a n d a m )
T r i c h m o i lo m o t i t l a m m a u thijf.
•
Ca(H2P04)2: Supe p h o t p h a t k e p ( p h a n I a n )
-
•
KRO'i.
[Ca(H2P04)2
•
m a n g t i n h k i e m n e n n g i / d i t a k h o n g t r o n p h a n d a m vcfi t r o bep.
Ong 2:
BaCl2 + K2CO3
M a u thCf cho ke't t i i a m a u x a n h la N a O H :
CUSO4 + 2 N a O H
kali nitrat (phan kali)
hoac K C l : k a h clorua.
4. Ong 1:
Cho d u n g d i c h CUSO4 I a n lifot vao cac m a u thuf t r e n :
•
2CaS0,
P h a n d a m de b i p h a n h i i y t r o n g m o i t r t f d n g k i e m , v i t r o bep
-
> B a ( N 0 3 ) 2 + C 0 2 t + H2O
BaCIa + Na^CO^
> BaCOgi + 2NaCl
2 H N O 3 + BaCOa
-> Ba(N03)2 + C02t + H2O
> C u ( 0 H ) 2 i + Na2S04
Cac m l u con l a i k h o n g c6 h i e n tu'ong.
D u n g N a O H l a m thuoc thiif, cho I a n liiot vao cac m a u ihxi con l a i :
•
Mau
t h L f CO
k e t tua mau t r S n g la MgClg:
2 N a O H + MgCla
> B a C O g i + 2KC1
2 H N O 3 + BaCOg
K N O 3 + K N O 2 + H2O
n a y cho t a t h a y r a n g t r o n g d u n g d i c h con A g N O s dif.
Cdch 4: Tac d u n g vcri muo'i cija c a n x i :
hoac supe pho't p h a t dcfn
-02t
2
— > 2Ag + 2 N 0 2 t + 0 2 t
2NO2 + 2 K 0 H
CaS04 + II2O
Cdch 3: Tac d u n g v6i bazcJ C a ( 0 H ) 2 :
H2SO4 + CaC03
NaNOa +
> CaS04 + H a t
H2SO4 + CaO
H2SO4 + Ca(0H)2
—>
2Mg(N03)2
2. D i e u che CaS04 t i f H2SO4:
Cdch 2: Tac d u n g vdri CaO:
Mg(N03)2
a: K h i d u n n o n g d u n g d i c h B :
NaNOs
) N 2 t - H ^ 0 2 t + 2H20
2
Cdch 1: Tac d u n g vdri c a n x i : H2SO4 + Ca
> 2 A g C U + Mg(N03)2
v a A g N O s dif.
> N a O t + 2H2O
>^°°°^
> A g C U + NaNOg
Sau k h i Ipc bo k e t t u a t h i d u n g d i c h (B) g o m : N a N O s ,
> N a N O s + N H a t + H2O
Na2S03 + H2SO4
hoac
5. P h a n ufng: N a C l + A g N 0 3
ZnS04 + S 0 2 t + 2H2O
-> FeS04 + H 2 S t
N a O H + NH4NO3
NH4NO3
mac dti l a a x i t H N O 3 v i t h e t r o n g o n g 3 con c h a t k e t t u a .
> ZnS04 + H 2 t
2H2S04dac
> Ba(N03)2 + 2 A g C l i
Trong 3 ong nghiem t h i ta thay ket tija AgCl k h o n g t a n trong axit
Zn + H2SO4 loang
hoac
BaCla + 2 A g N 0 3
A g C l + HNO3
1. T a d i e u che duac cac c h a t k h i sau:
CaCOa
3:
•
M a u t h i i tao d u n g d i c h t r o n g suot va toa n h i e t m a n h l a H2SO4.
2 N a O H + H2SO4
•
> Mg(0H)2 + 2NaCl
> Na2S04 + 2H2O
M a u k h o n g c6 h i e n t u g n g l a N a C l .
b ) T h e tich dung dich thu dUcfc chinh 1^ the tich cua N a O H tufc l a :
7. T a c h C u O , F e O , AI2O3 r a khoi nhau:
CuO
AI2O3
+C0,
^NaAlOg
N a O H da
CuO_
FeO
-> A l ( O H ) ,
+ HC1 du
FeO^
Fe(OH),
cliaii kliong
So' mol cua N a C l trong p h a n ufng (2): 0,1375 (mol)
+ N H . O H dil
FeCl,,
+HC1
[Cu(NH3)J(OH)2
CuCl^
Vduiigdich = 1 (lit)
->A1,03
->CuCL
+NaOH du
C^
-^Cu(OH),
-»CuO
AI2O3 + 2 N a O H
(mol)
> A l ( 0 H ) 3 i + NaHCOg
CuO + 2HC1
— > CUCI2 + H2O
F e O + 2HC1
-> FeCl2 + H2O
> Na2S04 + N a C l + H2O (2)
- > 0,1375
Ba(N03)2 + Na2S04
Txi (3)
^
^ ne^so, = ^i^.,so,
> [Cu(NH3)4](OH)2
(3)
+ 2NaN03
> Ba^O^i
0,1375
(mol)
(i)
0,1375
(mol) 0,1375 < - 0,1375
> Cu(0H)2i + 2 N H 4 C I .
Cu(0H)2 + 4 N H 3
^
2
SO2 + N a C l O + 2 N a O H
-> AI2O3 + 3H2O
CuClz + 2NH4OH —
> 2Fe203 + 8S02t
4FeS2 + I I O 2
> 2NaA102 + H2O
-
= 0,1375M
-^FeO
N a A l O z + CO2 + 2H2O
2A1(0H)3
=
c) P h a n ufng:
P h a n ufng:
•
=C„
0,1375
= 0.1375 (mol)
mg^so^ = 0,1375 x 233 = 32,04 (gam)
(xanh t i m , tan)
[Cu(NH3)4](OH)2 + 2HC1
CUCI2 + 2 N a O H
Cu(0H)2
Cdu
TO (1), (2)
+ 4 N H 3 T + H2O
> CuCOIDal + 2 N a C l
np^3
^so,
=^
2
_ 0,1375
=
(mol)
"
2
(gam).
-> F e ( 0 H ) 2 i + 2 N a C l
chankhong
DE SO 4
FeO + H2O
I I . Bai todn
T a c6:
:^
0,1375
=> mpes, = - ^ - r — X 120 = 8,25
> CuO + H2O
FeCl2 + 2 N a O H
Fe(0H)2
> CuCh
OE THI HOC SiNH GIOI HOA HOC 9 , CAP TP. HO CHI M I N H N A M HOC 1999 - ZOOO
n MnO.,
26,1
87
= 0,3 (mol): n^j^j =
M n O a + 4HC1
CI2 + 2 N a O H
36,5
0,55 (mol)
Cdu
(1)
0,1375
-> N a C l + N a C l O + H2O
(mol) 0,1375
I . Khi cho kirn
loai
vdo dung
Cdu
I I . Viet cdc phuang
(2)
Fe
0,1375
trinh
> FesO^
TCr p h a n ufng (1) t a c6 t i l e :
1
= 0,3>
Cdu
n HCl
4
0,1375
4
V a y l i r a n g H C l n a y k h o n g d i i de p h a n ufng h e t liicfng M n 0 2 da cho
h a y I I C l p h a n ijfng h e t v a M n 0 2 con d i i .
phdn
mud'i c6 the xdy ra nhitng
phan
I I I . Mot nhd hoa hoc dieu
dang
ben ngoai
vCng theo chuSi bien hoa sau:
FeCh
> Fe(0H)3
> FezOs
FeCl2
> Fe(0H)2
> FeO
<^
^
a ) L i f g n g H C l n a y c6 du de p h a n lifng h e t vcfi M n 0 2 k h o n g ?
n MiiO,
dich
ling gi? Cho vi du minh hoa.
> M n C l a + Cl2t + 2H2O
0,55
(mol)
20
che dugc
3 mdu
(mdu sdc) va da tim dugc phuang
chong.
dng lay cdc mdu
NaOH,
ket qua do dugc ghi tron^^M^:^,^^"^
L f l l R l A i n £ T u i u n n S I U H R i n i HflA H t l R 9
kim log.i cho tdc dung
—
kirn
loai gidng
phdp
phdn
nhau ve
biet
vdi^axitjm_d^g
nhanh
dich
6SNH THUAN
•
_
.
J O
17
Thuoc
Axit
HCl
Axit
HNO3
Dung dich
thii
Kim loai I
-
Kim loai II
Kim loai III
K i m loai I :
A g (Bac)
+
+
K i m loai I I :
A l (Nhom)
+
-
+
K i m loai I I I :
Zn (Kem)
-
+
+
NaOH
Trong do ddu (+) de chi trudng hap kim loai hoa tan, ddu (-) chi
Cdu I I I T h e o de b a i t h i :
•
triiang
ciiu, viet phuang
HCl, HNO3 ddc, AgNOa, KCl,
Vii't cdc phuang
trinh phdn
ling vd gidi
•
-
dung
dich A. Cho 1664 gam dung dich BaClz 10% vao dung dich A, xudt
hien
ket tua. Loc bo ket tua, them H2SO4 dU vdo nilac Igc thi tlidy tgo ra
ket tua. Xdc dinh nong do phdn
trong dung dich A ban
l a HNO3.
Cu + 4HNO3 dac
-
+) M a u CO k e t t u a m a u x a n h l a K O H .
Cu(N03)2 + 2 K 0 H
-
N e u k i m l o a i cho vao k h a c k i m l o a i t r o n g m u o i t h i x a y r a p h a n
Fe + CUSO4
nhiet la H C l
N e u k i m loai t r u n g v 6 i k i m loai t r o n g muoi t h i xay r a p h a n tfng tiT oxi hoa.
> SFeClg
Cdu I I . P h a n ufng:
3 F e + 202
^°
Fe304 + 8HC1
> Fe304
> K C l + H2O
KOH + HCl
M a u con l a i l a K C l .
Cdu V. G o i k i m l o a i hoa t r i I I l a A va c6 a m o l => o x i t l a : A O .
A O + H2SO4
(mol)
Theode:
a ^
2Fe(OH)3
a
« , , „ , ^ . = ^
" ^ d d Hi,S04
a ( A + 96)100
C%ddASO,
> FeCOIDgi + 3 N a C l
—>
FeCla + N a O H
FezOs + SHsO
> Fe(0H)2i + 2NaCl
TT-^rt
chan khong
> FeO + H2O
> ASO4 + H2O
a
> FeCla + 2FeCl3 + 4H2O
FeCl3 + 3 N a O H
•
Cho d u n g d i c h K O H vao 2 m a u con l a i , m a u nao c6 p h a n
uTng
> FeS04 + Cu
Fe + 2FeCl3
> 2KNO3 + C u ( 0 H ) 2 i
+) Cac m a u con l a i k h o n g c6 h i e n t u g n g .
K h i cho k i m l o a i vao d u n g d i c h m u o i c6 t h e x a y r a :
Fe(0H)2
> Cu(N03)2 + 2 N 0 2 t + 2II2O
Sau do cho d u n g d i c h viia t h u di/oc a t r e n I a n liicft vao cac m a u con l a i
ddu.
t h e - o x i h o a khijf.
-
> Cu(N03)2 + 2 A g
+) M a u thuf nao vifa t a o d u n g d i c h m a u x a n h va c6 k h i n a u do bay r u
tram cua A^a^SOj vd K2SO4
Ld\I
-
trUcfng
Cho b o t k i m l o a i Cu d t i I a n l u a t vao cac m a u t h t f t r e n :
Cu + 2 A g N 0 3
(Cho Na = 23; S = 32; K = 39; Ba = 137; CI = 35,5).
Cdu I .
A l k h o n g tac d u n g v d i HNO3 v i A l b i t h u d o n g t r o n g m o i
+) M a u t h i i nao d u n g d i c h id k h o n g m a u chuyen sang x a n h la AgNOs.
vd K2SO4 dilgc trgn Idn iheo ti le
I : 2 ve so mol. Hoa tan hon hap vdo 102 gam nUac thi thu dilgc
46,6 gam
v d i NaOPI v i k h o n g p h a i l a k i m
ufng
Cdu IV. T r i c h m o i lo m o t I t l a m m a u thijf.
KOH.
trinh phdn dug xdy ra trong qua trinh nhgn biet.
VI. Co mot hon hap gom Na2S04
hoat
HNO3 dac n g u o i .
day:
Cdu V. Hoa tan oxit cua kim logi hoa tri II trong mot lilgng vita dil dung
dich H2SO4 20% thi thu dicgc dung dich mudi c6 nSng do 22,69c.
Xdc dinh kim logi do.
Cdu
v d i H C l v i A g dufng sau h i d r o t r o n g day
l o a i iLfdng t i n h .
thich vi sao kim loai khong tdc dung vai cdc chat dd cho.
Cdu TV. Chi diing kim loai, hay nhgn biet cdc dung dich sau
uTng
dong k i m loai. A g k h o n g phan
hap kim loai khong tdc dung vai dung dich kicm hay axit.
Hay xdc dinh kim loai nghien
Ag khong phan
=
—
™ d d AS04
Ma:
<=>
=>
a(A + 9 6 ) x l 0 0
a x 98 x 100
•
= a(A + 16) +
22,6
20
A = 24: M a g i e (Mg).
ufng
toa
VI. T a c6:
Cdu
„,
0 3
Cdu
= 10 x 1664 ^ ^
100 X 208
^^^'2
'^NajSO,,
Gidi
Cdu
> BaS04i + 2NaCl
X
< -
X
- >
2x
<-
BaS04l
>
2x
+ 2KC1
(2)
2x
K l i i t h e m dung dich H2SO4 vao lo niidfc loc thi tao ket tua nCfa nen
BaCl2
+ H2SO4
>
BaS04i +
<-
2HC1
(3)
0,2
Theo de:
H
=
0,2
lugng
lugng
tinh
the
•
Tinh
•
Khi
trgn
50
ml
0,004M
Cdu
174
+
142 X 0,2
= —^——
200
NaaSO^
200
„„
=
174x0,4
^
m^^^^^^ + m^^^^^^ + m^^^
102
100 =
C % N a so
C%„
=
200
dich
gam
tdch
dich
dich
dung
ra
dich
khoi
20%
dcm
nung
10°C.
Tinh
khoi
II9SO4
den
dung
(a 20°C) vd khoi
dich.
Biet
do
tan
lugng
rieng
cua
dung
dung
dich
mol.
0,012M
CaCl2
thi c6 ket tila xudt
vai
hien
150
ml
khong?
10%.
trong
cua kim logi M bdng
mot
lugng
ling thu dugc dung
dich
A. Them
Sau phdn
lugng
vUa dii dung
dugc
so lieu
dich AgNOs
tUong
sau: ti so thdnh
20
—.
2 oxit do bdng
20%
iCng cila ciuig
phdn
vita
mot
% ve khoi
dii
vdo
thu dugc
Xdc dinli cong thvCc hidroxit
tich 2 oxit vd 2 hidroxit
hda hoc
X 100 = 3 4 , 8 % .
1 hidroxit
muoi c6 nong do 8,965%.
b) Khi phdn
14,2%
200
'^2^^i
dung
Iglml.
dung
todn
A mot
x
thi
V.
a) Hda tan hodn
md.ngdich =
0,4
trong
Id 0,2g
a 20°C
dich
+
HCl
iCng.
do tan cua CaSO^ theo nong do
dung
142
het
bdo hda coi bdng
HCl
X
dich
17,4g.
b) Clio biet do tan CaSO^
dich
0,2
tan
CUSO4.5H2O
dung
=
ket tila dugc chat rdn D. Cho H2
vila dii, sau do Idm ngugi
K h o i lugng
mdungdich
trinh phdn
CuO
TCr (1), (2), (3) => x + 2x + 0,2 = 0,8 =:> x = 0,2
=>
dagc
(mol)
ng^Q^ ban ddu = 0,8 (mol)
dung dich (A) l a :
thi thu
phdn.
mol
nong
NaoSO^
=
nBaso,„3,
vd CuSOj
dUgc chat rdn E. Hda tan E vdo dung
Viet phuang
0,25
dich
0,2
(mol)
nong
CUSO4 a 10°C Id
trong nude loc con dii B a C l 2 .
mud'i Al2(S04)3
IV.
a) Cho
X
BaCls + K2SO4
(mol)
(1)
thich.
dich
B vd ket tua C. Nung
thdy E tan mot
Cac p h a n ijfng:
(mol)
dich
di qua D nung
G o i X l a so' m o l cua Na2S04 => n^^g^^ = 2x ( m o l )
B a C l 2 + Na2S04
Na vdo 2 dung
khi A, dung
=• 1 • 2
• '^K2S04
I I I . Cho
dung
tren.
nguyen
lugng
ciia
to
0x1
,
2T
Tl
DE SO 5
I.
Cho chuoi
phuang
trinh plidn
+A,
ling
I ,
sau:
hidroxit
dung
dich
Cdu
II. Dung
mot
FeCh;
trinh
phdn
HCl;
nhau
vd khdc
ting.
kim loai de nhdn
FeCh;
khdc
BaCh;
CaCOs)-
dung
b) Neu
biet cdc lo dung
(NH^2S04;
dich
AlCh;
sau:
NH4CI
X tdc dung
hda V ml dung
CaCOs
cdc chat A, B, C, X, Y, Z (Id cdc chat
do bdng
khoi
lugng
. Hay xdc dinh
VI. Cho X, Y Id hai dung
a) Khi
Viet cdc phuang
% ve
ciia
nhdm
Iiidroxit
trorig
hai
nguyen
to' do.
135
Cdu
+B,
CaCOs
Tim
phdn
107
OE THI HOC SINH GlDl HOA HOC, QUAN 9 (VONG 2 ) TP. HCM NAM HOC 19S9 - 2000
Cdu
thdnh
so
trgn
dich
vai-AgNOs
Z. Tinh
dich
c6 nong
du tao thdnh
dich
dich
Tinh
35,876g
dich NaOH
Lay
V ml
ket tila. De
trung
nhau.
0,3M.
dich
Y thu dUgc 2 (lit^
Z.
X vd lay 100 nd dung
het vdi kim logi Fe thi lugng
(lit)unA(dktc).
u n ra
do khdc
X vol V (lit) dung
CM dung
lay 100 nd dung
LCII Riai niihau
c T u , „ „0,448
o
HCl
Y can 500 ml dung
V (lit) dung
dich
dich
hidro
thodt ra trong
CM dung
dich X, Y.
dich
Y cho tdc
hai trUdng
hgp
dung
lech
21
Cdu
VII.
Trgn
20 nd dung
Idn
dich
10 ml dung
dich
A de diCgc dung
dich
B can 8 ml dung
dich
tqo thdnh
dung
dich
dich NaOH
dugc
b) Dung
nhieu
CM dung
dich
l,365g
dich
1,N=
dich
vai 20 ml dung
dich A, pha
doi.
khan.
Neu
dich
HNO3
II2O
vdo
dung
hoa 25 nd
dung
Bern c6 can
dung
them
Trung
8% (D = l,25g/nd).
muoi
cho 40 tnl dung
dich BaCl-z thi thu dugc
axit ban
C chi'ia hon
ml dung
Cho: H=
HCl
dung
B c6 the tich gap
vai mot liCgng dU dung
a) Tinli
dich
thu dugc
H2SO4
dich
0,932 ket
vd
M & u nao cho k e t t u a n a u do l a FeCla:
3Ba(OH)2 + 2FeCl3
M a u CO k e t t i i a v a k h i m i i i k h a i bay r a l a ( N H 4 ) 2 S 0 4 :
B a ( 0 H ) 2 + (NH4)2S04
B tdc
tiia.
3Ba(OH)2 + 2AICI3
hap NaOH
0,8M
vd Ba(0H)2
hoa het 50 ml dung
0,2M.
dich
Can
bao
B?
B a ( 0 H ) 2 + 2NH4CI
B a ( 0 H ) 2 + 2HC1
GIAI
t-HCl
->Ca{OH),
lAl
'Cdu
->CaCl,
III. P h a n ufng:
(Bl
N a + 1120
CO,,
+ NH011
+CO.,*U.,0
-^Na^CO,
6 N a O H + Al2(S04)3
->NaHCO.,
CaC03
iooo"c
CaO + CO2T
(A)
CaO + I I 2 O
(X)
Ca(0H)2
—
CO2 + 2 N a O H
> Na2C03 +
> 2NaHC03
(Z)
—
-> C a C O y i + 2 N a N 0 3 + CO.T + II2O
Cho k i r n l o a i B a vao cac m a u ihii t r e n , dau t i e a c6 p h a n ufng:
B a ( 0 H ) 2 + H2T
M a u nao cho k e t t i i a t r f i n g x a n h la FeCl2:
Ba(01i)2 + FeCL —
-> CuO + H2O
^ — > Cu + H2O
AI2O3 + 6 H C 1
- > 2AICI3 + 3H2O
CuO + 2HC1 -
-> CuCla + H2O
CuO + H2SO4
(mol)
0,25 ->
—
> CUSO4 + H2O
0,25
0,25
K h o i iLfoag d u n g d i c h H2SO4 2 0 % :
II. T r l c h m o i d u n g d i c h m o t i t l a m m a u thuf.
B a + 2H2O
—
H2O
(Y)
2 N a I i C 0 3 + Ca(N03)2
Cu(0H)2
- > AI2O3 + 3H2O
V i E t a n 1 p h a n n e n E chufa: CuO, Cu, AI2O3
> CaCO^i + 2NaCl
NaaCOs + CO2 + I I 2 O
-
CuO + H2 -
> CaCl2 + 2H2O
CaCl2 + NaaCOa
2A1(0H)3
-> C u ( 0 H ) 2 i + Na2S04
AI2O3 + H2
(B)
C a ( 0 H ) 2 + 2HC1
> 2 A l ( O H ) 3 i + 3Na2S04
2 N a O H + CUSO4 —
P h a n ijfng:
22
-> N a O H + ^ H a t
CaCO;,
CaCOg
Cdu
> BaCl2 + 2H2O
C o n l a i l a BaCla.
ThiTc h i e n chuoi p h a n ufng:
CaO
> BaCl2 + 2 N H 3 t + 2H2O
M a u cho d u n g d i c h t r o n g suo't v a toa n h i e t l a H C l :
Cu = 64, Ag = 108, Ba = 137
Cdu I.
> 2 A l ( O H ) 3 i + SBaCla
M a u c h i CO k h i m t i i k h a i bay r a l a NH4CI:
14, O = 16, Na = 23, S ^ 32; CI = 35,5, Co = 40, Fe = 56,
LCfl
> B a S 0 4 ^ + 2NH3T + 2H2O
M S U cho k e t t i i a keo t r a n g la AICI3:
ddu.
C de trung
> 2 F e ( O H ) 3 i + 3BaCl2
> FeCOIDai + 2BaCl2
I rii p i A i n c T u i u n r Qtwu n n i unA u n r Q
0, 25
X
98
X
100%
20
= 122,5
K h o i l u g n g CUSO4: 0,25 x 160 = 40
(gam)
(gam)
K h o i lu'gng d u n g d i c h sau p h a n ufng:
' ^ d u i i g d i c h s a u p h a n iMig —
niCuO +
= 0,25
X
'^^ddn^SO^
80 + 122,5 = 142,5
(gam)
K h i h a n h i e t do:
K h o i lucfng d u n g d i c h t r o n g (2):
• C U S O 4 + 5H2O
(mol)
— > CUSO4.5H2O
a
->
mdu„gdich2
_
T h e o de, n o n g do cua d u n g d i c h m u d i t h u dirge sau p h a n ufng (2) l a :
17,4
8,965% =
4 6 9 6 - 18784a = 2479,5 - 4 3 5 0 a
«
2216,5 = 1 4 4 3 4 a
Cong thuc oxit:
20°C: 100 g a m H2O + 0,2g CaS04 - > 100,2 g a m d u n g d i c h CaSO,
CO d = I g / m l
(mol)
M2OX,
V^^^^^ =
0,0006
"caci, =
o
16y
= 20 : 27
2 M + 16x • 2 M + 16y
x
M + 8y
20
M + 8x
y
27
2 7 M x + 216xy = 2 0 M y + 160xy
2 7 M x + 56xy = 2 0 M y
> CaS04 + 2 N a C l
17y
17x
M a t khac:
M + 17x • M + 17y
X 0,012 = 0,0006 ( m o l )
o
^Na^so^ = O'^'^ ^ 0'004 = 0,0006 ( m o l )
(1)
= 107 : 135
M + 17y
:
107
y
M + 17X
135
135xM + 2295xy = 1 0 7 M y + 1819xy
d 2Q°C t r o n g 1 l i t d u n g d i c h CaS04 bao h o a c6 0,015 m o l CaS04
t r o n g 0,2 l i t d u n g d i c h CaS04 bao h o a c6 0,003 m o l CaS04
V i 0,6.10"^ < 3.10"^ n e n k h o n g c6 h i e n t i / g n g k e t t i i a .
Cdu
M 2 0 y
16x
= 100,2 ( m l )
C^.
=
x — - — = 0,015M
Mc,so,
136
0,1002
0,0006
M = 39a
C o n g thilfc h i d r o x i t : M ( O H ) ^ , M ( O H ) y
T h e o de b a i , t a c6:
+ Na2S04
<r>
b ) Goi nguyen to can t i m la M :
2 5 0 a = 250 x 0,154 = 38,5 (gam)
CaCla
100%
X
D o do c o n g thufc c a n t i m l a K O H .
Vay k h o i lifgng CUSO4.5H2O k e t t i n h :
Khi tron:
M + 62a
M+1232a-143,5a
100%
=^ N g h i e m h o p l i : a = 1; R = 39: k a l i ( K )
a = 0,154 ( m o l )
V a y thi t i c h d u n g d i c h CaS04 Ik:
X
M +1232a-143,5a
'3'a
<r>
Suy r a :
M + 62a
C% M ( N 0 , ) ,
142,5 - 2 5 0 a " 100 + 17,4
h)*a
+ 382a + 170a : 2 0 % - 143,5a
= M + 1232a - 143,5a ( g a m )
K h o i l i i o n g d u n g d i c h c o n l a i : 142,5 - 2 5 0 a ( g a m )
40-160a
=
ra^i^ + ra^^j^^o,
a
Kho'i liicrng C U S O 4 c6n l a i t r o n g d u n g d i c h : 40 - 1 6 0 a ( g a m )
.
=
135xM + 476xy = 107My
<=>
(2)
T i f (1) v a (2), t a c6:
5(20My - 56xy) + 476xy = 107My
V.
a ) G o i c o n g thufc cua h i d r o x i t k i m l o a i M : M ( O H ) a
M(OH)a + aHCl
> M C l a + aHaO
(I)
MCla + aAgNOg
> aAgCll + M(N03)a
(2)
<=>
lOOMy - 2 8 0 x y + 4 7 6 x y = 1 0 7 M y
<=>
7 M y = 196xy <=> M = 28x
Bang bien luan:
K l i o ' i l u g n g d u n g d i c h t r o n g (1)
X
1
2
M
28
56
3 "
84
V a y M = 56 (Fe).
Vay
= M + 17a + 36,5a x 10 = M + 382a ( g a m )
.
I rli n i l n c T U I u n n c i u u n i n i unA unp n
c o n g thijfc o x i t :
FeO; Fe203
cong thiic h i d r o x i t : F e ( 0 H ) 2 ; F e ( 0 H ) 3
25
Cdu
Trifffng hap 2: Cx < Cy
VI.
H C l + AgNOa
a)
(mol)
—> A g C U + HNO3
0,25
0,25
HCl + NaOH
(mol)
(1)
^ N a C l + H2O
(2)
0,15 <- 0,15
35,876
T a c6: n ,
143,5
= 0,25 (mol) wk n^^oH = 0'^ x 0,3 = 0,15 (mol)
TCf p h a n ijfng (1) v a (2), t a c6:
0,25 + 0,15
0,4
v +v
(mol)
= 0,2M (Vi V + V = 2)
(mol)
FeCl2 + H a t
0,025
0,025
V
2V
(3)
V
(mol)
2V' •
- n^^^.,^
= n^^^^^ = Cx-V = 0 , 2 5 ( m o l )
-• nHci/,2)
= "NaOH
= CY-V
0 , 1 h t d u n g d i c h H C l t h i c6
= 0,15
(mol)
molHCl.
0 , 1 l i t d u n g d i c h H C l c6
(mol)
<=>
V.V
<=>
3V - 10 + 5V
<^
8V - 8V - 10 = 0 =^
(mol)
+
0,005CB <-
8V^
V, = 1 , 7
(loai)
> N a C l + PI2O
HCl
0,0025CA
— > N a N O s + H2O
0,004
> Na2S04 + 2H2O
(3)
0,005Cc
> B a S 0 4 i + 2H2O
<-
(2)
0,005CB
0,005CB
—
(1)
0,0025CA
HNO3 -
0,01Cc <- 0,005Cc ->
(4)
0,004
Goi CA l a n o n g do m o l cua d u n g d i c h H C l .
So m o l cua H C l t r o n g 25 m l d u n g d i c h B :
0^.0,01x0,025
= 0,0025CA
G o i CB l a n o n g do m o l cua d u n g d i c h HNO3
c
^
V
V
=> So m o l ciia HNO3 t r o n g 25 m l d u n g d i c h B :
= 8
C B . 0 , 0 2 x b , 025
0,1
=> So m o l ciia H2SO4 t r o n g 25 m l d u n g d i c h B :
\5
Cg.0,02 X 0,025
V2 = 2,5 > 2 (loai)
I rtl n i f l l n p T H I H n r c i w u n i n i u n A u n p n
= 0,005CB
G o i Cc l a n o n g do m o l cua d u n g d i c h H2SO4
10 - 5V - 3V = 16V - SV^
0,1
V = 0,5 => V = 1,5 ^ Cx = 0,5M v a Cy = 0,1M
26
V:
T h e t i c h cua d u n g d i c h (B): 100 m l = 0,1 l i t
molHCl.
V
NaOH
0,0025CA <
= 8; V + V = 2 (V, V < 2)
8V^ - 24V + 10 - 0
16V
0,1
5(2 - V) - 3V = 8V(2 - V) «
«
3V - 5(2 - V) = 8V(2 - V)
= 0,02 (mol)
0,015^-^,32
2V
5V' - 3V
o
H2SO4 + BaCla
TriTdrng hdp 1: N e u Cx > Cy
0,025
2V
3V - 5V' = 8 W ' , V + V = 2
2 N a O H + H2SO4
T i i a n g tLf: T r o n g V l i t d u n g d i c h H C l c6 0 , 1 5 m q l H C l .
22,4
<=>
NaOH
T r o n g V l i t d u n g d j c h H C l t h i c6 0 , 2 5 m o l H C l .
T h e o de: n H„
V'
5
a ) Cac p h a n ufng:
(4)
0,015
0,448
2V
25
3
=
40
<
=
>
V
V
vn.
(mol)
-> FeCl2 + H 2 t
0,015
T h e o cau (a):
2V'
15
= 0,02
V i = 1,7 => V = 0,3 =^ C^ = 0,147 v a Cy = 0,5.
Cau
2HC1 + Fe
0,025
V2 = -0,72
b) P h a n ufng
2HC1 + Fe
0,015
"^LSIMI
THI HOC SINH GIOI HOA HOC 9
= 0,005Cc
27
Khoi
iLforng
d u n g d i c h N a O H = 8 x 1,25 = 10 ( g a m )
So m o l cua N a O H :
^
100
So m o l cua B a S 0 4 :
0 Q*^?
— — = 0,004 ( m o l )
233
X
40
G o i a, b , c I a n Ixxat Ik so m o l ciia H C l , HNO3, H2SO4 t h a m g i a p h a n
= 0,02 ( m o l )
L^ng (1), (2), (3).
a', b ' , c' I a n l i i a t l a so m o l cua H C l , HNO3, H2SO4 t h a m g i a p h a n
ang (4), (5), (6).
,
C A X 0,01x0,05
i i H c i = a + a' =
So' m o l cua H2SO4 t h a m g i a p h a n ufng (4):
X 0,02x0,04
= 0,004 =^ 0,008Cc = 0,004
0,1
^Cc
0 , 0 0 2 5 C A + 0 , 0 0 5 C B = 0,02 - O.OlCc =
2 5 C A + 5 0 C B = 150
«
CA+2CB = 6
Va:
5CA + I O C B =
0,015
0 , 1 3 2 5 C B = 0,1325 =^ C B = 1 (mol/1)
Suy r a :
= 6 - 2CB
CA
=
2
(0,8V - 2c - b) + a' + b + b ' + c' + c = 0,035
(1!
0,14625(6 - 2 C B ) + 0 , 4 2 5 C B = 1,01
o
^
0 , 8 V - c + (0,4V - 2c') + c' = 0,035
=^
1,2V - (c + c') = 0,035
1,2V - 0,005 = 0,035
=:>
4 (mol)
V - 0,03333 ( l i t ) = 33,33 m l .
DE SO 6
b ) Cac p h a n ufng t r u n g h o a :
NaOH
(mol)
a
+
<--
(mol)
+
2c
b
H2SO4
<-
NaOH
> N a C l + H2O
(1
a
2NaOH
(mol)
HCl
+
<-
> Na2S04 + 2H2O
(2
c
HNO3
> N a N O s + H2O
(3
> BaCla + 2H2O
(4
(mol)
b'
—
2
<-
c'
«-
> B a ( N 0 3 ) 2 + 2H2O
b'
B a ( 0 H ) 2 + H2SO4
(mol)
Cdu I.
Viet phan
c'
bien hoa
> Ca(0H)2
sau:
^
CaCOs
CaClo
Cdu
Ca(N03)2
11.
Tdch
phdp
hon
hap
hoa
Cdu in.
Hay
nhdn
gdm
BaCOs,
BaS04,
KCl,
MgCh
bdng
phuang
hoc.
b) Cho cac hoa
chat: Na, MgCh,
biet
Cho phuang
viet 4 phuang
xdy ra hodn
Ldl GlAl Oi THl HOC SINH Gidl HOA HOC 9
chuoi
CaCOs
hay
> B a S 0 4 + 2H2O
ling de hieu dien
CaO
a)
B a ( 0 H ) 2 + 2HNO3
(mol)
O E THl H O C SINK GIOI H O A H O C 9, C A P TP. H O CHJ M I N H N A M H O C 2 0 0 0 - 2 0 0 1
b
Ba(0H)2 + 2HC1
mol
= 0,01
2]n,^^ = a + a' + b + b ' + c + c' = 0,035
T h e ( I ) vao ( H ) t a c6:
0,8775 - 0 , 2 9 2 5 C B + 0 , 4 2 5 C B = 1,01
,
,
CcxO,02xO,05
,
= c + c =
-^-^1^
= 0,005 m o l
2
1,365 = 0 , 0 0 2 5 C A X 58,5 + 0 , 0 0 5 C B X 85 + 0,005Cc x 142
o
•
nH2S04
—
^ + K +c' = 0 , 2 V
(I
1,365 = 0 , 1 4 6 2 5 C A + 0 , 4 2 5 C B + 0,355
CB X 0,02x0,05
= b + b' =
a + b + 2c = 0 , 8 V ( v d i V l a t h e t i c h ciia d u n g d i c h C ) .
30
mniuo'i = n i N a C i + mNaNo, + ™Na,so,
<=>
, ,
nHN03
= 0,5 (mol/1)
So m o l cua N a O H t r u n g h 6 a t r o n g p h a n ufng 1 , 2, 3 1 ^ :
«
,
,
= 0,02 m o l
todn.
FeCls,
FeCk,
AICI3.
Chi dung
them
H2O
chung.
trinh
trinh
phan
phan
ling c6 dang
sau: BaCh
ling xdy ra. Biet
rdng
+ ? = NaCl
cac phan
+ ?
i2ng
deu
.au rV.d 25°C nguai ta da. hda tan 450 gam kali nitrat vao trong 500 gain
niiac cat (dung dich A). Biet rdng dp tan ciia nitrat kali la 32 gam „
Cau IIa) Scf do tach:
BaCOy
20°C. Hay xdc dinh khoi lugng kali nitrat tdch ra khoi dung dich khj
BaSO.
Idm lanh dung dich A den 20°C.
~lau y . Cho 3 gam
hdn hap hai kim loqi vun nguycn
Xdc dinh
BaSO^ +H;,o
KCl
cliudn.
MgCl^
thdnh phdn
phdn
tram
ve khoi lugng
ciia nlwm
vd magic
KCl
trong hdn hap.
Na + H2O
dich HCl thi sau kJii phdn ling ket thiic thu diigc 896 ml khi H2 (dicii
-
kiwi
hodn
M S U CO k e t tua mau trSng la MgCl2
-
> Mg(0H)2i + 2 N a C l
M a u nao c6 k e t tiia mau trang xanh, de lau hoa nau do la FeCl2.
I'ing vol axit thi Mij'
FeCla + N a O H
ling dcu xdy ru
> F e ( 0 H ) 2 i + 2NaCl
trdng xanh
todn).
Biet: H=1;N=
-> NaOH + - H a t
2
MgCl2 + 2NaOH
lugng cdc chat trong X, Y.
pJidn ling trade, het Mg mai den Fe. Cho biet cdc phdn
2Fe(OH)2 +
14; O ^16; Mg = 24, Al = 27; CI = 35,5; K = 39; Ca = 40; Fe = 56
-> 2 F e ( O H ) 3 i
+ H2O
ndu do
LCilGIAI
du I.
-
Chuoi bien hoa:
M a u cho k e t t u a m a u nau do l a F e C l s .
F e C l a + 3NaOH
CaCOs
^ - ^ ^ — > CaO + CO.T
CaCOa + 2HC1
CaO + 2HC1
CaO + H2O
Ca(0H)2 + CO2
CaCl2 + 2AgN03
Ca(0H)2 + 2HNO3
Ca(N03)2 + Na2C03
-
> F e ( 0 H ) 3 i + 3NaCl
M a u cho k e t t u a keo trang la AICI3.
AICI3 + 3NaOH
> CaCh + COgt + H2O
-> A1(0H)3^ + 3NaCl
Neu N a O H dii thi k e t tiaa t a n dan:
> CaCl2 + H2O
> Ca(0H)2
->MgCl,
+) Cho dung dich N a O H I a n liigft vao cac mau thtf tren t h i :
kien chuan) vd c6 can dung dich thi dugc 6,68 gam clidt rdn Y.
ling vai nitdc vd khi phdn
+ HC1
+) Cho ni/orc Ian liJOt vao cac mau thiJf tren. Mau nao c6 k h i bay ra la natri.
Ncu cho hon hap gom a gam Fe vd b gam Mg vdo trong 400 ml dun/-
sii Mg khong phdn
'Mg(OH)^ i
b) Trich mSi chat 1 i t l a m mau thuf.
ling ket thuc dem c6 can dung dich thu dicgc 6,2 gam chat rdn X.
Tilth a, b, nong do phdn td gam ciia dung dich HCl vd thdnh phdn
KCl
+K011
vL/a du
MgCl^
^du VL Khi cho a gam Fe vdo trong 400 ml dung dich HCl, aau khi phun
(Gid
BaSO^ i
IdC
BaCO,,
chat Id nhom v,i
magie tdc dung hct vai H2SO4 lodng thi thu duqc 3,36 lit mot chat khi a
dieu kien ticu
BaCl. - ^ M loc
i_^BaCO. i
+HC1
Cdu
-> CaCOsi + IlaO
> Ca(N03)2 + 2AgCU
III,
A1(0H)3 + N a O H
> NaAlOa + 2H2O
BaCl2 + Na2C03
> B a C 0 3 i + 2NaCl
BaCl2 + Na2S04
— > B a S 0 4 i + 2NaCl
3BaCl2 + 2Na3P04
B a C l a + Na2S03
^ Ca(N03)2 + 2H2O
(hoac BaCl2 + NaaSiOa
^ CaC03>l + 2 N a N 0 3
LOi
r.iAi,
> Ba3(P04)2i + 6NaCl
> B a S O g i + 2NaCl
> BaSiOgi + 2NaCl
Cdu IV. Bap so: Kho'i Itfang kali n i t r a t t^ch ra k h o i dung dich:
Tif (3)
290 (gam) K N O 3 .
Cdu V. Goi a va b Ian liigt la so' mol cua M g
Phan ijfng:
Mg + H2SO4
(mol)
->•
(mol)
Giai he ta dtfOc: a = 0,05 va b =
Cdu
%mMg =
^'
3
x
mcLatrdn = 0,04
Theo de bai:
nichat rin
Phan Lfng:
M g + 2HC1
(mol)
= 0,15
(mol)
CO
dich t h i thu difgc 6,2 gam chat r ^ n . Neu Fe phan ijfng het tufc chat rSn
=
0M^9, (mol)
(*)
Vay:
niMg = 0,02
"^Fecij
Thi nghiem 2: Cho a gam Fe va b gam M g vao 400 m l dung dich H C l , c6
can dung dich t h i thu diigc 6,68 gam chat r ^ n va 0,896 l i t H2 (dktc).
Phan
Lfng:
M g + 2HC1
(mol)
X
> FeCla + H a t
y
mpedLT
^
nj,
"2
+y=
22,4
« x + y = 0,04
(2)
0,04
0,08
0,04
^0,04
=
6,68
|x + y = 0,04
fx = 0,02
[95x + 71y = 3,32
l y = 0,02
x 24
= 0,48
X 127
= 3,36 - 0,02
(gam)
= 2,54
x
(gam)
56 = 2,24 (gam) ^ b = 0,48 (gam).
Cdu 1.
^) Chi dung nuac hay nhdn biet 3 bgt kim loai: Ba, Al vd Ag.
(**)
^) Til cdc chat sau: Na^O, HCl, H2O, Al c6 the dieu che dugc n/ulng clid't
fnai ndo md kJiong dung them phiCang tien ndo khdc. Viet phan ling
0,04 (mol).
(mol)
=
+ ^Fed,
NAM HOC 2000 - 2001
so mol H2 cf t h i nghiem 1 la:
> FeCla + H g t
+ y = 0, 04
Ki THI CHON HOC SINK GIQIHOA HOC 9 TP. QUY NHdN, TJNH MW MW
han d t h i nghiem 1 nhifng so mol H2 thu diioc l a i i t hon. Do do trong
Fe + 2HC1
= X
DES0 7
So sanh (*) va (**) ta thay: K h o i lugng k i m loai d t h i nghiem 2 nhieu
t h i nghiem (1) t h i Fe di/ va H C l het
nj^
y
95x + 127y + (s, 36 - 56y) = 6,68
y
= X
y
(5)
+ y = 0,04
(1;
(Vdi X , y Ian liTOt 1^ so' mol cua M g va Fe tham gia phan iJtng tren).
TCr (1), (2)
X
> FeCla + Hat
->
X
Fe + 2HC1
(mol)
> MgCla + H2t
(4)
n^ran = n^MgCl, + ^FeCl,
Thi nghiem 1: K h i cho a gam Fe + 400 m l dung dich H C l , c6 can dung
= ^
(gam)
+ H2t
x
Tif (4), (5) ta CO he phtfcfng t r i n h :
= 60%.
= 7,16
> MgCh
y
+) Tinh a va thanh phan cua (X).
np^^i,
+ 3,36
->
X
n^^ =
56 = 2,24 (gam)
= 6,68 < 7,16 => M g khong du nen Fe da phan ijrng.
X
VL
Chi la FeCl2 ^
X 95
Fe + 2HC1
—
3
%mAi
x
6,2 - 5,08 = 1,12 (gam)
Neu k h o i lircfng M g dung du t h i :
(2)
24a + 27b = 3
100 = 40% va
0,04
uTng:
(gam)
+) Tinh b va thanh phan cua (Y)
2
a .
= 5,08
Vay: a = 2,24 + 1,12 = 3,36 (gam).
3b
->
Theo de bai, ta c6 he phuong t r i n h :
Vay:
(1)
> Al2(S04)3 + 3H2t
b
x 127
va khoi lircfng Fe diT:
a
2A1 + 3H2SO4
= 0,04
Khoi lugng Fe phan
Al.
> MgS04 + Hgt
a
ra^^cA.,
• minh hoa.
2. Viet cdc phuang trinh phan ling tlieo sa do sau:
A
(3)
i
-> B
+Y
D
+z,t°
A
^'•^'t C Id chat ket tua mdu do ndu vd A, B, C, D, X, Y, Z Id ki hieu I'Cng
cong thitc 1 chat.
Cdu
3. Bern nij gam hon hap ZnCOs, Zn dun nong ngodi khong
' i dc
plidn ling xdy ra hodn todn, thu dugc nig gam chat rdn.
Biet nil = mo. Tinh % khoi lugng ZnCOs trong hSn hap ddu.
Cuu 4. Bem dung dich chiia 0,1 mol sdt clorua tdc dung vai dung dich
NaOH du thu dugc 9,05 gam ket tua.
Xdc dinh cong thiCc sdt clorua vd tinh hieu sudt phdn ling.
Cdu 5. Bem 46,4 gain Fe^Oy tdc dung vdi Ho dun nong thu dugc chat rdn B
gom Fe vd Fe^Oy du. Bem chat rdn B tdc dung het vai dung dich HNO.
lodng du thu dUgc dung dich C c6 chUa 145,2 gam mud'i Fe(N03)3 va
a mol NO thodt ra. Tat cd phdn Ung xdy ra hodn todn.
a) Xdc dinh cong thUc Fe^Oy.
h) Biet a = 0,52, tinh khoi lUgng tiCng chat trong B.
Ldi
Cdu
. 2Zn + O2
(1)
2ZnO
(2)
Goi X , y Ian lircft la so mol ZnCOg va Zn trong h6n hop dau. V i m j = m2.
K h o i liiong CO2 thoat ra 0 (1) = khoi lugng oxi tham gia a (2)
=>
Vay
I
Cdu
44x = 16y ^
- = ~
y
11
% khoi iLforng ZnCOg = 41,15%.
4. Neu la FeCl2: FeCl2 + 2NaOH
> F e ( 0 H ) 2 i + 2NaCl
Khi phan ufng xay ra 100% thi khoi lugng ket tiia Fe(0H)2 = 9g < 9,05g
GIAI
1
=i> V6 11, vay do la FeCls.
a ) Cho 3 k i m loai vao 3 coc niicfc.
-
Tan
CO
FeCla + 3NaOH
bot k h i bay len la Ba:
Ba + 2 H 2 O
> Ba(0H)2 + H a t '
-
Kliong t a n la A l va Ag.
-
Cho 2 k i m loai A l va Ag Ian liicft vao hai coc chufa dung dich
Ba(0H)2: K i m loai nao tan c6 bot k h i bay len la A l :
2A1 + Ba(0H)2 + 2 H 2 O
-
=>
Cdu
NasO + H2O
145,2
> 2NaOH
NaAlOa + H C l + H2O
> A l ( 0 H ) 3 i + NaCl
A l + 6HC1
= 84,58%
56
= 33,6 gam
K h o i iLforng oxi trong 46,4 gam Fe^Oy = 46,4 - 33,6 = 12,8 gam
,
X
y
> NaCl + H2O
> 2AICI3 + 3H2T
33,6
X
16
^ 1 2 : ^
3
=
1
Fe + 4 H N O 3
b)
2. Phan ijfng:
,
FeCl, + 3NaOH
2Fe(OH)3
Fe203 + 3H2
B: FeCla;
-> 2FeCl3
^
> Fe(N03)3 + N O t + 2H2O
> 9Fe(N03)3 + NOT +
UlUO
Goi X , y Ian liigt la so mol ciia Fe va Fe304 trong hon hop.
> F e ( 0 H ) 3 i + 3NaCl
V',-
So mol NO: x + ^ = 0,52
3
FeaOa + 3H2O
Bao toan Fe: 56x + 168y = 33,6
> 2Fe + 3H2O
C: Fe(0H)3;
„
- ^ " " ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^
3Fe304 + 28HNO3
2Fe + 3CI2
A: Fe;
X
242
> 2NaA102 + 3 H 2 t
NaOH + H C l
Hieu sua't (H) =
a ) K h o i liiang Fe c6 trong 145,2 gam FeCNOsJa:
> Ba(A102)2 + Sllat
2A1 + 2NaOH + 2 H 2 O
> F e ( 0 H ) 3 i + SNaCl
5.
Khong tan la Ag.
b)
Cdu
-> ZnO + CO2
ZnCOo
3.
D: Fe203;
=> X = 0,51 mol ^ m p e = 28,56 (gam)
Y: NaOH;
X: CI2;
Z: H2
A, B, D, X, Y, Z CO the khac, nhUng C phai la Fe(0H)3.
y = 0,03 mol ^
Wl
Rli, „S ....
m^^ ^
= 6,96 (gam).
L6\I
DE SO 8
flE T H I
HOC S I N H
GlDl H O A
HOC
9, C A P
T P . HQ CHI IVIINH N A M HOC 2 0 0 1 -
ZOOZ
Cdu 1: B o tuc v a c a n bkng:
3CaCl2 + 2Na3P04
Cdu 1: Bo tiic ud can bdng cdc phitang trinh phdn ling sau:
CaCh
+?
Ca3(P04)2^
Ba(HC03)2 + Ba(0H)2
+ ?
Ba(HC03)2 + ?
^ BaCOaJ-
+?
CaS03
+?
-> SO. f
+ ?+?
HCl
+ ?
-> NallCOs
+?
FeClo
+ ?
-> FeCh
FeCh
+ ?
FeCU
Cdu 2: Vict cdc pliUang trinh phdn I'tng de bleu dien chuoi bicn lioa sau:
FcS2 ^ SO2 -> SO3 ^ H2SO4 -> SO2
> C a 3 ( P 0 4 ) , i + 6NaCI
> 2 B a C 0 3 i + 2H2O
CaSOa + 2HC1
> S O a t + CaCl2 + H2O
H C l + NaaCOa
> NaHC03 + NaCl
2FeCl2 + CI2
> 2FeCl3
2FeCl3 + F e
> SFeClg
Cdu 2: V i e t phUcfng t r i n h
Na2S03 ^ BaSOa
4FeS2 + IIO2
2Fe203 + 8SO2
Cdu 3:
a) Cho 6 dung dich gom: NaCl, BaCU, CUSO4, NaOH,
KJioiig dung them hoa chat ndo khdc, hay nhan biet
b) Cho 3 dung
dich: BaCU, BafNOi).,
them mot hoa chat, hay nlidn bict
Ba(HC03)2.
MgCh,
2SO2 + O2
AgNO,.
chung.
SO3 + H2O
Chi diCgc si'i dung
cluing.
> CUSO4 + S02t + 2H2O
> Na2S03 + H2O
SO2 + 2 N a O H
rieng CO2.
Cdu 4: A vd B Id 2 loai clidt chi cliiia cdc nguyen to X, Y. Thdn.Ii
pJidn
Na2S03 + BaCl2
tram cua nguyen to X trong A vd B Idn liigt Id 30,4% vd 25,97/.
: i 2SO3
> H2SO4
2H2SO4 + C u
c) Cho hdn hap khi gom CO2, SOo. Bdng pliuang phdp hoa hoc, hay tdch
phdn
^
450*^ C
> BaSOsi + 2 NaCl
Neu cong thi'ic phdn til cua A la XY2, thi cong thiic phdn tii cua B Id gi?
Cdu 5: De gia tang nong do cila 50 gam dung dich CuSO^ 57c len gap luu
Idn, CO boil hoc sinh da thUc Jiien bdng bdn cdch khdc
nJiau:
Hoc sinh D: them 50 gam dung dich CuSO^ 157c vdo dung
Hoi hoc sinJi ndo da lam diing, gidi
nitac.
dich.
Hoc sinh C: them 4,63 gam tinh the CUSO4.5H2O vdo dung
dich.
dich.
thich.
Cdu 6: Hop chat A bi phdn hiiy a nhiet do coo thco phaang trinh plidn dug:
2A
B + 2D + 4E
San phdm tgo thdnh deu a the khi, khoi lugng mol trung blnh cua hon
hgp khi sau phdn ting Id 22,86 (gimol).
N h a n biet CUSO4: mau xanh
Cho mSu CUSO4 tac dung vdi cac mau c5n l a i c6 1 m a u cho ket tua
Hoc sinh A: dun nong dung dich de Idm bay hai phdn ni'ca lugng
Hoc sinh B: them 2,78 gam CuSO^ khan vdo dung
a)-
Tinh khoi lugng mol cua A.
Cho so lieu: H = 1; O = 16; S = 32; Cu = 6"-^.
Hoc sinh c6 the sii dung bdng do tan vd bdng he thong tudn hodn cdc nguyen to
hoa hoc.
x a n h l a m l a N a O H , mot mau cho ket tiia t r S n g l a BaCl2.
L a y mSu N a O H cho tac dung vdi cac mau con l a i : mau cho ket tua
t r S n g l a MgCla, mSu cho ket tua den 1^ AgNOa, mau khong tao ket
tua l a N a C l .
CUSO4 + 2 N a O H
-> C u ( 0 H ) 2 i + Na2S04
CUSO4 + BaCl2 MgCl2 + 2 N a 0 H
AgNOg + N a O H
2AgOH
•> B a S 0 4 i + CuCla
—
-
> Agp
- > Mg(0H)2i + 2NaCl
^ AgOH + NaNO,
+
up
b) -
D u n nong m a u cho k e t tua la Ba(HC03)2
160
mp^so^
them
2,5 + 2 , 9 6 3 2 = 5 , 4 6 3 2 g a m
BaCl2 mSu con l a i la Ba(N03)2.
c) -
niduag
[ C u S 0djch
4 ] %= =5 0- 5+ ^ 4^, 6^3 X
= 100
5 4 , 6=3 g10%
> Ba(N03)2 + 2AgCU
54, 63
Cho qua dung dich thuoc t i m SO2 b i hap t h u , con CO2 bay ra (cho
qua dung dich Br2 chi c6 SO2 b i hap thu).
-
Co the dot chay SO2
™ C u S 0 4 them vAo
SO3 hap thu vao Ba(0H)2 sau do cho H C l
™cuS04
= 7,5 gam
-
dung dich sau =
2,5
+
7,5
-
Cong thufc A: XY2
n i d u n g dich =
-
Cong thufc B: X a Y b
[CuS04]% = 10%) ^ dung
-
Trong A:
-
% X = 30,4%
X
_ 30,4
2Y
69,6
% Y = 100% - 30,4% = 69,6%
_
bY
TCrd), (2)
X
Y
60,8
69,6
_
74,1
60,8
25,9b
69,6
74,1a
X
25,9b
Y
74,1a
Dat
5
50
+
50
100
10
gam
gam
B + 2D + 4E
a
2a (mol)
=
=
2a
4a
Neu lay 2a mol A nhiet phan se tao t h a n h 7a mol k h i .
D i n h luat bao to^n khoi liicfng cho:
M A X 2a = 22,86 x 7a
a _ 2
b
5
=>
M A = 80.
OE SO 9
BE THI HOC SINH GIOI HQA HOC 9, TP HAI PHONG (BANG A) NAM HOC Z001 - Z002
Cong thufc cua B la: X2Y5.
Cdu J.-
>
2A
Cdu 6:
%Y = 74,1%
%X = 25,9%
Trong B:
aX _ 25,9
2,5
.50 = 2,5 gam (n,„so_ = ^
Cdu I
= 0,015625 mol)
1. Cho bang phdn loai cdc chat:
^H,o
= 50 - 2,5 = 47,5 gam
Nong do can dat di/cfc la 10%.
Hoc sinh A lam bay hoi phan nijfa luong niidc
m^^Q
biioai =
m d u n g d . c h sau
= 23,75 = m^^o
Hoc sinh B:
cbn
X
100 = 9,52% ^ 10%
' 5,28 '
52,78 J
3
4
5
HI
NO
CO
O2
Fe
H2SO,
Na20
NO
SO2
H2S
CO2
CH4
6
7
8
Cu(0H)2
CH4
KOH
N2
KOH
C6H22O6
Ba(0H)2
Br2
NaOH
CCI4
NaOH
a ) Nhung dinh sdt da cqo sach gi vao dung dich
sai
Sue khi SO2 vao dung dich
sau:
CuSOj.
Ca(HC0a)2.
^) Dan khi etilen qua dung dich nUac brom.
^' Cho day chuyen hoa sau:
= 50 + 2,78 = 52,78 gam
[CuS04]% =
2
•2. Neu hien tUgng, viet phUpng trinh phdn ling cho cdc thi nghieni
mC U S O 4 = 2,5 + 2,78 = 5,28 gam
mdungdich
1
Hay cho blet vi tri (1), (2), (3), (4)., (5), (6), (7), (8) la cdc ti( gi.
= 50 - 23,75 = 26,25 gam
[CuS04]% = 126,25y
I T^M^rJ
•
diing
- ii_ .50
Hoc sinh D:
vao de giai phong CO2.
Cdu4:
gam
= 2,9632
250
I ^ C u S O . dung dich sau —
H a i mau con l a i cho tac dung vdri dung dich AgNOs. Mau cho i
BaCl2 + 2AgN03
4,63
X
=
^ B a C O g i + CO2T + H2O
Ba(HC03)2
-
Hoc sinh C:
Fe ->A
100 = 10% => dung
-^B
->C-^Fe
->E->-F-^D
dinh A, B, C, D, E, F. Viet cdc phiiang trinh phdn ring.
-.2.
11
I un/t u n r a
39
V Doi c/'idy hodn todn 1,344 lit (dktc) hon hgp 3 hidrocacbon
Cau II.
1. Dung dich Boocdo dung chong nam cho cay dugc pha theo tl le:
1kg CUSO4.5H2O
+ 10kg voi song (CaO) + 100 lit nuac
C H'>n + 2! CinH2m', CkH2k - 2- Sau phdn Ung, dan lion hgp sdn piidm Idn
lugt Q'^'^ H2SO4
fdgc),
dung dich NaOH
(dd^^) ^'^"^ ^'^^
Boocdo.
1 Tinh thdnh phdn % theo the tich hSn hgp 3 hidrocacbon,
2. Tie glucoza
va cdc chat v6 ca can thiet, viet cdc phuang
ling dieu che:
trinh phcn,
etylaxetat.
hidrocacbon
1. Ba khi A, B, C c6 phdn ti2 khoi bdng nhau vd bdng 28 dvC. A, B cu
the hi dot chdy trong khong khi, sdn phdm sinh ra deu c6 khi COo, li
CkH2k-2 trong hon hgp gap 3 Idn the tich
CO so nguyen
cua hidrocacon
cdc phuang trinh phdn iCng.
trong chuang trinh hoa hoc pho thong cap trung hoc ca sd. (P) vd (Ni
CO cung cong thvCc phdn tit.
con Igi.
Cdu I.
(1): axit;
(2): oxit;
(3): oxit khong tao muoi; (4); chat k h i ;
(5): dcfn chat;
(6): baza;
(7): chat hau ccf;
Fe +
a)
(M); klpt (X) = 3klpt (R) - Gklpt (Q)
(8): bazcf kiem.
- Co ket tiia cua. dong xuat hien.
SO2 + Ca(HC03)2
cJio the tich H2 Ian nhdt la 3,36 lit (dktc), chi c6 (R) phdn itng dugc
hay
vai dung dich NaOH. Tit (X) cd the dieu che ra (N), (R), (Q) c6 phdn
thich.
Cdu IV. Hoa tan mudi nitrat cHa mot kim logi hoa tri 2 vao nudc dugc 200 lu'i
dung dich (A). Cho vao dung dich (A) 200 ml dung dich K3PO4, phan iintxay ra vita du, thu dugc ket tua (B) vd dung dich (C). Klioi lugng ket tua (I'
2SO2 + Ca(HC03)2
> Ca(HS03)2 + 2CO2T
(c6 khi bay ra)
c)
'
CH2 = CH2 + Br2
> CH2Br-CH2Br
(mat mau ndu do dung dich nifdc brom)
3. Thiic hien day chuyen hoa:
Pe
y i _ > FeClg
vd khoi lugng mudi nitrat trong dung dich (A) khdc nhau 3,64 gam.
1. Tim nong do mol/lit cua dung dich (A) vd (C), gid thiet the tich dwni
> CaSOgl- + 2C02t + H2O
(cd ket tua va c6. khi)
Ve tinh chat: (M), (N), (R) cd phdn itng vai Na, tic 0,1 mol (M) c6 the
ling vai CI2 (chieu sang). Xdc dinh cong thitc cdu tgo cua (M), (N), (P'.
> FeS04 + C u i
CUSO4
- Dung dich mau xanh bi nhat dan.
b)
kien.
(Q), (R) vd cong thUc phdn tii cua (X). Gidi
tit cacbon
L d l GIAI
2,3 gam (N) hay 1,5 gam (Q) c6 the tinh bdng the tich cua 1,6 gam 0^
-
so nguyen
hidrocacbon
2. Viet phan ijfng:
Ve khoi lugng phdn tit (klpt):
cung dieu
CJl2„+2-
1. Ten goi cac v i t r i :
2. (M), (N), (P), (Q), (R), (X) Id nhUng hap chat hUu ca dugc biet den
klpt (N) = -klpt
2
biet the tich
biet rhng cd 2
tit cacbon bdng nhau vd bdng
CO the kill} dugc CuO a nhiet do coo, C la thdnh phdn quan trgiii:
trong phdn bon hoa hoc. Xdc dinh cong thilc phdn ti2 cua A, B, C, vii')
H2SO4
khoi lugng dung dich NaOH tang 7,04 gam.
2 Xdc dinh cong thiic phdn ti2 3 hidrocacbon,
Cdu III.
-
(du) thdy khoi lugng
Hay tinh thdnh phdn % theo khoi lUgng cdc chat c6 trong dung die/,
Viet cdc phiiang trinh phdn ling.
the khi:
:
> Fe(0H)3
Fe(0H)2
FeCl2
Fe203
> FeS04
Fe
> FeCl2.
Phan ling:
dich khong thay doi do pha trgn vd the tich ket tua khong ddng ke.
2. Cho dung dich NaOH (lay du) vao 100 ml dung dich (A) thu dugc kH
tua (D), Igc lay ket tua (D) roi dem nung den khoi lugng khong do'
1) 2Fe + 3CI2
2FeCl3
2) FeClg + 3NaOH
> Fe(OH)34 + 3NaCl
can dugc 2,4 gam chat rdn. Xdc dinli kim logi trong mudi nitrat.
40
1 ^ 1 niAi np T M i u n r c i u u mni un/t unr Q
nau do
THI HQC SINH Gini
HfiA
H n n
Q
"
41
Fe + 2HC1
•
> 2Fe + 3CO2T
4) FeaOg + SCO
5)
Cac p h a n ufng d i e u che:
-> FeaOa + SHgO
3 ) 2Fe(OH)3
.
FeCl2 + H2T
-
6) FeCl2 + 2NaOH
•
> F e ( 0 H ) 2 i + 2NaCl
tvang
7) Fe(0H)2 + H2SO4
xauh
m
> FeS04 + 2H2O
8) FeS04 + BaCla
10*
10
T h e o de: ncao = — x 1000 = - — (mol); n„
56
56
=
1000
10^
250
250
- C l a t h a n h p h a n q u a n t r o n g t r o n g p h a n b o n => (C): N2.
- ( A ) k h i c h ^ y tao CO2 ( M A = 28) => (A): C2H4: ( e t y l e n ) .
C"(o")2
250
(1)
n
m
250
X
56
250
10*
10^^
56
CalOH)^ d i i
100 = 0,49%
%m„
250
12918
Ca(OH).^dLJ
^
=
1 0 0 %
2,3
0,05
1,5
0, 05
= 46
(gam)
= 3 0 (gam)
7 4 *
(0,353%
3,36 l i t H2
3,36
22,4
= 0,15
(mol)
= 2 : 3 => ( M ) l a ri/gu 3 I a n r u g u (ri/gu 3 chiirc)
=^ ( M ) : CH„ - C H - C H ,
I
I
IOH
OH OH
(mol)
X
mol M + N a
> nM :
12918
P CO cong thufc C2H6O n h i f n g k h o n g p h a n ufng v d i N a v a N a O H .
(gam)
=^ (P): C H 3 - O - C H 3 (ete).
R viia p h a n ufng v d i N a v a viTa p h a n ufng v d i N a O H .
X
1 0 0 =
=> C o n g thufc (R): CH3COOH ( M R = 60g)
1 1 , 6 4 %
1 1 1 x 1 0 0 0
-
(mol)
T h e o de: p h a n ttf k h o i cua ( N ) = - k h o i liJgng p h a n t i i cua M
2
p h a n tijf k h o i ciia ( M ) l a : 2.46 = 92 (gam)
0,1
- 121 _ 1^
%m
p h a n ttf k h o i cua ( N ) :
= 0,05
(gam)
1 1 1 x 1 0 0 0
Ca(OH)^ d.r -
Do do:
32
M a t k h a c , ( N ) tac d u n g vdi N a n e n ( N ) l a : C2H5OH.
•
544
CaS04
n CO,,
( Q ) tac d u n g v d i CI2 chieu s a n g n e n ( Q ) l a : C2H6.
136 = 544
" 250
-> Cu + C O 2 T .
p h a n ttjf k h o i cua ( Q ) :
(mol)
10^
1,6
+ 2H2O
-> 2CO9T
T a c6:
(gam)
100 = 0,3539fc.
X1000
111
10'
CaSO,
%m
Tir(l)
0,392.10'^
Cu(OH),
2.
(mol)
X 98 = 0,392.10^
250
n CaSO,,
m
200,1
CO + CuO
> C u ( 0 H ) 2 i + CaS04
_ 10^
axetat)
III.
> Ca(0H)2
•^4
TCrd)
Cdu
(mol)
K h o i iLTcfng d u n g d i c h Boocdo t h u dirge l a : 100 + 1 + 10 = 111 ( k g )
.
> 2C2H5OH + 2C02t
len men giam
C2H5OH + O2
> CH3COOH + H2O
^ ^ ^
dam dac
C H 3 C O O H + C2H5OH
^ \
= =
^ CH3COOC2H5 + H2O
2 C 0 + O2 -
C a ( 0 H ) 2 + CUSO4
%m
riiau
P h a n ufng: C2H4 + 3O2
P h a n ijfng: CaO + H2O
m
len men
(etyl
(mol)
18
„
1. - A , B , C deu CO M = 28 dvC.
- B CO the k l i L f dagc CuO d n h i e t do cao v a k h i chay tao CO2 => (B): CO.
> FeCl2 + B a S 0 4 i
10^
,y
L6ni2*-'6
V a p h a n tuf k h o i cua (X) l a : M x = 3 M R = 3 X 6 0 = 180
+ 0,49%
+
11,64%)
=
87,517%.
(gam)
v a ( X ) d i e u che r a C2H5OH v a C H 3 C O O H => (X): C6H12OC (glucozcf).
Cdu
rv.
1. P h a n l i n g : 3 M ( N 0 3 ) 2 + 2K3PO4
> M 3 ( P 0 4 ) 2 i + 6KNO3
(li
Ta co:
B i n h d i i n g H2SO4 t a n g c h i n h la k h o i l u g n g cua I-I2O.
(Vdi M la k i m loai hoa t r i I I )
TU
m
p h a n ufng (1): Su k h a c n h a u ve kho'i l i j g n g la do t h a y 6 N ( )
(1)
ri ,
>
= —
372-190
M3(PO.,)2
^M(N03)^
=
^''M,(^O,).^
-
=
=
3
X 0,02
=
0,06
lirgng b a n g 2 I a n so m o l cua A .
(mol)
'KNOS
HA
= 0,3M
0,02
0,2
+
X
6
0,2
M(N03)2 + 2NaOH
(mol)
> M(0H)2l + 2NaN03
0,03
—>
0,03
Tii (2), (3)
=:>
nc = 0,03 ( m o l )
=>
nB = 0,06 - 0,04 = 0,02 ( m o l )
(2)
%VA =
Vay:
0,03
M(0H)2
= 0,01 ( m o l )
=
Vi la chat k h i nen % V = % n
2 . P h a n iJfng:
(mol)
M O + H2O
%VB
,
(3)
X 100 = 16,67%;
0,06
0,02
=
0, 06
X 100 = 33,33%
%Vc = 50%.
. T i m C T P T cua 3 h i d r o c a c b o n :
0,03
TCr (1), (2), (3) :=> ^ n^o^ = 0,01n + 0 , 0 2 m + 0,03k = 0,16
muo = 0,03 x ( M + 16) = 2,4
=> M = 64; D o n g (Cu).
n + 2 m + 3 k = 16
Cdu V.
T a c6: n hon hep 3 khi =
V i m , n , k la n g u y e n v a chSn n e n n g h i e m h g p l i l a : n = 2; m = 4; k = 2
^
" ^'^^
3 h i d r o c a c b o n c a n t i m l a : ( A ) : C2H6; (B): C4H8 v a (C): C2H2
1. T i n h p h a n t r a m the" t i c h m o i hidrocacbon.
DE SO 10
Cac p h a n ufng:
C„H2n.2 +
O2
\
'—^
CkH2k-2 + [ ^ ^ J
nC02 + ( n + D H g O
flE THI CHQN HOC SINH GIQ! HOA HOC 9, TJNH BJNH OjNH NAM HOC 2001 - 2002
(D
J
a
C„,H2:n + —
2
an
2.
(21
Sue a (mol)
Tinh
^
kC02 + ( k - D H s O
(3,*
3a
.
3ak
3a(k - 1)
bleu dien
® " 3. iV/ue^ phdn
ling xong
KNO3, BaSO^.
COo vdo dung
so mol CaCOs
Ve duang
O2
CO2 vd HoO, hay trinh
^ chdt rdn: K2CO3, BaCOj,
a ( n + 1)
mCOs + m l l a O
O2
1- Chi dugc dung
phdn
(mol)
(*)
Do la c h a t k h i n e n n , m , k < 4 va m , k > 2.
D a t : A : C„H2n+2 ; B : CniH2,n ; C: CkH2k;-2-
(mol)
= 0,16 ( m o l )
44
Theo de: nc = 3nA, t h i so m o l H2O g i a m v a n h o h a n so' m o l CO2 m o t
0,16-0,14
Vay:
7,04
= 7,04 g a m => n CO,
m.
0,02 ( m o l )
=
= 0,14 ( m o l )
18
B i n h d u n g N a O H t a n g c h i n h la k h o i l i f g n g ciia CO2.
( M = 372) b a n g 2PO4 ( M = 190).
^
2,52
= 2,52 g a m => n H . , 0
H,0
tgo thdnh
m 1 gam
ta thu dugc
I
Viet phdn
biet 4 Ig
iCng de niinh
chila
hga.
chi'ia 1 mol CafOWs.
-
ting vai gid tri a = 0; a = 1; a ^ 2.
so mol CaCOs
tgo thdnh
hSn hgp Mg, MgCOs
7712 ga77i
lugng Mg t7-o7ig hdn hap
: THI
dich
bay each phdn
ddu.
theo so mol CO2 da
ngodi
mot chat ra'n. Biet
khong
cho.
khi den
7711 = 77x2.
Tinh
khi
%
If) Chon 3 diem:
Cdu 4. Hoa tan hon hap Na20, NaHCOs, BaCl2, NH4CI c6 cung so mol vaa
nuac du, dun nong nlie thu dugc dung dich A vd ket tiia BaCOs- Hoi
dung dich A chiia gi? Viet phdn iCng ininh hga.
Cdu 5. Trgn 11,2 gam bgt Fe
khong CO khong khi de phdn
chat rdn tim dugc trong chen
HCl IM, thodt ra a (mol) hon
vd 4 gam bgt S trong chen sii dem ?iuiig
iCng xdy ra tao FeS vai hieu sudt 80%. Lay'
sii cho tdc dung viia du vai V lit dung dich
hap khi vd m (gam) chat rdn khong tan.
a) Viet tat cd phdn ling xdy ra.
C02
n C02
^' "^CaCO.j
= 0
•'
'•'CaCO;,
= 1
~ ^
^CaCOg
= 0
"
C&u3.
Phan ufng xay ra:
b) Ttnh gid tri V, a, m.
6. Dem hSn hgp gom
Cdu
lugng H2SO4
d4c,
nong
Mg+
0,1 mol Mg vd 0,2 mol Al tdc dung
viia du thu dugc hon hgp mudi,
0,075 mol S va
MgCOg
thdnh.
a
b
1
Lay moi lo mot it cho vao 4 co'c.
-
Che niidfc vao 4 co'c, phan diioc 2 nhom: nhom (I) t a n : dung dich
m •Mg
m^MgCOg
.
11.24 _ 264
4.84
NaaO + H2O
Sue k h i CO2 vao 2 coc nhom ( I I ) :
NaliCOg + NaOH
NH4CI + NaOH
Lay i t dung dich Ba(HC03)2 nho vao 2 coc cua nhom (I).
Neu coc nao tao ra ket tua trSng do la coc chufa K2CO3, phan ufng tao
raBaCOg.
Cdu
> NagCOg + H2O
-> BaCOgi + 2NaCl
-
NaCl + H2O + N H g t
(1)
(2)
(3)
(4)
'°
FeS
(1)
Vi CO hieu suat 80% nen chat r a n gom: FeS, Fe di/, S diT
2.
FeS + 2HC1
a) T i n h so mol CaCOg: K h i a = 0 khong c6 phan ufng, so mol CaCOg = 0.
K h i a = 1 phan ufng xay ra:
> CaCOg + H2O
(D
=> So mol CaCOg = 1.
Fe + 2HC1
> FeCl2 + HaSt
> FeCl2 + H2t
(2)
(3)
^ ) Theo (1), neu hieu suat bkng 100% t h i S het, Fe dif.
K h i hieu suat 80% t h i : S d i i = 0,8 gam = a; Sd mol S phan ufng = Sd
oiol Fe phan ufng = So mol FeS sinh ra = 0,1 mol.
K h i a = 2 phan ufng xay ra:
2CO2 + Ca(0H)2
Klio'i liigng M g = 44%
Vi so mol cua 4 chat: NagO, BaClg, NaHCOg, NH4CI bSng nhau, nen
theo (1), (2), (3), (4) dung dich A chi chila NaCl.
5.
a) Phan ufng: Fe + S
> BaCOg + 2KHCO3
Coc khong ket tiia chufa KNOg.
CO2 + Ca(0H)2
~ 336
2NaOH
BaCl2 + NaaCOa -
> Ba(HC03)2
Coc khong tan chufa BaS04.
K2CO3 + Ba(HC0g)2
(2)
Cac phan ufng xay ra:
Neu tan la coc chtTa BaCOs, phan tfng xay ra tao ra dung dich BadlCOa)^:
Cdu
11
K2CO3, KNO3; nhom (II) khong tan; BaCOg, BaS04.
CO2 + H2O + BaCOg
MgO + C02t
Can 4.
-
-
—
Vi m i = m2 nen khoi lirgng O2 (1) = khoi liigng CO2 (2) =^ 16a = 44b
LCil GIAI
-
(1)
>oi a, b Ian iLTcft la so mol cua M g va MgCOg trong hSn hgp dau.
h) Tinh sd mol H2SO4 phdn iJCng viia du.
Cdu
MgO
vdi mot
0,175 mol SO2.
a) Tinh khdi lugng hdn hgp mudi tgo
^02
V a y s d m o l Fe = 0,1 mol.
> Ca(HC03)2
(2'
(2) va (3)
^
Sd mol HgS = Sd mol FeS = 0,1 mol.
=> So mol CaCOg = 0.
Sd mol H2 = Sd mol Fe = 0,1 mol =:> a = 0,2 mol, sd mol H C I dung la
Triicfng hop 2 c6 the viet phan ufng hoa tan het CaCOg.
0.4 mol
V = 0,4 l i t .
Co 6 to mat nhan diCng cdc dung dich khong
Cdu 6.
mdu Id: iVa^SOj (1);
MasCOs (2); BaCls (3); BafNOaJs (4); AgNOa (5); MgCls (6). Bdng
a) T i n h khS'i Itfgng hon hcfp muol:
phuang
CiJ 0,1 mol M g tao ra 0,1 mol M g S 0 4 (bao toan k h o i lirgng)
plidp hoa hoc vd khong dung them cdc hoa chat khdc hay trinh bay each
Cur 0,2 mol A l tao ra 0,1 mol Al2(S04)3 (bao toan k h o i lifgng)
nhan biet cdc dung dich tren, biet rdng chung deu c6 nong do du Ian de
Vay k h o i li/crng hon hgp muoi = (120.0,1 + 342.0,1) = 46,2 gam
cdc ket tua it tan cUng c6 the tao thdnh. (Khong
b) T i n h so mol H2SO4 dac n6ng da dung vCra du.
phdn
Phan ufng xay ra:
^°<^ dinh nong do cua cdc muoi NaHCOa vd Na2C03 trong mot
> 3MgS04 + S i + 4 H 2 O
(1)
dung dich
M g + 2H2SO4
> MgS04 + S 0 2 t + 2H2O
(2)
nghieni nhu sau:
2A1 + 4H2SO4
> Al2(S04)3 + S i + 4H2O
(3)
Thi nghiem
2A1 + 6 H 2 S O 4
> Al2(S04)3 + 3 S 0 2 t + 6 H 2 O
(4)
dich HCl IM (du) dun nong hdn hgp, sau dd trung hoa lugng axit du
Theo (1) va (3)
^
^H^SO,
dung
= ^.ng = (0,075 x 4 ) mol = 0,3 (mol)
Theo (2) va (4) => n^^^gg^ ^^^^ = 2n^^^ = (0,175 x 2) mol = 0,35 (mol)
Vay so' mol H2SO4 da dung viTa du 1^ 0,65 (mol).
D E S 0 11
D E THI HOC SINH Gidl HOA HOC 9 (VONG 1), TJNH KHANH HOA N A M HOC 2001 - 2 0 0 2
Cdu I.
hdn hgp cua chung
(dung
dich A), nguai
1: Lay 25 ml dung dich A cho tdc dung
bdng lugng vita du la 14 ml dung dich NaOH
ta lam cdc thi
vai 100 nd dung
2M.
Thi nghiem 2: Lai lay 25 ml dung dich A, cho tdc dung vai lilang du dung
dich BaCl2. Loc bo ket tua mdi too thdnh, thu lay nUdc Igc vd nuac nla gap
Igi roi cho tdc dung vai lugng vica du la 26 ml dung dich HCl IM.
1. Viet cdc phuang
trinh phdn ling xay ra vd gidi thich vdn tdt.
2. Tinh nong do mol cua moi muoi trong dung dich A.
Cdu 4. Mot dung dich axit axetic CH3COOH
cd C% = 10%. Lay 300 gam
dung dich axit nay cho tdc dung vai 300 ml dung dich NaOH 2M tao ra
1. Hay thuc hien
phep tinh):
cdc phep tinh sau (yeu cdu ghi day du dan vi troii;;
a) Tinh so mol do trong 7,19 gam do;
b) Tinh
trinh
ling).
Cdu 3-
3Mg + 4H2SO4
can viet phuang
so mol O vd O2 trong
8 gam oxi;
dung dich A. Dung dich A cd tinh axit hay baza?
Tinh nong dd % cdc chat tan trong dung dich A, biet rdng dung
NaOH2M
cd d = 1,2
dich
giml.
c) Tinh khoi lugng cua 0,05 mol kem;
Ghi chu: Hoc sink dugc phep svt dung bdng tudn hodn, bdng tinh tan, gido vien
d) Tinh khoi lugng cua 0,75 mol nudc;
coi tin khong gidi thich gi them.
d) Tinh so nguyen
g) Tinh so phdn tvC CO2 trong 1,1 gam khi CO2;
h) Tinh so g cua 1 nguyen tii Na;
i) Tinh so g cua 1 phdn tii SO2.
2. Mot hdn hgp X gom FeO vd Fe203
hon hgp nay trong
LCll GIAI
tii C trong 0,02 gam C;
cd khoi lugng Id 30,4 gam. Nun.^
mot binh kin cd chica 22,4 lit CO (dktc). Kho'
lugng hon hgp khi thu dugc sau khi nung la 36 gam.
a) Hay xdc dinh thdnh phdn hon hgp khi. Biet rdng hdn hgp X b\
khv[ hodn toan thdnh Fe.
b) Tinh khoi lugng Fe thu dugc vd khoi lugng cua mSi oxit sdt trong X.
Cdu 1
7
1
1. a) nci = —-— (mol)
35,5
'
8
8
b) no = — (mol); n^^ = — (mol)
c ) mzn = 0,05 X 65,38 = 3,27 gam
d ) mj^^Q = 0,75 mol x 18 g/mol = 13,5 gam
iSlSi^li^ THI HOC SINH G|6'HOA
HQC 9
d ) nc =
Tii bang t r e n t a thay:
= 1,66.10^^ (mol)
D u n g dich n a o cho v a o t a o r a 4 I a n ket t u a l a d u n g dich Na2C03 va
So n g u y e n tuf C = 6,02.10^^ n g u y e n ti:f/mol.l,66.10^ m o l
• AgNOa (cap d u n g dich 1). D u n g dich n a o cho v a o t a o r a 3 I a n k e t t u a
= 9,993.10^° n g u y e n tuf.
g) n
la dung dich Na2S04 v a BaCla (cap dung dich 2). D u n g dich n a o cho
1,1
= 0,025 (mol)
44
CO2
vao t a o r a 2 I a n k e t t i i a l a dung dich M g C l s v a Ba(N03)2 (cap dung
dich 3).
So p h a n td CO2 = 0,025 x 6 , 0 2 . 1 0 ' ' = 1,505.10^^ p h a n tit.
h ) MNa = 23 g/mol
2,
niNa.-
MgQ = 64,06 g/mol =>
a)
Ta
nco
( b a n ddu) =
=
1 mol
=>
64,06
-,23
6,02.10'
mco
+) L a y m o t trong h a i chat a c&p dung dich 3 I a n lirgt cho v a o 2 dung
= 3,82.10"^^ g/nguyen t t f
6,02.10'23
i)
c6:
23
dich 6 cap 2, n e u c6 t a o r a k e t t i i a : thi chat cho v a o l a Ba(N03)2, con
l a i l a MgCl2.
C h a t t a o r a ket t u a a cap 2 l a Na2S04, c o n l a i l a BaCl2.
= 1 0 , 6 4 . 1 0 ' " ' g/nguyen tuf
+) L a y Ba(N03)2 d a tim diTOc a cap 3 cho vao h a i d u n g dich d cap 1,
( b a n d s u j = 28
neu CO ket t u a t h i : C h a t tao r a ket tiia vcfi Ba(N03)2 l a Na2C03 con
gam.
Do t a n g k h d i liiOng: 36 - 28 = 8 gam = mo => n^^ = 0,5 mol.
lai l a AgNOa.
V a y CO 0,5 mol C O ket h o p v d i 0,5 mol O cho r a 0,5 mol CO2.
Cdu 3
T h ^ n h p h a n h o n hap k h i l a : 0,5 mol C O v a 0,5 mol CO2.
Bat so mol Na2C03 v a NaHCOa trong 25 m l dung dich A I a n liTot l a x, y.
b ) T a c6: n^^ (c6 t r o n g FeO v a Fe203) = n^^ l a y r a ( d t r e n ) = 0,5 m o l .
Doi v(5i t h i nghiem 1, t a c6:
Na2C0a + 2HC1
G o i a = npeo v a b = np^^p^. V a y t r o n g X c6 (a + 3 b ) m o l O.
D o do:
a + 3 b = 0,5
(1)
Va:
72a + 160b = 30,4
(2)
G i a i (1) v a (2) dirge: a = 0,2 m o l FeO v a b = 0 , 1 m o l FeaOg.
Vay:
mpeo = 0,2 x 72 = 14,4 g a m
"^Fe.,03
= 0,1
X 160
=
16
(mol)
1^
( t h u dLToc)
nipe
( t h u duoc)
x
k
> N a C l + H2O
(3)
0 , 1 x 1 = 0 , 1 mol
So mol H C l da tac dung vdri dung dich A l a :
(4)
Boi vdfi t h i nghiem 2:
Cdu 2
BaCl2 + NaaCOa
1. L a y m o t d u n g d i c h b a t k i cho v a o 5 d u n g d i c h c o n l a i , t a c6 b a n g sau:
50
y
2x + y = 0,1 - 0,028 = 0,072
56 = 22,4 g a m .
Na2S04
Na2C03
BaCl2
Ba(N03)2
AgNOa
MgCl2
Na2S04
-
-
i
i
i
Na2C03
-
-
i
i
i
-
BaClz
i
i
-
-
i
-
Ba(N03)2
i
i
-
i
i
-
-
AgNOa
-
MgCl2
-
i
-
i
-
-
(2)
So mol H C l dir sau p h a n ufng (1) v a (2) l a : 0,014 x 2 = 0,028 mol.
= npe ( t r o n g FeO) + npe ( t r o n g Fe203)
= 0,4
> N a C l + CO.T + H2O
So mol H C l trong 100 m l dung dich l a :
= a + 2b = 0,4 mol.
Vay:
y
H C l (dii) + N a O H
H||
T h e o d i n h l u a t bao t o a n n g u y e n t o t h i :
npe
(mol)
(1)
2x
NaHCOa + H C l
I
gam.
X
> 2 N a C l + COgt + H2O
i
LOI GIAI at THI HOC SINH GIQI HOA HOC 9
> BaCOai + 2NaCl
(5)
Sau k h i loc bo ket tua, lay niidfc loc, ni/dc riia chufa N a H C O a cho tac dung
vefi dung dich H C l .
||^|^'
^
•
NaHCOa + H C l
(mol)
y
> N a C l + CO2 + H2O
y
^ i a i r a t a c6: y = 0,026 x 1,0 = 0,026 mol
x = 0,023 mol
nong do mol cua NaHCOa l a :
0,023 : 0,025 = 0 , 9 2 M .
nong do mol cua N a H C O g l a :
0,026 : 0,025 = 1,04M.
i J ^ i i l i l n f ....
(6)
Cdu
IL
4
300 g a m C H 3 C O O H
^
300 m l d u n g d i c h N a O H 2 M chufa
nNaOH
P h a n l i n g : CH3COOH + N a O H
0,5
= 0,3
x
icng dieu che
CaCOs.
vdi Mg thi khi dot chdy hap kim do, oxlt tqo nen c6 khoi
2 = 0,6 m o l .
doi khoi lugng
> C H a C O O N a + H2O
0,5
tri/ih phan
Cho 4 kim loai sau: Be, Na, Al, Ca. Hoi kim loai ndo khi tqo hap kim
\
So m o l C H 3 C O O H = 0,5mol.
(mol)
5 phuang
a) Viet
V
i
1 0 % chOra 30 g a m C H 3 C O O H
Cho
hgp kim ban ddu.
Kim
T i le p h a n lifng l a 1 : 1.
loai
Hda
V a y sau p h a n ufng con N a O H = 0,6 - 0,5 = 0 , 1 ( m o l )
tri.
Khoi
lugng
nguyen ti'i
Be
Na
Al
Ca
H
I
HI
H
9
23
27
40
d u n g d i c h A c6 t i n h bazcf.
c) Dot chdy
phdm
m ( d u n g d i c h A ) = m (dung d i c h CH3COOH) + m ( d u n g d i c h N a O l I
= 300
C% ( C H g C O O N a ) =
+ Vd,.>gdkh
0,5
X
82
X
X d = 300
100%
660
X 1,2
= 660
du
gam
Tinh
= 6,21%
Cdu
660
0,6%.
todn
tqo thdnh
khoi lugng
HI. Dot chdy
dich
5,6 lit (dktc)
dich
NaOH
inol trung
2M de tqo thdnh
cua mpt kim loqi
chi cd hai hda' tri 2 vd 3) trong
tqo nen c6 nong
Idng
Lam Iqnh
xud'ng
vd dung
muoi Id 23%. Xdc dinh cong thUc cua tinh the
DE THI HOC SINH GIQI HOA HOC 9, CAP TP. HO CHJ MINH NAM HOC 2002 - 2003
Cdu
muoi
trung
hda.
MeS (kim loqi
rV. Cho m gam hon hgp CaCOs
Me
oxi du. Hda tan chdt
viia du dung dich H2SO4 29,4%.
dp 34,5%.
the hidrat
khi nay tdc dung viCa
binh ciia hon hgp khi ban ddu.
ilng bdng mpt lugng
2,9 gam tinh
hdn hgp khi gom H2 vd H2S. sdn
la nuac vd mpt chdt khi. Lugng
1,76 gam sunfua
cdc hgp chat
rdn sau phan
DE SO 12
Cdu
hodn
vdi 100 ml dung
trong
= 2 i i L f ^ ,
«(NaOH)
+ 300
gap
biet:
0,5
T i n h k h o i lufOng d u n g d i c h A :
lugng
dung
dich
dich
Dung
thi thu
dugc
con Iqi cd nong
dp
hidrat.
vd FeS tdc dung
vUa du vdi V ml
dung dich HCl (D ^ 1,1 g/ml) thi thu dugc a lit hon hgp klii X (dktc) c6
I.
a) Cho biet dp tan mot loai muoi
neu lam Iqnh
thi phqm
nhiet
dung
c6 nong
dp % ciia muoi
dp theo bang sau
do trong
khodni^
do ndo?
dp
Dp tan (trong
b) Co 6 dung
lOOg
dich
nicac)
dugc ddnh
mpt chdt gSm: BaCh,
n,3g
10° C
20°C
30°C
40°C
15,2g
18,9g
24,8g
36,7g
so ngdu nhien
H2SO4,
luat thuc hien cdc thi nghiem
NaOH,
tic 1 den 6, moi dung c/?'
HCl, MgCl.,
NazCOa.
i
nghiem
1: Dung
dich (2) cho ket tua vai cdc dung dich (3) vd (4'i'
Thi
nghiem
2: Dung
dich (6) cho ket tua vai cdc dung dich (1) vd ^4'
Thi
nghiem
dung
dich
(4) cho khi bay len khi tdc dung
dich (3) vd (5).
Hay xdc dinh s6 cua cdc dung
dich.
vai
mol trung
binh
Id 40,67
(gam/mol)
vd dung
dich
Y cd khoi
ugng la b gam.
<.) Tinh
a theo m, V, b.
b) Ap dung:
Cho m = 1,44 gam; V = 400 ml; b = 440,83
mpt dU kien
^^0,67 (gam/mol),
La>-
ud dugc ket qua sau:
lugng
c) Neu chi sii dung
Thi
3: Dung
khoi
do Id 22% tie 50°Cl
vi de bdt ddu xudt Men ket tinh cua muoi
Nhiet
chiia
dich
bii'n doi theo nhiet
Cdu
khoi
hay tinh % theo khoi lugng
V. Mpt hon hgp A gom Na2C03
(iich
HCl du thu dugc
56 (gam/mol).
^"(0H)2du
Tinh
vd NaoSOs
hon hgp khi X cd klidi
C/io 0,224 tit (dktc)
•Sau thi nghiem
lugng
phdi
diing
gam. Tinh
mol trung
dich
ciia X la
cua hdn hgp ban ddu.
cho tdc dung
lugng
vdi
mol trung
khi X di qua 1 lit dung
50 ml dung
binh
a.
dich
dung
binh Id
Ba(0H)2-
HCl 0,2M de trung
hda
'
% so mol mdi khi trong
X vd % khoi
lugng
moi chdt
trong
hon
hgp A.
Tinh
nong
dp mol/lit
''^'10 sd lieu: H= 1,0=
VTHI HOC
S I N H Rini HOA
HOC
cua dung dich Ba(0H)2
ban ddu.
16, Na = 23. S = .32, CI = 35,5, Ca = 40, Fe = 56, Ba = 137
9
53