Phân Dạng Và Phươg Pháp Giải Bài Tập Hoá Học 11 Phần Vô Cơ Phần 3

hSn dgng vi phaong ph^p
gii\a hpc 11 VP co - D5 Xuan Hung
;:au 13. Cho quy tim h1n
lu'cJt
vao tiTng mau thuT:
Mau
lam quy hoa xanh la K2CO3 va NaOH
MSu lam quy hoa do la HNO3 va
(NH4)2S04
Ca(N03)2
khong doi mau quy tim
Cho
Ca(N03)2
vao 2 mau lam quy tim hoa xanh, mau tao kct tua la
K2CO3
mau con lai la NaOH. Lay NaOH cho vao 2 mau lam quy tim hoa do. Man
giai
phong khi c6
miii
khai la
(NH4)2S04,
con lai la HNO3.
Hoac
lay
K2CO3
cho vao 2 mau Ircn, mau c6
giiii
phong khi la HNO3
=> Chon A
;:au 14. Goi cong thtfc ciia muoi nitrat la .M(N03)n
2M(N03)n '—^ MzOn +

2nN02
+
n/202
a a/2 na na/4
Ta CO a(M + 62n) = 9,4 (1)
0,5a(2M+16ii)
= 4 (2)
(M+62n)
9,4
Lay (1): (2) ta du^dc:
M
= 32n.
M
+ 8n 4
Khi
n = 2 thl M = 64.
Vay muoi
nilral
la
Cu(N03)2
=> Chon C
:au 15. Goi X, y la so mol ciia KNO3 va
Cu(N03)2
trong hon hdp
PhiTcfng
trinh
phan uTng xay ra:
2KNO3.—
2x
2Cu(N03)2

2y
-> 2KNO2 + O2
2x X
-i^2Cu0 +
4y
Ta
CO
he phiAJng
trinh:
4NO2 + O2
y
10L2x
+
188.2y
= 95,4
32(x +
y)4-46.4y
x + 5y
=
37,82
Giai
he tim x, y tiT do chpn ke't qua dung. => Chon B
'au 16. Phiin ufng nhict phan:
Pb(N03)2
—^ PbO + 2NO2 +
I/2O2
X 2x x/2
Goi
X la so mol
Pb(N03)2

da nhiet phan:
Khoi
iMng
khi thoat ra =
2x.46
+
0,5x.32
= 66,2 - 55,4 = 10,8
=> X = 0,1 mol
Vay hieu sual phan
iJng
la: H = 50% => Chon D
on
:&u 17. The tich khi thoat ra = 2.5x x 22,4 = 5,6 lit
Chon A
18. Cac phu'cfng tnnh phan tfng xay ra:
2NaN03 —^ 2NaN02 + O2 (1)
2x X
2CuO + 4NO2 + O2 (2)
4y
(3)
2Cu(N03)2
|.2y
4NO2 + O2 +
2H2O
-> 4HNO3
4y y 4y
So mol O2 tao ra d (1) va (2) la: x + y, sau phan
tirng
(3) con lai la x mol

Ta
CO
X =
1,12/22,4
= 0,05 mol
=> khoi
li/dng
NaNOa =
2.0,05.85
= 8,5 g i=> Khoi
lifdng
Cu(N03)2 = 18,8 g
=> Chon A.
&u 19. Phi^dng
trinh
phan
iJng
xay ra:
2N0 + O2 -> 2NO2
Trirdc
pif 3 2 (lit)
Phantfng 3 1,5
Sau phan iJug 0 0,5 3
Hon
hdp sau phan uTng gdm O2 dirO,5 lit, N2 8 lit, NO2 3 lit
The tich hon hdp
lii:
11,5 lit
=> Chpn A.
-Su 20. Xct cac qua

irinh
cho-nhan electron:
Qua
trinh
cho e:
Al-
3e->Ar'
39x 13x
2N^'
+ 8e-> 2N*'
16x 2x
Qua
trinh
nhan e
N^'
+
.
3e ^ N^'
3x x
2N^'
+ lOe^
20x 2x
N2
Pa c6: 5x = 2 mol => x = 0,4 mol
mAi
=
13.0,4.27
= 140,4 g => Chon C.
C3u 21. Phi/dng
trinh

phan (Jng:
3M
+ 4nHN03 -> 3M(N03)„ + nNO +
2nH20
=> Chon C
Cau 22. Lap
lujin,
rut difdc M = 32n M la Cu
=> Chpn C
191
Phan
djing
phuBng ph^p
gijii
H6a hgc 11 VP co - D5 Xuan Himg
Cfiu
23. Cu -> Cu(N03)2 ^
Cu(OH)2
CuO
Ta
c6:
So mol dong ban dau = 3.2/64 = 0,05 mol
=>
ncu =
nc„c)
= 0,05 mol
=>
X = 0,05.80 = 4 gam
=>
Chon A

Cfiu
24. Phifdng
irinh
phan
uTng
xiiy
ra:
3Cu
+ 8HNO,
-)•
3Cu(NO,)2
+ 2NO +
4H2O
X 2 x/3
Cu
+ 4HN0,-> Cu(N03)2 +
2NO2
+
2H2O
y
- 2y
-^ 30 + 2y.46
Ta
c6: x + y = 0,2 mol,
mill
khac
= 38
=>
X = 3y
=>

X = 0.15.y = 0,05 mol
=>
Vkhi
= (2x/3 +
2y).22,4
= 4,48 lit
=>
Chon C
Cau 25.
PhUitiig
liinh
phan u^ng:
2x8Al
+
.30HNO.,
->
8AI(NO.,).i
+
3N2O
+
I5H2O
3xlOAl
+ 36HNO., ^
10A1(NO.,).,
+
3N2
+
I8H2O
46A1
+ 168HNO, ->

46Al(NOj).,
+
6N2O
+
9N2
+
84H2O
Vijy
li Ic
Al:
N:0 : NO = 46 : 6 : 9 => Chon B.
Cau 26. So mol HNO, phan
iJ-ng
=
3 X .so mol
Al
+ so mol NO + 2 x so mol
N2O
+ 2 x so mol
N2
=
3.
13. 0,4 + 0,4 +
2.2.0,4
+
2.2.0,4
= 19,2 mol
=>
The
lich

axil
la 38,4
lil
=> Chon A.
Cau 27.
Tinh
difdc so mol =
0.1
mol
so mol
NO.r
= 0,04 mol va so mol Cu =
0,0375
mol
Phifdng
irinh
phan i?ng dang ion
nil
gon:
3Cu
+
8H^+2NOr-»3Cu'^+2NO
+
4H20
Tri/ckphanifng
0,0375
0,1 0,04
0,0375
0,1 0,04
Xcll

Ic ~ —
3
8 2
Qua li Ig lhay NO,'
diT,
Cu va
H^phiin
iJng hcl
so mol NO =
0,025
mol ^ V = 0,56
lil
=^ Chon D.
Cfiu
28. Ctic phu-cJng
Irinh
phan
I'rng:
3M
+
4nl
INO,,
->
3M(N03)„
+ nNO + 2nH20
a mol a a n/3
2M
+ mH2S04
M2(S04),„
+ mHj

a
mol a/2 am/2
Theo de
ra:
an/3 = am/2 => n : m = 3 : 2.
V4y
kirn
loiii c6 hoa trj lhay doi
a(M+62.3)
159,21
100
,
giai
ra ta CO M = 56 M la Fe
-(2M+
96.2)
2
=^
Chon A.
iu
29.
Tinh
so mol cac chat:
nNaOH=«'7.0,2
= 0,14 (mol)
nH,po4
=0,1.1 =
0,1
(mol)
X^ttllc

-11N20H_
=
0:14^1
4
nH3i'04 O'l
NaOH
+
H3PO4
NaH2P04 +
H2O
2Na0H
+ H.,P04 -> Na2HP04 +
2H2O
=>Chpn B.
Cfiu
30. So mol ciic chat:
44
nNaOH=
—-l'l(mol)
•
tao 2 muo'i
n
39,2
H,P04
=
0,4 (mol)
K
98
Xet
ti Ic:

-^^^^^
= — = 2,75
=>
tao 2 mucn
nH.,P04
2NaOH
+
H3PO4
-> Na2HP04 +
2H2O
2x
X X
3NaOH
+ H.,P04 -> Na,P04 +
3H2O
3y
y y
Ta
CO
2x
+ 3y = 1,1 va X + y = 0,4 X = 0,1 va y = 0,3.
Tinh
khoi
Ming cac muo'i la chon
diTctc
kcl
qua
=>
Chon D
-fiu

31.
Xcl
qua
trinh
Ca3(P04)2
1
mol
2551 mol
.SiO,
,C,1
->2P-
-^PA
-).2H3P04
2 mol
5102 mol
Khoi
li/c.»ng
bol quang la: 2551.310.
100 100 1
90 73 1000
-'
= 1203,6 kg ^ Chon C.
Phan dgng
va
phuong ph^p
giai
H6a hpc
11
VP
cd

-
D8
Xuan Hi^ig
Q£Mng3.
CACBON
-
SnJC
A. T6M TAT Li THUYET
Khai
qu^t ve
nhom cacbon:
Nhom cacbon
gom
:
cacbon
(C),
silic
(Si),
gemani
(Ge),
thiec
(Sn)
(Pb)
deu
thuoc nguyen
to
p,
cau
hinh electron ngoai ciing :
ns^

npl
I.
CACBON
-
Cau
hinh electron
:
ls^2s^2p^
-
Cac so oxi hoa cua
C
: -4; 0; +2 va +4.
-
Cac
dang
thu
hinh
:
kirn
cifdng, than chi, fuleren
* Tinh chat
hoa hoc
:
The hien tinh khuT hoSc tinh
oxi h6a
nhiftig
chu yeu
la
tinh khuT.
1.

Tinh
khur
a)
Tac
dung
vdi
oxi:
C
+
O2
>
CO^
0
„ +2
CO2
+
C 2CO
b)
Tac
dung
vdi hdp
chat:
C +4HN03dac
—^
CO2
+
4NO2
+
2H2O.
2.

Tinh
oxi hoa
a)
Vdi
hidro :
C
+
2H2
'"
>
CH4
0
„ -4
b)
Vdi kim
loai:
3C
+
4A1
—^
AI4C3
*
PhUcmfi
trinh dieu
che :
CH4
'" '
C +
2H2.
II.

H0P CHAT CUA
CACBON
1.
Cacbon
mono
oxit
(CO) : C^O
*
La mot
oxit khong
tao
muoi (oxit trung tinh)
*
CO the
hi?n tinh
khijr:
+2
,„ +4 •
2C0
+O2
—^
2C.O2
+2
„ +4
4C0
+
FejO*
—^
3Fe
+

4CO2
-
* Dieu
che
:
- Trong phong
thi
nghiem
:
HCOOH
>
CO +
H2O
(axit fomic)
- Trong cong nghiep
:
C
+
H2O
,
-'"^""^^
CO +
H2
CO2
+ C
2CO
+4
2.
Cacbon
dioxit

(COg) :
0=C=0
-
La
mot
oxit axit.
-
Tan
trong niTdc dung dich axit cacbonic.
CO2
+
H2O
^
H2CO3
- Tac dung diTdc
vdi
oxit bazd, bazd
->
muo'i cacbonat.
CO2
+
2K0H,
>
K2CO3
+
H2O
-
Tac
dung
vdi

chat khuf manh
:
0
+4 , +2 0
2Mg
+
CO2
—^
2MgO
+ C
* Dieu
che
:
CaCOj
+
2HC1:—>
CaCb
+
C02t
+
H2O.
3.
Axit
cacbonic
(H2CO3) va
mu6li
cacbonat
a) Axit cacbonic
(H2CO3):
^

C=0
H-0
-
La
axit raft yeu,
kem ben.
-
La mot
diaxit,
tao ra hai
loai muoi:
+ muoi cacbonat (COf")
+ muoi hidrocacbonat (HCOJ).
b) Muoi cacbonat:
-
Tac
dung
vdi
axit
->
C02t
NaHC03
+
HCl
>
NaCl
+
CO2T
+
H2O

Na2C03
+
2HC1
>
2NaCl
+
C02t
+
H2O
-
Tac
dung
vdi
dung dich kiem
:
KOH +
KHCO3
>
K2CO3
+ H2O
- Phan iJng nhi$t phan
:
BaC03 BaO
+ C02t
2NaHC03 Na2C03
+
COzt
+
H2O.
III.

SILIC
VA H0P CHAT CUA
SILIC
1.
Silic
(Si = 28)
(
-
Cau
hinh electron :
1
2s^
2p'*
3s^
3pl

Cd
so
oxi hda : -4; 0; +2
vi
+4.
- Cdc dang thO hinh : silic tinh
th^
va
silic
v6
dinh hinh.
J
Phan
danp

vS
phuonfl
ph^p
giai
H6a hgc 11 VP
CO
- B6
Xufln
Hung
0
+4
a) Tinh khuf:
Si +
O2
>
SiOj
Si
+
2NaOH
+
H2O
>
Na^SiOj
+
2H2t
b) Tinh oxi
h6a :
Tac dung vdi kirn loai
->
silixua kim loai.

0
-4
2Mg
+ Si >
Mg2Si
(magie
sihxua)
*
Dieu
che:
- Trong phong thi
nghiem
:
Si02
+
2Mg
2MgO
+ Si
- Trong
cong
nghiep
:
Si02
+ 2C Si +
2CO.
2. H^p
chS't
cua
silic
+4

a)
Silic
dioxit
(SiOg)
-
La
oxit axit:
SiOz
+
2NaOH
NajSiOj
+
HjO
SiOz
+
NazCOj
NajSiOs
+
CO2
-
Si02
tan trong axit flohidric (dCing kh^c thiiy tinh)
Si02
+
4HF
).SiF4 +
2H20.
b) Axit silixic muoi
sihcat:
- Axit silixic

(HzSiOs)
:
chS't
dang
keo, khong tan trong ntfdc,
de
ma't
niTdc
khi
dun
n6ng.
HzSiOj
-A
SiOa
+
H2O
- Muoi
silicat
:
trong dung dich,
silicat
cua
kim loai kiem
bi
thuy
phan
manh
tao ra moi trtfdng kiem.
Na2Si03
+

2H2O2NaOH
+
H2Si03.
B. PHAN
LOAI
VA
PHiTdNG
PHAP
GIAI
CAC
DANG
BAI TAP
Dang
1.
- L^p phi/cfng trihh h6a hoc
-
Viet chudi phan
CTng
-
Di4u
ch§' c6c chfi't
BAI TAP
MAU
VA BAI TAP AP
DUNG
Bai
1.
Lap phtfdng trinh h6a hoc cua
cic
phan

tfng sau dSy
:
a) C +
H2S04d5cCO2
+
SO2+?
b)
C +
HNO3 dj. NO2
+
CO2
+?
c)
C +
CaO
CaCz
+
CO
126
I
d)
C +
SiOj
>
Si
+
CO.
Gidi
a) C +
2H2SO4,,,.

b)
C +
4HN03,^,
c)
3C +
CaO
—^
d)
2C +
SiOz
—^
+4
COj +
2SO2 +2H2O
4N02+
C62 +
2H20
CaC2
+ CO
Si
+
2C0.
Bai
2.
Viet cac phiTdng trinh phan iJng
theo
scJ do sau
:
CaCl2-^
CaS04

^^^CaO
(9)
(7?
(8)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
CO2
Ca3(P04)2
Giai
->
CaCOj
+
2KC1
Ca(HC03)2
CaCl2
+
K2C03
CaCOj
CaO
+
CO2
CaO
+

CO2
>
CaCOs
Ca(HC03)2
CaCOj
+
CO2
+
H2O
CaC03
+
CO2
+
H2O
>
Ca(HC03)2
CaCOj
+
2HC1 >
CaClz
+ C02t +
H2O
3CaC03
+
2H,P04
>
Ca3(P04)2
+
SCOjT
+

3H2O
3CaO
+
2H3P04
> Ca3(P04)2
+
SHjO
Ca(HC03)2
+
2HC1 >
CaCb
+
2CO2
+
2H2O
(10) CaC03
+
H2S04 > CaS04
+
CO2
+
H2O.
B4i3.
a) Lam the nao de chuyen NaHCOj thknh Na2C03,
Ca(HC03)2
thanh CaC03 va
ngi/dc lai.
b) Khi nung ndng kem oxit vdi than
c6c
thi

tao
th^nh mot chat khi chdy di/dc.
Viet phiTdng trinh phan tfng hoa hoc cua phan tfng.
Giai
a)* chuyen NaHCOj thanh Na2C03
ta
cho NaHC03
tic
dung vdi dung dich
NaOH.
NaHC03
+
NaOH >
Na2C03
+
H2O
*
Dl
chuyen
Ca(HC03)2 thanh
CaC03
ta dem
muoi Ca(HC03)2 khan nung
i
nhi^t.
127
Phan
dgng
phJong ph^p giSi H6a hgc 11 va
cO

- D8
Xufln
Hung
Ca(HC03)2
CaCOj +
CO2
+
H2O
* NgiTdc
lai :
chuyen NajCOj thanh NaHCOj
va
CaCOj th^nh
Ca(HC03)2
ta
cho hai muoi Na2C03
va
CaCOs Ian liTcJt vao dung dich
CO2
bao hoa.
Na2C03 +
CO2
+
H2O
>
2NaHC03
CaCOa
+
CO2
+

H2O
>
Ca(HC03)2.
b) ZnO
+ C
Zn
+
CO
2C0
+ 02-^
2CO2.
Bai 4. Hoan thanh can hkng cac phan
uTng
oxi hoa
khuT
sau
:
a)
Ca3(S04)2
+
CI2
+ C >
POCI3
+
CO
+
CaCl2
b) CS2
+ 02^
CO2

+?
c)
FexOy
+ CO FeO +
CO2
d) CuO
+ C ^ ? +
CO2
e)
Ca3(P04)2
+
Si02 +
C
P + CaSiOj + CO
f) Mn04
+C6H|206
+ H^
g) C,2H220n
+
H2S04dac-
> Mn^*
+
CO2
+
a)
Ca3(P04)2
+ 6CL + 6C
-> COzt
+
H2O

+ SOzt.
Gidi
->
2POCI3
+ 6C0 +
3CaCl2
+2
-I
0
b)
CSj +
3O2
-^-^
+2y/x
+2
c)
Fe^O,
+(y-x)CO
+4-2
+4-2
CO2
+
2SO2
+2
+4
xFeO +(y-x) CO^
+2 0
d)
2CuO
+ C

+4
2Cu
+ CO,
e)
Ca3(P04)2
+
3Si02+
5C
0
24MnO;
+
5QH,206
+
72H*•
2P
+
3CaSi03+ 5 CO
24Mn^++
3OCO2
+
66H2O
24
X
5
X
+7
Mn
+5e-
0
6C

+2
->
Mn
+4
-> 6 C
+ 6 X 4e
hay 24KMn04
+
SCfiHuOe
+
36H2SO4
•
-> 24MnS04
+
I2K2SO4
+
+
30CO2t
+
66H2O
g)
C,2H220„
+
24H2SO4
+4
-> 12 CO
J
+35H2O+
24SO2T
1 X

24
X
0
12c —
+6
S
+2e-
+4
->
12c +
12x4e
+4
->
S .
128
ai 5. Bo tuc chuSi phan ufiig, viet cdc phtfdng trinh phan
uTng
(m6i mui ten
1^
mot
phan tfng). Cho biet B Ik khi dting nap cho c^c binh chi?a lufa (dap t^t
luTa).
A
Ik
khodng san phd bien thUcJng dilng
6i
san xuS't voi song.
A,
^'
NaQH

D
HCl
(Trich TS DHQC TP.HCM)
Gidi
A:CaC03
B
:
CO2
C
:
NaHC03
D
:
NazCOj
E:Ca(0H)2
F: CaCb
CaCOj CaO
+
CO2
C02
+
NaOH >NaHC03
C02
+
2NaOH >Na2C03 +
H20
NaHC03
+
NaOH
>

Na2C03 +
H2O
Na2C03
+
HCl
>
NaHC03 + NaCl
NaHC03
+
Ca(0H)2
>
NaOH + CaCOj
+
H2O
NasCOj
+
CaCla
>
CaCOj + 2NaCl.
Bai 6.
Tur
nguyen lieu chinh Ih muoi an, dd voi, ntfdc, khong khi va cha't xuc
tic
c6 du. Viet cac phiTdng trinh dieu che cac cha't sau:
a)
NH4HCO3
b)
(NH4)2C03
c)
NajCOj.

Gidi
a)
2NaCl
+ 2H2O —. > 2NaOH + Hzt + C^t
^
CO
mang
ngSn
^
CaC03 CaO +
CO2
(H6a long khong khi sau do chiftig
cS't
phan doan Ian
luTdt
ISy
N2
roi den
O2).
N2 + 3H2 , • 2NH3
p
CO2 +
NH3
+
H2O
>
NH4HCO3
b)
NH4HCO3
+

NH3 > (NH4)2C03
c)
(NH4)2C03
+
2NaOH
>
NazCOj
+ 2NH3 + 2H2O
hoac
C02
+
2NaOH >Na2C03 +
H20.
Bai
7.
Cho
cic
cha't CaC03, dung dich NaOH, dung dich NaHCOj, dung dich
HCl. Vid't
cdc
phiTdng trmh phan tfng
hoa
hoc
xay ra khi
cho
cic
chS't
tac
dung vdi nhau tilfng d6i mot.
(TNTHPT)

129
Phan
d?ng
va
phuong
ph&p
giai
H6a hpc 11 V6 cd - D8
XuSn
Hung
Gidi
CaCOj + 2HC1 >
CaCb
+
COzt
+ H2O
NaOH + NaHCOj > NaiCOj + H2O
NaOH + HCl > NaCl + H2O
NaHCOj + HCl > NaCl +
COst
+ HjO.
Bai 8. Co cac
chat
sau : CO2,
Na2C03,
C, NaOH, NazSiOj,
HjSiOj.
Hay lap
thanh
mot day

chuyen
hoa
gifl"a
cac
chat
va
vie't
cac phiTcJng
trinh
hoa hpc.
Gidi
Day
sd
do
bien
hoa :
C CO2 N&iCOi NaOH NazSiOj HiSiOj
t
<5>
(1) C + 02-^C02
(2) CO2 + 2Na0H >
NaiCOj
+ H2O
(3)
Na2C03
+
Ba(0H)2
>
BaCOj
+ 2NaOH

(4) 2NaOH + Si02 >
NazSiOj
+ H2O
(5)
NazSiOj
+ 2HC1 > 2NaCl +
H2Si03
(6)
Na2Si03
+ CO2 + H2O >
NazCOj
+
H2Si03.
Bai 9. Nhiet phan mot
lirpng
CaC03,
sau mot
th6i
gian
di/cJc
chS't
r^n A khi
B. Cho khi B hap thu
hoan
toan vao dung dich KOH, thu difdc dung dich D.
Dung dich tic dung dU'dc vdi dung dich
BaCl2
va vdi dung dich NaOH. Cho
r^n
A tac dung vdi dung dich HCl di/, diTdc khi B va dung dich E. Co can

dung dich E difdc muoi khan F. Dipn phan muoi F nong
chay,
di/dc kim loai
M.
Viet phiTdng
trinh
hoa hoc cua cac phan tfng xay ra.
Giai
KHCO,
A
: CaO; CaC03
duf
B : CO2 Dung dich D
[K2CO3
Muoi
khan F :
CaClj
Dung dich E : CaCl2 M : Ca.
CaCOj
CaO + CO2
CO2 + KOH
)-KHC03
CO2 + 2K0H —y K2CO3 + H2O
Ca0 + 2HC1 >CaCl2 + H20
CaC03
+ 2HCl >
CaClz
+
COzt
+ H2O

K2CO3 +
BaClz
>
BaCOjJ-
+ 2KC1
2KHGO3 + 2NaOH
>
K2CO3
+ Na2C03 + 2H2O
CaClj
-^2SL>
Ca + Cbt.
'Bai 10. Viet cdc
phiTcfng
trinh
h6a hoc
theo
cdc set do sau :
a) Silic
dioxit
>
natri
silica^—-+^aA^^
dioxit
magie
silixua < silic
b) CO2 > CaC03 > Ca(HC03)2
>
CO2 > C > CO
>•

CO2.
Gidi
a)
Si02
>-Na2Si03 •
HzSiOs
>•
Si02
Mg2Si < Si
Si02
+
2NaOH
> NazSiOj + H2O
Na2Si03
+ CO2 + H2O >
Na2C03
+ H2Si03
H2Si03
-A
Si02
+ H2O
Si02
+ 2C Si + 2C0
Si
+ O2 SiOz
Si
+ 2Mg Mg2Si
Si
+ 2NaOH + H2O >
Na2Si03

+
2H2t.
b) CO2 + Ca(OH)2 >
CaC03
+ H2O
CaC03
+ CO2 + H2O >Ca(HC03)2
Ca(HC03)2
CaC03
+ CO2 + H2O
CO2 + 2Mg 2MgO + C
C
+ CO2 2C0
CO + O2 CO2.
Bai 11.
a) Vie't
phuTdng
trinh hoa hoc cua phan
iJng
m6 ta thuy tinh bi
axit
HF an m6n.
Biet r!ing thanh phan chu yeu cua thiiy tinh Ik
NaiSiOj
(Na20.Si02)
CaSi03 (CaO.Si02).
b)
Tilf
Si02 vk c&c hda
cha't

can
thiet
khac hay vi^t phifdng trinh hda hpc cia
cic phan
urng
dieu
ch^ axit silixic.
Gidi
a) Na2Si03 + 6HF -
-
->
SiF4t
+
2NaF
+
3H2O
CaSiOa + 6HF > SiF^ +
CaF2
+ SH.O
b) Si02 + 2NaOH >
NajSiOj
+ H2O
Na2Si03
+ CO2 + H2O )-Na2C0, +
H2Si03.
Bai 12. Bo tuc va hoan thanh cac phan iJng sau
:
Cacbon
+
HNO3

jsc
>
A +
NO2
+
HjO
A + cacbon
>
B
B +
CI2
photgen
HCOOH B
+
H2O
B
+
O2
A
A
+
NaOH >C
+
H20
C +
Si02
>
D + A
D
+ A +

H2O
>C + E
E Si02
+
H2O.
Gidi
A:C02
B:CO C:
Na2C03
D
:
Na2Si03
E
: H2Si03 Photgen : COCI2.
C +
4HNO3
JSC
> CO2
+
4NO2
+
2H2O
CO2
+
C-A2CO
CO +
CI2
^
COCI2
HCOOH

"a^Q^dac^
CO
+
H2O
2C0
+ 02-^
2CO2
C02
+
2NaOH
)•
Na2C03
+
H2O
Na2C03
+
Si02 Na2Si03 +
CO2
Na2Si03
+
CO2
+
H2O
>Na2C03
+
H2Si03
H2Si03 Si02
+
H2O.
Dang

2.
Nhcjn
bi§'t,
tach
rdi, tinh
chd' c6c chd't: CO, CO2, SiOa,
muoi
cacbonat
BAI TAP MAU VA BAI T^P AP
DyNG
Bai 1. C6
ba
khi gom CO, HCl va
SO2
difng trong
ba
binh rieng bi$t. Trlnh bay
phUdng phdp hoa hoc de phan bi$t tiTng khi. Viet c^c phiTdng trinh h6a hoc.
Giai
- Dan
ba
khi trdn Ian lifdt vao dung dich ntfdc voi trong dir. khi n^o tao kd't tua
tr^ng
la
SO2,
con lai
la
CO va HCl.
SO2
+

Ca(0H)2
>
CaS03i +
H2O
- Cho quy tim am Ian liTcJt vao hai khi con lai, khi n^o lam cho quy tim am h6a
do
la
HCl. Con lai
la
CO.
132
jgifAMfivnar
ki
2.
Trinh
bay
phifdng phap
h6a hoc de
tach rieng
cac
chat
:
AI2O3, SiOa,
MgO ra khoi hon hdp ciia chung.
Gidi
Cho hon hdp
ba
oxit AI2O3,
Si02
va

MgO v^o dung dich axit HCl duf
ta
t^ch
diidc SiOz khong tan c6n lai dung dich gom AICI3, MgCh
va
HCl dU.
AI2O3
+ 6HC1
> 2AICI3
+
3H2O
MgO
+
2HC1
>
MgClj
+
H2O
_ Cho dung djch NaOH diT vao dung dich
thu
diTdc
ta
tach diTcJc
Mg(0H)2
va
dung dich
g6m
NaA102, NaCl, NaOH duf, sau
do lay
Mg(OH)2

dem nung
ta
diTdc MgO.
AlCl3
+
3NaOH
>
3NaCl +
Al(0H)3i
Al(OH)3
+
NaOH
>
NaAIO2
+
2H2O
MgCl2
+
2NaOH > 2NaCl
+ Mg(OH)24'
Mg(OH)2
MgO +
H2O
- Thdi
CO2
v^o dung dich thu
dmc ta
difdc ket tua
A1(0H)3
sau

do
dem nung
ra thu diTdc AI2O3.
CO2
+
NaA102
+
2H2O
>
NaHCOj
+
Al(OH)34'
2Al(OH)3
AI2O3
+
3H2O.
Bai
3.
a)
Lam the
n£lo
de
loai
hdi
nirdc
va khi
CO2
c6 Ian
trong
khi

CO? Viet
cac
phU'dng trinh hoa hoc.
b)
Hay
phan biet
khi CO va khi
H2
b^ng phiTdng phap
hoa
hoc. Viet phiTdng
trinh hoa hoc cua phan itng de minh hoa.
Gidi
a) Dan hon hdp khi qua nuTdc v6i trong thi khi
CO2
bi giff lai
CO2
+
Ca(0H)2
>
CaC03>l
+
H2O
Tiep tuc cho hon hdp khi con
lai di
qua H2SO4 dam dSc thi H2O bi giff lai c5n
CO thi khong, do do thu dffdc CO tinh khiet.
b) Cho hai khi Ian Iffdt dem dot chay, sau do dan san pham di qua nffdc voi trong
dir, thay nifdc voi trong hoa due thi khi ban dau
la

CO, con lai
la
khi hidro.
2CO
+ 02-^
2CO2
2H2
+ 02-^
2H2O
CO2
+
Ca(OH)2
>
CaCOji
+
H2O.
Bai
4.
Co mpt hon
hdp
khi
gom cacbon dioxit
va
Iffu huj'nh dioxit. Trinh
bay
phifdng phdp hoa hoc de chffng minh sif c6 mat cua moi khi trong hon hdp.
133
Phan
dgng
va

phuong
ph^p
giSi
H6a hpc 11 VP co - D5
Xuan
Hung
Gidi
Dan
h§n
hcJp
hai khi
CO2
SO2
di qua dung dich brom c6 miu da cam, thS'y
khi
nao l^m cho dung dich brom mat mau d6 la
SO2.
SO2
+
Br2
+
2H2O
>
H2SO4
+
2HBr
Dan
khi c6n lai qua dung djch midc voi trong dif thay xua't hien ket tua thi do
la
CO2.

CO2
+ Ca(OH)2
>
CaCOji +
H2O.
Bai
5. L^m the nSo de tach rieng tCfng khi CO va
CO2
ra
khoi
hon hcJp
cfia
chung :
a)
Bing
phiTdng
phdp v$t
li.
b) B^ng
phiTdng
phdp hoa hoc.
Gidi
a) B^ng phiTdng phap vat li:
Ncn
hon hcfp hai khi di/di dp suat cao, khi
CO2
se hoa
long
diTcJc iich
ra ta

thu
dtfcfc
CO2,
khi con lai 1^ CO.
b)
B^ng phiTdng phdp h6a hoc :
-
Dan hon hdp hai khi di qua dung djch Ca(0H)2
AM,
khi
CO2
bi gii? lai vl tao
kct
tiia,
c6n lai la khi CO.
CO2
+ Ca(OH)2 > CaCOji +
H2O
-
Loc lay ket tua cho tdc dung vdi dung dich
HCl
thu dufdc khi
CO2
bay ra.
CaCOj +
2HC1
>
CaClj +
CO2T
+

H2O.
Bai
6, B^ng phtfdng phap nao c6 the nhan ra cdc cha't tin sau day : Na2C03,
MgCOj,
BaCOj.
(Trich
TSDH
Y Dmc
TP.HCM)
Giai
*
Cho niTdc Ian
liTdt
vao ba mau cha't r^n.
-
Mau nao tan la Na2C03.
-
Hai mau con lai khong tan la MgCOs va BaCOa.
*
Cho dung dich
H2SO4
loang Ian
liTdt
v^o hai mau c6n lai BaCOs va
MgCOj.
-
Mau nao tao ket tua tr^ng v^ c6 khi bay len 1^ BaCOa.
BaCO,
+
H2SO4

> BaS04i +
COjt
+ H.O
-
Mau
CO
khi bay len la
MgCOj.
MgCOj
+
H2SO4
> MgS04 +
CO2T
+
H2O.
B^i
7.
a)
PhSn biet cdc khi
H2S,
SO2.
CO2
bkng phiTdng phdp hoa hoc
Vieft
cdc phifdng
trinh
phan drng
tifdng
thig.
b)

Tinh
che
CO2
c6 Ian mOt it
SO2
v^
H2.
Vid't
cdc phifdng
trlnh
phan
uTng
xay ra.
(Trich
DHMii
TP.HCM)
134
Giai
)
Dan ba khi lln
Itfdt
v^o dung djch Pb(N03)2.
Khi
nko tao ket tua den la
H2S.
Con lai la
CO2
va SO2.
H2S
+ Pb(N03)2

>
PbSi +
2HNO3
Dan
hai khi con lai Ian
Itfdt
qua dung dich brom, khi nao lam ma't mau dung
dich
brom ^
SO2,
c6n lai
1^
CO2.
SO2
+
Br2
+
2H2O
>
H2SO4
+
2HBr
b)
Cho hon hdp khi v^o dung djch niTdc brom
diT
loai
di/dc
SO2,
sau do dan hon
hdp khi con lai di qua dung dich ni/dc voi trong ta thu difdc kct tua, cho kct tun

tac dung vdi dung dich
HCl
thu
diTdc
CO2
tinh
khiet.
SO2
+
Br2
+
2H2O
>
H2SO4
+
2HBr
CO2
+ Ca(OH)2
>
CaC03i +
H2O
CaC03
+
2HC1
>
CaCl2
+ C02t +
H2O.
Bai
8. Chi c6 nufdc va khi cacbonic c6 the phan biet difdc nam chat bot trang sau

day hay khong : Na2S04, BaC03,
NaCl,
Na2C03, BaS04. Ncu diTdc
hiiy
trinh
bay
each
phan biet. ; ,
Gi&i
*
Cho nirdc Ian
liTdt
vao ndm mau diTng nam chat hot
trcn.
-
Mau nao tan la :
NaCl,
Na2S04,
NajCO,.
-
Mau khong tan la BaC03 v^ BaS04.
*
Thoi
khi
CO2
vao hai mau
thuf
khong tan.
-
Mau nao tan tao thanh dung dich la BaCOj.

-
Mau nao khong tan la BaS04.
BaC03 +
CO2
+
H2O
>
Ba(HCO,
)-
*
Lay dung dich thu diTdc d tren cho vao ba mau tan, mau nao tao ket tua la
Na2S04 va Na2C03, c6n lai la NaCl
Na2S04 + Ba(HC03)2
>
^aSO^i
+ 2NaHC03
Na2C03
+
Ba(HC03)2
>
BaCOji
+ 2NaHC0,
*
Loc lay ket tua thu
difdc,
sau do
thoi
khi
CO2
v^ cho ntfdc vao, kcl lOa nao

tan la
BaCOs
-> mau ban dau la Na2C03, mau con lai la Na2S04.
Bai
9. Cho dung djch A c6 chtfa cac ion: Na*. NH^',
HCO3.
SO; , CO^
(khong
k^ cdc ion
H*
v^
OH'
cua
H2O)
chi c6 quy tim vS cdc dung dich
HCl.
Ba(OH)2
c6 the nhan bie't di/dc cdc ion nao trong dung djch A?
(Trich
DH
Nfiocii
Tliunna)
Giai
Chi
CO
quy tim vi cdc dung dich
HCl,
dung dich
Ba(OH)2
thi nhan

biet
difdc
13?
Phan dgng phiiong phap
glii
H6a hqc 11 VP cd - D8 Xuan Hang
cdcion: NH+, HCO", SO^", COl~, c6n lai Na^ khong nhan
biet
dtfcJc
.
HCl—>H*
+ cr
Ba(OH)2
>Ba'*
+ 20H-
* + HCOJ >
C02t
+
H2O
* Ba^* + CO^ >
BaC03^
Ba^^ + SO^ >
BaS04i
* NH; + OH- > NHjt +
H2O
BaCOj
+ 2HC1
>
BaCh
+

CO2T
+ H2O
Ba(0H)2 +
CO2
>
BaCOji
+
H2O.
Bai 10. Co bon lo ma't nhan, moi lo di/ng mot trong cac dung djch sau :
NaHCOj,
Na2S04,
Ba(HC03)2 va Mg(HC03)2. Hay
trinh
bay
each
nhan
biet
tiTng
lo, chi
dU'Oc
dung thdm
each
dun nong.
Gidi
* Trich moi lo mot it dc lam mau
thiir.
* Dem dun nong tifng mau thuT.
Hai
mau tao ket tua va c6 khi bay ra la Ba(HC03)2 va Mg(HC03)2, ton lai la
NaHCOj va

Na2S04.
Mg(HC03)2
MgC03
+ COjt +
H2O
Ba(HC03)2
BaC03
+
C02t
+
H2O
* Lay v^i giot dung dich mot trong hai Ip vuTa tao ket tua cho vao hai dung djch
e6n lai thay mau nao tao ket tua tr^ng thi mau thilr nho vao la
Ba(HC03)2,
mau c6n lai la
Mg(HC03)2
ta
biet
diTdc ong
diTng
Na2S04.
Con lai la
NaHCOj.
Ba(HC03)2 +
Na2S04
>
BaS04i
+
2NaHC03.
Dang

3.
Bai
t^p ve
tinh
chdt
CO2,
mudi
cacbonat
LiAi
y : Tinh
cha't
CO2
tac dung vdi dung dich kiem :
*
CO2
phan urng vdi dung dich NaOH (dung djch KOH )
C02 + NaOH
>NaHC03
(1)
CO2
+
2NaOH
>
Na2C03
+ H2O (2)
De xac dinh muo'i tao thinh, phai lap ti le so moi NaOH va
CO2
(NaHCOj) 1 < ^^^^ < 2
(NnjCOd
"C02

(NaHCOj v^
Na2C03)
*
CO2
phan urng vdi Ca(OH)2 hay Ba(OH)2:
CO2
+ Ca(OH)2
>
CaC034'
+
H2O
(1)
2CO2
+ Ca(OH)2 y Ca(HC03)2 (2)
(CaCOj)
1 < -^^^ < 2 (Ca(HC03)2)
'lCa(0H)2
(CaC03
va Ca(HC03)2)
Neu nco2 > T^k^t tua
^^i
xay ra ck hai phan tfng (1) v^ (2).
BAI
TAP
MAU
Bai 1. Cho
224ml
khi
CO2
(dktc) hap thu het trong

100ml
dung dich KOH 0,2M.
Tinh
khoi li/dng cua nhiifng
cha't
eo trong dung dich tao thanh.
Gidi
n^o^ =^^1:^ = 0.01 (moi); n^oH =
0,2.0,1
= 0,02 (moi)
22,4
Ta c6 ti 16 : = — = 2 thu di/dc muoi
K2CO3.
• nco, 0.01
CO2
+
2K0H
>
K2CO3
+ H2O
0,01 moi 0,01 moi
mK.co,
=0,01.138
= 1,38 (g).
Bai 2. Cho mot coc ni/dc c6
chufa
cac ion : Na* (a moi); CI" (b moi);
CO3"
(c moi);
NH+

(d moi) va
HCOJ
(e moi).
a) Tim bieu thufc lien he
givta
a, b, c, d, e.
b) Viet
cong
thuTc
tinh tdng khoi liTdng mu6'i trong dung dich.
Gidi
a) Ap dung dinh luat bao to^n di^n tich ta c6: a + d = b + 2c + e
b) Cong thuTc tinh tdng khoi lifdng mudi:
m„.,i = m^^, +
m^,
+
m^^_
+ m^^, + m^^^.
,
m^ufl-i
= 23a +
35,5b
+ 60c + 18d + 61e.
I
Bai 3. Dan khi
CO2
dieu che dUdc b^ng
each
cho 50g da voi tac dung vdi axit HCl
du"

di qua dung dich c6 chiira 32g natri hidroxit. Tinh khoi liTdng mudi thu
diTdc
trong dung djch.
Gidi
"caco,
= ^ = (mol); n,,oH = ^ = ^'^ (moi)
137
Phan
d^ng vk phuong phip giai H6a hpc 11 V6 co - 05
Xuan
Himg
CaCOj + 2HC1
0,5 mol
CaClz
+ COat + HjO
0,5 mol
Tacotile: 1< i^ = M = i,6<2
0,5
•
thu diTdc hai
muoi
NaHCOj va NazCOs.
C02 + NaOH ).NaHC03
XX X (mol)
CO2 + 2NaOH >
Na2C03
+ H2O
y 2y y (mol)
Ta CO he
phuTdng

trinh
:
x = 0,2
y = o,3
(mol)
x + y = 0,5
x + 2y = 0,8
mNaHco, =84.0,2 = 16,8(g); mr,,^co, = 106.0,3 = 31,8 (g).
Bai 4. Nung 52,65g CaCOj d 1000"C va cho loan bo liTcJng khi thodt ra hap thu
het vao 500ml dung dich NaOH 1,8M. Hoi thu drfdc nhffng mu^i nao? khoi
lufdng Ih bao nhieu? Biel rang hieu suat cua phan tfng nhiet phan CaCOj la
95%.
Giai
-I^^^^CaO
+
C02t
0,5265 mol
CaCOj
0,5265 mol
52,65
n
CaCO,
100
= 0,5265 mol
Hi^u suat phan lirng 95% :
"co2
=^.0,5265 = 0,5 (mol)
lUU
nNaOH=
1,8.0,5

= 0,9 (mol)
Tac6til?:
1 < ilNii2H.
=
M = l,8<2
ncoj 0,5
=>thu difdc hai muoi NaHCOs \h Na2C03.
CO2
+ NaOH >NaHC03
a a a
CO2 + 2NaOH > Na2C03 + H2O
b 2b b
a + b = 0,5
Ta
CO
h$ phifttng trtnh :
a + 2b = 0,9
a = 0,1
b = 0,4
(mol)
mNaHco, =84.0,1 = 8,4 (g);
mN.^cp,
=106.0,4 = 42,4 (g).
138
ai 5. Dan 2,36 lit
CO2
(15"C 2 atm) qua 40ml dung djch
Ba(OH)2
2M. Muoi
sinh ra la muoi gi? Tinh nong dp mol cua dung dich muo'i thu duTdc (biet the

tich dung dich khong thay doi).
Gidi
PV 2.2,36
nco2 = — -•
= 0,2 (mol)
2,5 > 2 => thu dircfc muo'i
Ba(HC03)2.
RT 0,082.(15 + 273)
nBa(OH), =0,04.2 = 0,08 (mol)
Tacotile:
J^
=
-^
=
"Ba(OH)2 0,08 .
2CO2
+ Ba(OH)2 >Ba(HC03)2
0,08 mol 0,08 mol
N6ng
dp mol cua Ba(HC03)2: CM= ^ = ^ = 2 (M).
Bai 6. Nung m gam hon hcfp X gom hai muoi cacbonat trung tinh ctia hai kim loai
A, B deu c6 hoa tri 2. Sau mot thcfi gian thu diTdc 3,36 lit
CO2
(dktc) \h con lai
hon hdp r^n Y. Cho Y tac dung vdi dung djch HCI dit roi cho khi thoat ra hap
thu hoan toan bdi dung dich
Ca(0H)2
di^, thu diTdc 15 gam ket tua. Phan
dung djch dem c6 can thu dtfcJc 32,5 gam hon hdp muoi khan. Vie't cac
phiTcJng trinh phan iJng va tinh m?

(Trich TS DHQG TP. HCM)
Gidi
Dat cong thiJc cua hai muoi cacbonat \h
ACO3
va BCO3.
ACO3
—^ AO +
C02t
BCO3 —^ BO +
C02t
3,36
= 0,15 (mol)
"CO2-224
Cho Y + dung dich HCI:
A0 + 2HC1
>ACl2
+ H20
B0 + 2HC1
>BCl2
+ H20
ACO3
+ 2HC1
>
ACI2
+ COzt + H2O
BCO3 + 2HC1
>
BCI2 + COzt + H2O
C02
+ Ca(OH)2 >

CaCOji
+ HzO
0,15 mol 0,15 mol
'CaCO,
15
100
= 0,15 (mol)
139
Phan
d?ng vi phuing ph^p
giai
H6a h^c 11 VP CO - D5
Xuan
Hung
Ta CO
tdng so mol CO2 :
Yl^co^
= 0,15 + 0,15 = 0,3 (mol)
"co?- "coj =0,3 (mol); n^|_ =0,3 (mol)
Khoi
liTdng hai kirn loai:
mh„ikin,ioai
= 32,5 - m^__ = 32,5 -
0,3.2.35,5
= 11,2 (g)
nihh
X
=
mhai
kim

io,i
+ m^Q2_ = 11,2 +
0,3.60
= 29,2 (g).
Vay m = 29,2 gam.
Bai
7. Cho 10 lit khi (dktc) gom N2 va CO2 di qua 2 lit
dung
dich Ca(0H)2
0,02M thu
duTdc
1 gam ket tua. Hay xdc dinh %
theo
the tich cua
CO2
trong
honhdp.
Gidi
nca(OH,,
=
0,02.2
= 0,04 mol;
nc^co,
= 7^ = 0.01 mol
lOU
"caco, < nca(OH)2 nen xay ra hai triTdng hdp :
*
THl :Ca(0H)2dir.xayraphani?ng:
CO2
+ Ca(OH)2 > CaCOji +

H2O
0,01 mol 0,01 mol
0
224
=>Vco^
=
0,01.22,4
= 0,224(lit) %Vcoj =^^^.100% =
2,24%.
*
TH2 : Ca(OH)2 khong dir, xay ra hai
phan
urng :
CO2
+ Ca(OH)2 > CaCOji +
H2O
0,04 0,04 0,04
CO2
+
CaCOj
+ H2O >Ca(HC03)2
(mol)
0,03
(0,04-0,01)
=>S6 mol
CO2:
ncoj = 0,04 + 0,03 = 0,07 (mol)
=>
Vcoj
=0,07.22,4

= 1,568
(lit)
^ %Wco, =-^^.100% =
15,68%.
Bai
8. Hap thu
hoan
toan
3,36 lit khi CO2 (dktc) vao 125 ml
dung
dkh
Ba(0H)2 IM, thu
diTcJc
dung
djch X. Coi the tich
dung
dich khong
thay
do'i,
Tinh
nong
do mol cua
chat
tan trong
dung
djch X.
Gidi
Cach
1;
Ta c6: n^Oj = 0,15 mol;

n
Ba(OH)2
= 0,125 mol n^^_ = 0,25 mol
[40
Ta
thay:
1 < = = 1,67 < 2 => tao 2 muoi
ncoj 0,15
Ptptf:
CO2
+ Ba(OH)2
>
BaCOji
+
H2O
XX X
2CO2
+ Ba(OH)2 )-Ba(HC03)2
2y y y
Ta
c6 hfe
rx
+ 2y= 0,15
x + y = 0,125 ly =
0.025
x = 0,l
=>[Ba(HC03)2]= ^ =0,2M
Cach
2;
Khi

cho
CO2
vao
dung
dich Ba(0H)2 l^n liTdt
x&y
ra cdc phan iJog:
CO2
+ Ba(OH)2 >
BaCOji
+ H2O
0,125 0,125 0,125
=:>
ncoj
dir
= 0,15 - 0,125 =
0,025
mol
Theo d^: Hap thu ho^n
toan
khi
CO2
=>
CO2
het, m^ sau (1)
CO2
diT
=>
tiep
tuc xay ra phan uTng:

CO2
+
BaCOj
+
H2O
-> Ba(HC03)2
i
0,025 0,025
>[Ba(HC03)2]= ^=0.2M
a 9. Hap thu ho^n to^n
2,688
lit khi CO2 (d dktc) vSo 2,5 lit
dung
djch
Ba(OH)2
nong
dp a mol/1, thu diTdc 15,76 gam ket tua. Tim a.
Gidi
rTac6:
neo, = ^ = 0,12 mol ; Oe^eo, = ^ = 0,08 mol
Do Ucoj nflacoj
ngoai
BaCOs con c6 Ba(HC03)2 diTdc tao th^nh.
[ 2CO2 + Ba(0H)2 -> Ba(HC03)2
0,04 0,02 0,02
CO2
+ Ba(OH)2 -> BaC03>l + H2O
0,08 0,08 0,08
Theo
phkn

v(ng:
EnB,(OH)2 = 0.08 + 0,02 = 0,1 mol
a = — = 0,04M
2.5
141
Phan
dgng
va
phuang
phip
giai
H6a hpc 11 VP co - P8
Xuan
Hung
Bai 10. Nung 13,4 gam hon hcfp 2 muo'i
cacbonat
cua 2 kim loai hoa tri 2, thu
diTdc 6,8 gam
chS't
tin va khi X.
LUcJng
khi X sinh ra cho hap thu vao 75 ml
dung dich NaOH 1M thu dU"dc m gam muo'i khan. Tinh m.
Gidi
Goi
cong
thtfc
chung
cua hai muoi
cacbonat

kim loai hoa tri II la
RCO3.
RCO3
—^ RO + CO2
Ap
dung dinh luat bao
toan
kho'i Itfdng ta c6:
i"coj
=
niRco,
-
niRo
= 13,4 - 6,8 - 6,6 (g)
Ucoj
= 0,15 mol
Tac6:
UNaOH =
0.075
mol
^ ^
%aOH.
^ MZ^ = 0,5 < 1 tao muoi
NaHC03
va CO2 dir.
Ucoj
0,15
CO2
+ NaOH ^ NaHCOj
0.075 0,075

m„„.ti =
0,075.84
= 6,3 (g)
Bai 11. Cho hon hdp K2CO3 va NaHC03 (ti 16 mol 1 : 1) vao binh dung dich
Ba(HC03)2 thu di/dc ket tua X va dung dich Y.
Them
tur tCr dung djch HCl
0,5M vao binh den khi khong con khi
thoat
ra thi het 560 ml.
Biet
toan bo Y
phdn tfng vifa du vdi 200 ml dung djch NaOH IM. Tinh khoi
liTcJng
ke't tua X.
Gidi
Goi s6' mol K2CO3 bd =
NaHCOs
= x mol
So mol
Ba(HC03)2
= y mol
Dung dich Y tac dung vdi
NaOH:
OH* +
HCO3'
C03^" + H2O
0,2 0,2
Tac6:x
+ 2y =

0,2(1)
Dung dich Y tic dung vdi HCl: H* +
HCO3"
->C02
+ H2O
0,2 0,2
2H*
+ C03^ ^ CO2 + H2O
2x X
Suy ra: 2x + 0,2 = 0,28 x = 0,04 the v^o (1) ->• y = 0,08.
Phin tfng t?o k^t tiia: Ba^* + COj^' ->
BaCOa
So mol
BaC03
= mol COj^' = x = 0,04 kh^i lUtfng k^t tua = 7,88 gam
Bai 12. Hap thu ho^n to^n 2,24 lit CO2 (dktc) v^o
100ml
dung dich g6m K2CO3
0,2M vi KOH X mol/lit, sau khi cic ph^n tfng xdy ra ho^n
toiln
thu dtfdc
dung dich Y. Cho
tokn
bp Y tic dyng vdi dung djch
BaCU
(dtf), thu dtfcfc
11,82 gam k^t tOa. Tim gii
tri
ciia
x.

142
Gidi
11 82
Ta c6: nco, = 0,1 mol;
nB.co,
= ~[^^
"xzco,
= 0'02 mol
Khi
sue CO2 vao dung dich hon hdp gom K2CO3 va KOH, gia sur chi xay ra
phan
uTng:
CO2
+2KOH^ K2CO3 +H2O
0,1 0,1
=>
n^^j^Q^
(irong dung djch) = 0,1 + 0,02 = 0,12 mol
BaCl2
+ K2CO3
BaCOji
+ 2KC1
0,12 0,12
Ta thay:
nj,
= 0,12
?t
nj,
a
,ho

= 0,06 mol
Vay U-ong phan ufng CO2
voti
KOH
ngoai
muoi K2CO3 con c6 muoi
KHCO3.
Ap
dung djnh luat bao tojln nguyen to C ta c6:
(IrongCOj) "c (Irong
KjCO,)
"c
(iR)ng
BaCO,) + "c (Irong KHCO3)
0,1 + 0,02 = 0,06 + X (X la so mol
BaC03)
=> x = 0,06
CO2
+ KOH
KHCO3
0,06 0,06 0,06
CO2
+ 2K0H -> K2CO3 + H2O
0,04 0,08
I =>
UKOH
= 0.14 mol => [KOH] = ^ = 1,4M
Bai 13. Dung dich X
chtfa
hon hdp gom NaiCOj 1,5M va

KHCO3
IM. Nho
tiir
tuf
tirng
giot cho den het
200ml
dung dich HCl IM vao
100ml
dung dich X, sinh
ra V lit khi (d dktc). Tim gia tri cua V.
Gidi
Ta c6:
n^^^^co^
= 0,15 mol =:> n^^j. =
0,15mol
"NaHCOj
= """^ ^"HCXJJ =
nHci=
0,2 mol => n^, =
0,2mol
Khi
cho lit tir dd HCl v^o dd X (HCOj',
C03^")
phan tfng xay ra
theo
trinh
ttf:
H*
+ COj^" -> HCOj'

0,15 0,15 0,15
=> n^^ c6n = 0,2 - 0,15 = 0,05 mol vk n^^^_ = 0,1 + 0,15 = 0,25 mol
Do H*
c6nsau
phan ufng (1) ndn tiep tuc xdy ra pi?:
H*
+
HC03"->-C02t
+ H20
0,05 0,05
143
Phan d;nq phoang pMp
glil
Hito hpc 11 V6 co - Bfi Xuin Hung
=>
Vcoj
=
-0,05.22.4 = 1,12 (1ft)
Bki
14. Dun n6ng 116 gam qu$ng
xiderit
(chtfa
FeCOs
\k tap
cha't
trd) trong
kh6ng
khi cho d^n khi
]sh6i
liTdng

khdng ddi. Cho hSn hdp khi sau khi phin
Ung
h^p thv vko binh ditog dung
d}ch
niTdc v6i c6 h6a tan 0,4 mol Ca(OH)2,
trong
blnh c6 20 (g) k^t tda. N^u dun n6ng phin dung dich sau
khi
loc k^t tua,
thl l^i
th^y c6 k^t
tOa
xu^t
hi$n.
Tinh
% kh^i
liTdng
FeCOa
trong quSing
xiderit.
am
20_
100
Ta
c6: ncco, =
T::^
= 0.2 mol
2FeC03
+ -O2 —^
FezOa

+
2CO2
(1)
2
DSn
h5n
h<?p
khi sau phin
tfng
vko dung djch
Ca(0H)2
th^y c6 kd't
tfia,
dun
philn
nirdc Ipc, lai xuS't hi$n k^t tOa, chd'ng t6
CO2
t^c dung vdi dung dich
Ca(OH)2
t?o 2
mu^i.
CO2
+
Ca(0H)2
->
CaCOsi
+
H2O
0,2 0,2 0,2
2CO2

+
Ca(0H)2
Ca(HC03)2
„
0,4 0,2
Encoj = 0,4 + 0,2 = 0,6 mol
Theo phSn
tfng
(1):
npeco,
= ncoj = 0,6 mol
%FeC03
trong
qu|ng
xiderit
= "'^'^^^100% = 60%
116
m 15.
TrOn
100ml dung dich A g6m
KHCO3
IM vk
K2CO3
IM \ko 100 ml
dung dich B g6m NaHCOs
IM
vk
Na2C03
IM
thu dtf^c dung dich C. Nh6

tilf
tilf
100ml dung dich D
gfim
H2SO4 IM
vk HCl
IM
\ko dung dich C thu diTdc
V lit CO2
(dktc) vk dung djch E. Cho dung dich
Ba(0H)2
tdi diT
\ko dung dich
E
thu
dir<?c
ra
gam
k^t
tiia.
TInh
m\kW.
cm
•KHCO3:
0,1 mol =>
HCO3"
: 0,1
mol
.K2CO3:
0,1 mol => C03^'

:
0,1 mol
•NaHCb3:
0,1 mol =>
HCO3"
: 0,1 mol
.Na2C03:0,1 mol =>
COi^'
: 0,1 mol
=> Khi
trOn
dung dich A
vdi
dung dich
B
diA^c
dung dich C c6:
rHCO3-:0,2mol
lCO3^-:0,2mol
100ml
dung dich A c6:
100ml
dun^ dich
B
c6:
144
100ml
dung dichDcd:
SO4^:0.1
mol

LH*:0,3
mol
•H2SO4:
0,1 mol
l.HCl:0,l
mol
Khi
cho
tir tilf
dung dich D v^o dung dich
C,
diu
ti6n
xdy ra phan
uTng:
+
CO3'- ->
HCO3-
(1)
0,2 0.2 0.2
Sau phdn
tfng
(1):
H* diT:
0.3 - 0.2 = 0,1 mol
Wk
trong dung dich E:
HC03'
: 0.2 + 0.2 = 0,4 mol
H*

+
HCO3- -)• CO2
+
H2O
(2)
),1
0,1 0,1
=>
Sau phan
tfng
(2):
HCO3' dif:
0.4 - 0,1 = 0,3 mol;
ZO2
dtfdc
tinh
theo
H*.
leo phdn
iirng
(2) ta c6: s^
mol CO2
=
0,1
mol
Vcoj
=
0,1.22.4
= 2.24
lit

:ho
Ba(OH)2
dur
\ko dung dich E:
ICOj-
+ OH" C03^" +
H2O
).3 0,3
Ba^*
+ C03^" ->
BaCOji
0,3 0,3
Ba^*
+ S04^-
BaS04
4'
0,1 0,1
mi =
0,3.197
+ 233.0,1 = 82,4 (g) => Ddp
&n
B.
BAIT^SIPAP
DMNG
1. Cho CO2 h^p thu ho^n to^n \ko dung
dich
chtfa 14,8g
Ca(0H)2
thu
diTdc

10
;gam k^
tiia
\k dung dich
A.
Lpc b6 k^
tiia
r6i 1^ dung dich A cho tic
dung
vdi
dung dich
Ba(OH)2
diT thu
dvKfc
m gam k^t
tiSa.
Tinh
m?
cm
ncoHh
=
= 0'2
nccoj ==
0.1 mol
C02
+
Ca(OH)2
• CaCOsi +
H2O
(1)

0,1 mol 0,1 mol
C02
+
Ca(OH)2
>Ca(HC03)2
(2)
0,1 mol 0,1 mol
Ca(HC03)2
+
Ba(OH)2
•CaC03i+
BaC034+
H2O
0,1 mol 0,1 mol 0,1 mol
nc.(OH)j
a
pMn
tfng
(2)
= 0,2 - 0,1 =
0,1
(mol)
145
Phan
djing
va
phuong
ph^p
giai
H6a hgc 11 VP

CO
- P8
Xufln
Hiing
^Vco^
=0,05.22,4=
1.12 (lit)
Bai 14. Dun
n6ng
116 gam
quSng
xiderit
(chufa
FeCOs
tap
chat
trd) trong
khong khi cho den khi
khfi'i
liTdng khong ddi. Cho hon
hcJp
khi sau khi phan
tfng hap thu v^o binh diTng dung dich nrfdc voi c6 hoa tan 0,4 mol
Ca(OH)2,
trong binh c6 20 (g) ket tua. Neu dun
n6ng
phan dung dich sau khi Ipc ket tua,
thi
lai
th3y

c6 ket
tiaa
xu§'t
hi$n. Tinh % khoi li/cfng
FeCOj
trong
quSng
xiderit.
Giai
20
Ta c6:
nr.^^n.
=
2FeC03
CaCOj
100
= 0.2 mol
FesOj
+
2CO2
(1)
Dan h6n hdp khi sau phan
iJng
v^o dung dich
Ca(0H)2
thay
c6 ket tua. dun
phin
nirdc loc, lai
xuS't

hi^n ket tua.
chi?ng
to CO2 tic dung vdi dung dich
Ca(OH)2
tao 2 mu6'i.
CaCOji
+ H2O
0.2
Ca(HC03)2
CO2
0.2
2CO2
0.4
+
Ca(OH)2
0.2
+
Ca(0H)2
0,2
=^
Encoj
= 0.4 + 0.2 = 0.6 mol
Theo
phan drng (1):
np,.co3
"coj = 0.6 mol
=!> %FeC03 trong
quSng
xiderit =
^^^^100%

= 60%
116
Bai 15. TrOn
100ml
dung dich A g6m
KHCO3
IM va
K2CO3
IM
v£lo
100 ml
dung dich B gom
NaHCOs
IM vk Na2C03 IM thu
diTdc
dung dich C. Nho
tiir
tit
100ml
dung dich D gom
H2SO4
IM va HCl IM v^o dung dich C thu
diTcJc
V lit CO2 (dktc) \h dung djch E. Cho dung dich
Ba(0H)2
tdi
diT
v^o dung dich
E thu
diTdc

m gamkd't tua.
Tinhm\kW.
Giai
•KHCO3:
0.1 mol => HCOf : 0.1 mol
.K2CO3:
0.1 mol => C03^- : 0.1 mol
NaHC03:
0.1 mol =>
HCO3"
: 0.1 mol
.Na2C03: 0.1 mol => CO^^- : 0.1 mol
=> Khi
trpn
dung dich A vdi dung dich B dufdc dung dich C c6:
HCO3":0.2mol
l(X)ml dung djch A c6:
100ml
dung
dichBc6:
C03^ : 0.2 mol
144
100mldung
dich Deo:
SO/": 0.1 mol
-H*
: 0,3 mol
'H2SO4:0,l
mol
I.HC1:0.1

mol
Khi
cho tif tCf dung djch D vao dung dich C. dau tien xay ra phan
iJng:
+ CO3'- ->
HCO3-
(1)
0.2 0,2 0,2
Sau phan
vCng
(1):
di/:
0,3 - 0,2 = 0,1 mol
Va trong dung dich E:
HCO3"
: 0,2 + 0,2 = 0,4 mol
+
HCO3"
^ CO2 + H2O (2)
0,1 0,1 0,1
=> Sau phan ufng (2):
HCO3"
diT: 0,4 - 0,1 = 0,3 mol;
CO2
diTdc
tinh
iheo
H"^.
Theo
phan

i?ng
(2) ta c6: so mol CO2 = 0,1 mol
=:>
Vcoj
=
0,1.22,4
= 2,24 lit
Cho
Ba(0H)2
du"
vao dung dich E:
HCO3-
+ OH-
0,3
Ba
2+
C03^-
Ba
>
COi^- + H2O
0,3
BaC03i
0,3 0,3
+ S04^" ->
BaS04i
0,1 0,1
m^
=
0,3.197
+

233.0,1
= 82.4 (g)
2+
Dap an B.
BAI TAP AP
DUNG
B&i 1. Cho CO2 hap thu ho^n
toan
v^o dung dich
chuTa
14,8g
Ca(0H)2
thu diTdc 10
gam ket tua dung dich A. Loc b6 ke't tua roi ISy dung dich A cho tdc dung
vdi
dung dich
Ba(0H)2
diT thu diTpc m gam ket tua. Tinh m?
Giii
• = 0,2 mol; 0^00, = — = 0,1 mol
"Ca(OH)2
- ^4
CO2
+
Ca(0H)2
0,1 mol
CO2
+
Ca(OH)2
0,1 mol

100
C&COii
+ H2O
0,1 mol
(1)
(2)
>Ca(HC03)2
0.1 mol •
Ca(HC03)2 + Ba(0H)2
>
CaCOji +
BaC034'
+
H2O
0,1 mol 0,1 mol 0,1 mol
"ca(OH)2 a pMii
rfng
(2)
= 0,2 - 0,1 = 0,1 (mol)
145
Phfln
dgng
va phuano phip
giai
H6a hpc 11 VP cd - B8 Xuan Hunp
m,^. .a. =
mB.co,
+
nic,co,
=

197.0.1
+100.0.1
= 29.7 (g).
B^i
2. Din 11.2 lit
CO2
(dktc) \ko 500ml
dung
dich NaH 25% (D = 1.3 g/ml).
H6i
thu
diftjc
muoi gl? Tinh
nong
dp %
dung
dich mu6'i thu
dtfcJc.
Gidi
=^
= 0'5(mol);
n,.oH
=^^x^ = 4.0625(mol)
Ta
c6 ti
1§
: ^^^^^ = = 8.125 >
21:>
thu dtfdc mudi
Na2C03.

•
ncoj
0,5
C02
+ 2NaOH
^NazCOs
+ HzO
0,5 mol 0,5 raol
niNa.co,
=
0.5.106
= 53(g); mjd = d.V = i,3.500 = 650 (g)
Nong dO'%
dung
dich mu^i thu dtf^c: C% = —
x
100% =
8.15%.
650
B^i
3. Nung
33,8g
hon hdp hai mu6'i NaHCOj
NaaCQs
cho den khi khoi
Itf^ng
h5n hdp
khftiig
ddi thl thu difdc
29,15g

chat
r^n. Tinh th^nh
phan
%
khoi
liTdng mu6^i
NaHCOs
trong hon hdp dau.
Gidi
2NaHC03
Na^COs
+
CO2
+
H2O
X
mol
—
mol
2
Goi X,
y Ian
lvt(ft
\h s6 mol ban dau ciia
NaHCOa,
Na2C03.
X
29 15
Tac6:
- + y =

^^^-0,275
(mol)
2 ^ 106
Ta c6 h? phtfdng trinh :
84x
+
106y
=
33,8 rx = 0.15
X
=> i ' (mol)
|
+ y =
0.275
ly = 0.2
Thinh phin % NaHCOj trong h5n hdp:
%NaHC03
=
"'^^'^'^•100%
=
37,28%.
33,8
Bai
4. MOt hSn
h<?p
X gdm
ACO3 BCO3.
Phan
trim khoi Itfdng cia A trong
inn

- ACO3
m =j-% vjl cua B
trong
BCO3
m 40%.
a) Xic djnh mu^i
ACO3
vJl
BCO3.
b) LS'y 31,8 gam h5n hdp X cho v^o 0,8 lit
dung
dich HCl IM thu
di/dc
dung
djch
Y.
Hay chtfng t6 hSn hdp X bi h6a tan h^t, cho v^o
dung
djch Y m^l
Itfdng
thira NaHCOj thu difdc 2,24 lit khi CO2 (dktc). Tinh kh6'i iMpng moi
muoi
cacbonat.
146 - '
a)
%A=—^.100%
= ^^
A +
60
B

.100% = 40%
GOti
A
= 24 : Mg
B
=
40:Ca
B +
60
Vay hai muoi m MgC03 va CaCOj.
b) nHa = 0,8.1 =0,8 (mol)
MgC03 +
2HCl
*
MgCl2
+
CO2T
+
H2O
(1)
X
2x
CaC03
+
2HCl
>
CaCl2
+
C02t
+

H2O
(2)
y
2y
^
= 0,318<n,,,,,„„,,<^ =
0,3785
Theo
phrfdng trlnh
:
2.0,318
=
0,636
<
UHCI
< 2.0,3785 =
0,757
Ma
nHcibandlu = 0,8 mol >
0,757
mol
=:>HCldtf
=> hon hdp X
b} h6a
tan h^t.
*
ChoY
+NaHCOj:
NaHCOj + HCl
>

NaCl +
C02t
+
H2O
(3)
0,1 mol 0,1 mol
2,24
"002
=
•
=
0,1
(mol)
Tac6
h$ phiTdng trlnh :
(mol)
22,4
nHci(i).(2)
= 0,8-0,l=0,7(mol)
=>
2x + 2y = 0,7
f84x + 100y = 31,8 fx = 0,2
l2x
+ 2y = 0,7 ^|y = 0,15
mMgcoj =
84.0.2
= 16,8
(g);
m^coj =
100.0,15

=
15
(g).
Bai
5. DSn 1,792 lit
CO2
(dktc) qua 100ml
dung
dich
Ba(0H)2
0,5M. Tinh khSl
Irfdng
mu^i sinh ra sau
phdn
tfng.
Gi&i
1
792
Tac6:
nco,
=-!r:J^
= 0,08 (mol)
nBa(0H)2
=0.5.0,1
= 0,05 (mol)
Tac6til?:l<-^ = ^ = 1,6<2
nBa(OH)2
0,05
=>hai mu6'i BaCOj
Ba(HC03)2.

PhSn
djing
va
phuang
phap Qiai H6a
h<?c
11 VP
co
-
Bfl
Xuan
Hung
CO2
+
Ba(OH)2
>
BaCOsi
+
H2O
X
X X
»
2CO2 + Ba(OH)2 > Ba(HC03)2
2y y y
Ta
CO
h§ phUdng
trlnh
^ 2y = 0,08 ^ fx = 0.02
[x

+ y = 0,05 [y = 0,03
mBaco,
=0,02.197 = 3,94 (g);
mB.^Hco,,,
=0,03.259 = 7,77 (g).
Bai
6. Cho V lit
CO2
(54,6"C 2,4 atm) hap thu ho^n to^n v^o 200ml dung
dich
hon
hcJp
KOH IM va Ba(0H)2 0,75M thu difdc 23,64 gam ket tua.
Tinh
V
lit?
(Trich TS
DHSP
TP.HCM)
Giai
Tac6:
nKon
= 0,2.1 = 0,2 (mol)
KOH
>K^+ 0H\
0,2 0,2
nBa(OH),
=
0,75.0,2
= 0,15(mol)

Ba(0H)2
> Ba'^ + 20H-
0,15 mol 0,15 mol 0,3 mol
=>
2"oH-=0'2
+ 0,3 = 0,5(mol)
23,64
197
*
XdtTHl
:OH-dir
CO2
+ 20H- > COj^- + H2O
0,12 mol 0,12 mol
Ba^* + CO^- > BaCOji
0,12 mol 0,12 mol
=> Hcoj =0,12 mol
^y^aRT^0.12.0.O82.(54.6.273)^,3^3^„,^
"
^
•^i
4
*
X6t TH2 : OH" khong drf
CO2
+ 20H- > CO]- + H2O (1)
0.12 0,24 0,12
CO2
+ OH" • HCO; (2)
0,26 0.26

148
CO^-
+ Ba'^ > BaCOji (3)
0,12 mol 0,12 mol
n
_ =0,5-0,24 = 0,26 (mol)
OH
ncoj
=0,12+
0,26 = 0,38 (mol)
^^Jo,38.0,082.(54,6
+ 273)^^^^^3^^.^^
2,4
Bai
7. Cho CO2 can thiet hap thu
hoan
toan vao dung dich chtfa 0,2 mol
Ca(OH)2
thu di/cfc 10 gam ket tua va dung dich A. Loc bo ket
tiia
roi lay dung
djch
A dem dun nong thl thu diTdc m gam ke't tua nffa.
Tinh
m?
Gidi
CO2
+ Ca(0H)2 > CaCOji +
H2O
(1)

0,1 mol 0,1 mol
CO2
+ Ca(OH)2 > Ca(HC03)2 (2)
0,1 mol 0,1 mol
Ca(HC03)2 CaCOsi +
CO2
+
H2O
(3)
0,1 mol 0,1 mol
ncacov<>(»=
]^ =^'^
=>
nca(OH)2
0(2)=0,2-0,l
= 0,l(mol)
Theo
phan
urng (3), khoi
liTdng
ket tua la : m =
0,1.100
= 10 (g).
Bai
8. Dung dich A
chiJa
a mol Na", b mol
NHJ,
c mol
HCO3.

d mol
CO3"
va
e mol sol' (khong ke c^c ion va OH" cua
H2O).
Neu them (c + d + e)
mol
Ba(0H)2 vao dung dich A dun nong dxTdc dung djch X. khi Y duy
nha't
c6
mui
khai v^ ke't tua B.
Tinh
so mol cua cac
cha't
trong B. khi Y vi moi ion
trong dung dich X
theo
a, b. c, d, e.
Giai
Ap
dung dinh luat bao toan dien
tich
ta c6: a + b = c + 2d + 2e (I)
Khi
them Ba(0H)2 vSo dung dich A :
Ba(0H)2
> Ba^* + 20H-
(c + d + e) (c + d + c) 2(c + d + e)
Ba'^+CO^- >BaCOii (1)

d
d d mol
Ba'^ + SO^
>
BaS044
(2)
e e e mol
149
PhSn
dgng
phaong
phip giSi Hda hpc 11 VP cO - Qg
Xufln
Hung
Ba'* +
HCO3
+ OH-
c c c
NH:
+
OH-
—^ NHjt +
H2O
b b b mol
>
BaCOji +
H2O
c mol
(3)
(4)

.
fBaCOj:(c + d) mol
K6ttuaBg6m \ '
[BaS04
:
e mol
Theo cic phi/dng trinh thl n
OHdi/
= (c + 2d + 2e - b) mol => n^H, = b mol
Khi y la NHj: b mol
Tfif
(I)
=> a = c + 2d + 2e -
b
=
Vay trong dung djch X c6n lai
Na*
^"OH-
Na*: a mol
OH"
dir:a mol
=> dung dich X chd'a a mol NaOH.
Hay Na* :c + 2d + 2e-b = a (mol)
OH"
:c + 2d + 2e-b = a (mol).
Btki
9. Cho tir
tCr
dung djch X chi?a a mol HCl v^o dung dich Y chd'a b mol
NazCOs. Sau khi cho het X vao Y ta difdc dung dich Z. Hoi trong dung dich Z

CO
nhumg
chS't
gi? bao nhieu mol? (tinh theo a, b).
Gidi
Khi cho tilf
tir
dung djch HCl v^o NajCOj:
NazCOj + HCl >NaCl + NaHC03 (I)
NaHCOj + HCl
>
NaCl + COzt +
H2O
(2)
Dung djch Z gom nhOhg chat gi ta c6 cac triTdng hdp sau:
*
Neua<b=>chic6phanurng(l)nenNa2C03dir.
Vay dung djch Z gom :
NaHCOj:
a mol
NaCl:a mol
NajCOj dir:(b-a) mol
Neu a = b thl phan tfng (1) xay ra ho^n toan nen dung dich Z g6m:
jNaHCOj :a hoac b mol
|NaCl:a hoic b mol
Neu a > b thl phan tfng (1) xay ra xong tiep den phan tfng (2).
^
fb mol NaCl
dphanurng(l)tac6: , ur^r^
[b mol NaHCOj

Va HCl dir: (a - b) mol phan ufng vdi b mol NaHCOj theo phan iJng (2) ndn
CO
cac trirOng hdp xay ra:
a - b = b =^ a = 2b thl phan tfng (2) xay ra hojkn to^h nen dung dich Z chtfa
.
NaCl: 2b mol.
b>a-b=>a<2b=>
phin
tfng (2) c6 (a - b) mol NaCl.
NaCl =>
SO
mol NaHCOj
diT:
b - (a - b) = 2b - a (mol)
,
rNaHC0j:(2b-a)
mol
=>
dung djch Z gom ^. _, , . ,
[NaCl:b + a-b
= a
mol
b<a-b=>a>2b
fb mol NaCl
=>phanurng(2)c6
j^^^
d^,(a-b-b).(a-2b) mol
•' ,1.,
„ ,:i ii.^\
'/,L.iL',.i]«fc,i,iii.,a^UJtfc':i,'Aiiitir<'t,ik,t;ulkAi;

rNaCl:b
+
b = 2b mol
'[HCl:(a-2b) mol.
Bfki
10.
m(g) hon hcJp muoi vao
H2O
diTdc
dung dich A chtfa cic ion: Na*; NH,*;
COj^~;
S04^". Khi cho A tac dung vdi dung dich
Ba(OH)2
di/ va dun n6ng thu
duTdc 0,34(g) khi lam xanh quy tim am v^ 4,3(g) ket tua. C6n khi cho A
lie
dung vdi dung dich H2SO4 dif thl thu di/dc 0,224 lit khi (dktc). Tinh gia tri
cua m.
cm
0,34
>
dung dich Z gom
Ta c6: n^H, =
D5t: CO
17
i^-: X mol
S04^-
: y mol
= 0,02 mol
Dung dich A tac dung vdi dung dich

Ba(0H)2
di/:,
NH4*
0,02
Ba^*
Ba'* +
NHj + H2O
X
4
BaS04i
OH"
0,02
COj^-
X
S04^-
y y
197x + 233y =
4,3
(1)
Dung djch A tac dung vdi dung dich H2SO4 diT:
+ 2H* -> CO2 + H2O
0,224
COj^-
0.01
=> x = 0,01
Tir(l) => y = 0,01 mol
22,4
= 0,01 mol
151
Phan

djing
vh phuong phAp giai H6a hpc 11 VP cO - Bfi Xufln Hang
Ap dung dinh luat bao to^n dien tich cho dung dich A
ta
diTOc:
nNa+
=0,01.2+ 0,01.2-0,02
=
0,02
mol
Vay: mn,urfi =
nicaiion
+ m^nkm = m„ * + rn^^* + m^^2- + m,^2-
INa INMjj
^"[l
= 0,02.23
+
0,02.18 +0,01.60+ 0,01.96 = 2,38
(g)
Dang
4.
- Bai tqp v4 C, CO, Si va licJp chdt cua silic
-
Bai tap tdng Mp cacbon - silic
BAI TAP IVlAU VA BAI TAP AP D^NG
Bai
1.
Mot
loai quSng
sat

dung
de
luyen gang, thep
c6
chtfa 10%
Si02 va 80%
Fe304,
con lai la tap
chat. Tinh
ham
li/dng
cua Si va Fe c6
trong loai quSng
tren.
Gidi
28
10
* Ham liTcfng Si trong quSng
:
%Si = —x
x
100%
=
4,67%
^
• ^ 60 100
* Ham liTdng Fe trong quSng
:
%Fe
=

^^x —x
100%
=
57,93%.
232
100
Bai
2.
Do't
mau
than
da
(chiJa
tap
cha't khong chay)
c6
kho'i liTdng 0,6kg trong
oxi
dir,
thu
dufdc l,06m^ (dktc) khI cacbonic. Tinh thanh phan
%
khoi liTdng
cua cacbon trong mau than da U-en.
Gidi
C
+ O2 —^ CO2
47,32 mol 47,32 mol
1,06.10^
„

ncoj =
—^77-
=
47.32 (mol)
% khoi liTdng cua
C
trong mau than da:
%C
=
100%
=
94,64%.
600
Bai
3.
Khuf
m gam mot
oxit
s^t
bang
CO d
nhi^t
do cao thu
difdc 11,2g
s^t va
6,72 lit khi
CO2
(dktc).
a) Xac dinh cong Ihtfc cua oxit
s^t.

b) Tinh
the
tich dung dich
HCl 0.4M can
diing
de h6a tan het m gam
oxit
s^t
noi tren.
Gidi
a) Oxit
s^t
c6 dang FCxOy. 152
iciiATgfiirwiyr
np,
=
= 0,2 (mol)
; nco^ =
= 0,3 (mol)
56
FexOy + yCO
Ta
c6 ti le :
22,4
+ yC02
>
xFe
0,2
mol 0,3 mol
y

X 0,2 2
3
0,2
0,3 y 0,3
>C6ng
thtfc oxit
s^t la
FejOj.
) Fe203
+ 3CO 2Fe + 3CO2
nFejO,
=~'^Fc =0,1 (mol)
FejOs
+
6HC1
>
IFeCh + SHjO
0,1
mol 0,6 mol
n
0,6
-M
" 0,4
'
dd
HCl
-
=
1,5
(lit).

Bai
4. Cho hon hdp
silic
va
than
c6
khoi lifdng
20 gam tac
dung
vdi
lifdng
diT
dung djch NaOH
dac,
dun
nong. Phan ufng giai phong
ra
13,44
lit khi
hidro
(dktc).
Xac
dinh thanh phan
%
khoi li/dng
cua
sihc trong
hon hdp ban dau,
biet r^ng phan tfng xay
ra

vdi hieu suaft 100%.
Gidi
Si
+ 2NaOH + H2O >
Na2Si03
+
2H2t
0,3
mol 0,6 mol
13,44
22,4
= 0,6 (mol) => msi = 0,3.28
= 8,4 (g)
%Si= —.100%
=
42%.
20
Bai
5. X Ih hdp
chat
hoa hoc tao ra
trong
hdp kim gom Fe va C
trong
do c6
6,67% cacbon
ve
khoi li/dng.
,
a)

Lap cong
thiirc
cua
X.
|b)
H6a tan X
&-ong
HNO3
6&c nong
thu
dUdc dung dich
A va hon hdp khi B.
Cho A, B
Ian
li/dt tac dung vdi dung djch NaOH diTthi A tao ra
ket
tua Ai,
B
tao
ra
hon hdp Bi gom
ba
muoi. Nung Ai
va Bi d
nhiet do cao A| tao
ra
oxit
A2,
B, tao ra hon hdp
B2

g6m hai
muoi. Cho
CO
khuf
A2
d
nhi?t
do cao thu
drfdc
hon hdp A3 gom bon
cha't
r^n. Cho B2 tac
dung
vdi H2SO4
loang
thu
dufdc
khi
B3
va
axit
B4.
Cha't
B4
lam mat mau
dung dich KMn04 (trong
moi
trU"dng axit). Viet cac phiTdng trinh phan tfng.
(Trich TSDHNaoai thum^ TP.HCM)
153

Phan
djing va
phuong
phAp
gai H6a hpc 11 VP CO - D8
XuSn
Hung
Giai
a) Dat cong thtfc ciia X
la
Fe^Oy.
%Fe = 100% - 6.67% =
93,33%
Tacdtil?: x:y= ^^:^ =
1.67:0,56
= 3:1
56 12
=> X c6 c6ng thurc phan tur la FcjC.
b) PhiftJng trlnh phan tfng :
FejC +
22HN03d^c
> 3Fe(N03)3 + BNOzt + COzt +
1IH2O
+ dd A :
Fe(N03)3;
hon hdp khi B: |
[CO2
Fe(N03)3
+
3NaOH

>
Fe(OH)3>l'
+
SNaNOj
+ ket tua A, : Fc(0H)3
2NO2
+ 2NaOH
>
NaNOz + NaN03 + H2O
CO2
+ 2NaOH
>
NazCOj +
H2O
+ hon hdp B| :
NaNOj
NaNO,
Na^CO,
2Fc(OH)3
Fe203 + SHjO
NaNOi
NaN02+
-O2T
+ oxil
A2:
Fe203;
hon hdp Bj:
NaNOj
Na^COj
>

3CO2
+ 2Fe
>
2Fe304 +
CO2
FejOj + 3C0
3Fe203 + CO —
FciOi
+
CO
>
2FeO +
CO2
fFc
+ hon hdp A3 gom:
FejO^
FeO
FejOj dif
NazCOj + H2SO4 > Na2S04 +
C02t
+ H2O
2NaN02
+ H2SO4 > Na2S04 +
2HNO2
+ khi B3:
CO2;
axit B4:
HNO2
5NO-
+ 2MnO; + 6H* >2Mn^*+ 5NOJ +

3H2O.
6. Cac hdp cha't canxi silicat la hdp chS't chinh cua xi mang. Chdng c6 thanh
ph^n nhir sau : CaO : 73.7%; SiOj : 26.3% va CaO :
65,1%;
SiOz : 34,9%.
H6i
trong moi hdp chat canxi silicat tren c6 bao nhieu mol CaO kd't hdp vdi
1 mol
Si02.
Gidi
Dat
s6'
mol CaO.
Si02
Ian lufdt 1:1 a, b mol.
THl :Tac6til^: x:y = ^:^ =
1,31:0,44
= 3:1
=>Thanh phan hdp chat silicat:
3CaO.Si02.
TH2:Tac6tne:x:y = ^:^ =
1.16:0,58
= 2:1
56 60
=>Thanh phan hdp chat silicat:
2CaO.Si02.
Bai 7. Nung hon hdp gom l,5g
Si02
va 3,6g C trong 16 di?n d nhi?t do cao thu
difdc r^n A va khi B (khi B chay diTdc trong khong khi).

'
a) Xac dinh thanh phan djnh tinh va dinh lifdng cua A.
b) Tinh the tich khi B thu diTdc (dktc).
Gidi
60
a) Si02 + 2C -
0,025 0,05
=0,025
(mol);
nc = ^ = 0,3 (mol)
12
Si + 2CO
0025 0,05
Vay r^n A gom
(mol)
Thco phi/dng trinh
:
nc = 0,05 mol => ncdu = 0,3 - 0,05 = 0.25 mol
Si:0.025 mol
C dir:0.25 mol
Khi B : CO : 0.05 mol
msi = 0.025.28 = 0,7 (g)
mc = 0,25.12 = 3(g)
mA = 0,7 + 3 = 3,7 (g)
=> %Si = —.100% = 18,92% => %C = 81,08%
3,7
l>)
Vco = 0,05.22.4= 1.12 (lit).
Wi
8. Khi dot chdy hon hdp khi

SiH4
va
CH4
thu di/dc mpt
s&n
pham r^n cSn
nang 6g
va
san pham khi. Cho san pham khi d6 di qua dung dich
Ca(OH)2
lay dir thu diTdc 30g ket tua. Xdc djnh thanh ph^n % the tich cua hon hdp
khi.
1SS
Phan dgng phDOng ph^p
giai
H6a hpc 11 V6
CO
- D5 XuSn Hung
Gidi
SiH4 + 2O2
SiOj
+
2H2O
0,1
mol 0,1 mol
CH4
+ 2O2 CO2 +
2H2O
0,3
mol 0,3 mol

CO2
+
Ca(OH)2
>
CaCOji
+
H2O
0,3
mol 0,3 mol
%VsiH,
=
—.100%
=
25%
=^
%VcH^
=75%.
"sioj
=-^ =
0,1 (mol); nc.co, ^^^^'^^^'""^^
Tong
so mol hon hdp
khi:
nnh
=
0,1
+ 0,3 = 0,4 (mol)
0,1
0.4
Bai

9. Xac
dinh thanh phan
% (ve the
tich)
cua hon hdp khi gom c6
N2,
CO va
CO2,
bi6't rkng
khi cho 10 lit
(dktc)
hon hdp khi do di qua mot
Itfdng
AM
niTdc
voi trong,
roi qua
dong
(II)
oxit
diT
dot
nong
thi thu
diTdc
lOg ket tua va 6,4g
dong.
Neu
cung
lay 10 lit

(dktc)
hon hdp khi do cho di qua ong
diTng dong
(II) oxit
dir dot
nong,
roi di qua mot
lufdng diT nuTdc
vol
trong,
thi thu
dU'dc
bao nhieu
gam ket
tiia?
Giai
CO2
+
Ca(OH)2
>
CaCOji
+
H2O
0,1
mol 0,1 mol
"caco,
=-j^ =
0.1(mol)
CO
+ CuO Cu + CO2

0,1
mol 0,1 mol
Hcu
=
—
=
0,1
mol
64
%Vcoj
=
%Vco
=
"'^^^^'"^.100%
=
22,4%
%VNJ
=
100%-(%Vco^ + %Vco) = 55,2%
Neu
hon hdp khi di qua CuO
triTdc
roi den di qua
Ca(0H)2
diT
thi ta c6 cac
phan tfng:
CO
+ CuO Cu + CO2
0,1

mol 0,1 mol 0,1 mol
CO2
+
Ca(OH)2
>
CaCOa
+
H2O
0,1
mol 0,1 mol
=>
So
mol
CO2
:
nco2
= 0,1 + 0,1 = 0,2 (mol)
=:>
ricacoj =0,2 (mol)
Vay khoi
liTdng
ket tua : =
0,2.100 = 20 (g).
•
10.
Mot
loai thuy tinh
c6
thanh phan
hoa hoc

dU'dc
bieu dien bang cong
thtfc
K20.Pb0.6Si02.
Tinh khoi
liTdng K2CO3,
PbCOj
va
Si02
can
dilng
de c6
the
san
xuaft
diTdc
6,77 ta'n
thuy tinh tren.
Coi
hieu suaft ciia
qua
trinh
la
100%.
Gidi
Tac6:
MK.o.Pbo.asiOj =677
Khoi Itfdng
moi
chat trong thuy tinh

:
6,77
m
K2CO3
m
617
_6,77
Pbco,
-
6,77
.138
= 1,38 (tan)
.267
= 2,67 (tan)
mc;^
=
.60.6 = 3,6 (tan).
Bai
11. Dan 8 lit hon hdp khi
CO2
va CO
chtfa 39,2%
CO2
(ve th^
tich)
cho tit
tCr
qua dung dich chtfa
17,lg
Ba(OH)2.

Tinh li/dng
ket
tiia
c6 mat sau khi
phan
tfng ho^n
tat.
Giai
CO2
+
Ba(0H)2 >
BaCOji
+
H2O
(1)
0,1
mol 0,1 mol 0,1 mol
3,136
V
=^:^ =
3,136
(lit)
*C02
n
'Ba(OH)2
100
^17,1
171
"002
=

=
0,14 (mol)
= 0,1 (mol)
22,4
nco2 J.
=0.14-0,1
= 0,04
(mol)
C6
phdn tfng
xdy ra
tiep
:
CO2
+
BaCOj
+
H2O
0,04
mol 0,04 mol
-> Ba(HC03)2
(2)
V$y liTqJng
ket
tiaa
c6 mat sau khi
phan tfng hokn tS't
1^
:
mcco,

=(0.1-0,04).197
=
11,82 (g).
Bai 12. D6't
chiy
6,8g hon hdp X gom
hidro
va
cacbon mono oxit
c^n 8,^6 lit
oxi (dktc).
Xic
dinh th^nh phan phan tr3m theo
the
tich
va
theo kh6'i Ix/dng
cia
hSn hdp X.
Gidi
Phufdng trlnh phSn tfng
:
157
Phan
dpng \ii
phuong
pWp
giai
Hoa hyc 11 VP CO - DO Xufln
Hiftig

2H2
+02-^
2H2O
X mol
—
mol
2
2C0 + O2 —
2CO2
y
mol ^ mol
n„ =-?l^ =
0,4mol=>-
+ ^ = 0,4
22,4 2 2
Ta c6 h?
phiTcJng
trinh
:
2x + 28y = 6,8
^+y=o.4
^
2 2
* Thanh phan %
theo
the tich hon hdp :
x = 0,6
y
= 0.2
(mol)

%V
=M.i00%
= 75%
0,8
%Vco=25%.
* Thanh phan %
theo
khoi lifdng hon hdp :
%m^^^ = 100% =
17,65%
%mco =
82,35%.
Bai 13. Nung
n6ng
7,2g
Fe203
vdi khi CO. Sau mpt
thdi
gian thu drfdc m(g)
chS't
rin X. Khi sinh ra hip thu het bdi dung dich Ba(OH)2
dtf(?c
5,91(g)
ket tua,
tiep tvc cho Ba(0H)2 du vao dung dich tren th^y c6
3,94(g)
ket tua nifa. Tinh
m.
Giii
HS'p thu

CO2
vao dung dich Ba(0H)2 thl c<3 k^t tiia
BaCOs
xuSft
hi^n, cho
tigjp
dung dich Ba(0H)2 du vao lai c6 ket tiia, chtfng t6
CO2
tic dung vcti
dung djch Ba(0H)2 sinh ra '2 mu^i.
CO2
+ Ba(0H)2 ->
BaCOji
+
H2O
(1)
5,91
0,03 = 0,03 mol
(2)
197
2CO2
+ Ba(0H)2 -> Ba(HC03)2
0,02 0,01
Ba(HC03)2 + Ba(0H)2 ->
2BaC03
i +
2H2O
(3)
• 0,01
TheophSntirngd), (2), (3)

3 94
-
i^
= 0,02 mol
197-—^
In^^
= 0,03 + 0,02 = 0,05 mol
CO2
Trong
phin
iJng
khi3r cic oxit b^ng CO, ta lu6n c6:
158
no
(trong
oxit)
=nco=
n^oj
=0,05
mol
. m = mp^^oj - mo = 7.2 -
0.05.16
= 6.4 (g)
Bki 14. Cho lu6ng khi CO di qua m (g)
Fe203
dun n6ng. thu dir^c 39,2 gam hon
hdp gom 4
chat
r^n la s^t kim loai \k ba oxit ciia n6, dong
thdi

c6 hSn hdp
khi
thodt ra. Cho hon hdp khi nay hS'p thu vko dung dich niTdc v6i u-ong c6 du
thi
thu du^c 55 gam ket tfia. Gid
tri
cda m la:
am
55
Tac6:
nr
= 0,55 mol
•ceo,
-
CO2
+ Ca(OH)2 ->
CaCOji
+
H2O
0,55 0,55
Trong phdn tfng khijf cic oxit b^ng CO, ta lu6n c6:
no
(trong
oxit)
=nco=
Hcoj
=0,55
mol
=> m = 39,2 + mo = 39,2 +
16.0,55

= 48 (g)
Bki 15.
KhiJr
hoan
toan 8,72 gam hSn hdp X g6m
Fe203
va FeO b^ng CO thi thu
dtfdc m gam
chS't
r^n Y va khi
CO2.
HSp thu
hoan
toan khi
CO2
b^ng
nir<5c
v6i
trong dvt thu
diiigJc
6 gam k6't tua. Tinh gid tri
ciSa
m.
Gidi
HSfp thu
CO2
vao Ca(OH)2 dvt => chi tao mu^i
CaC03.
ncaco3=j^
= 0.06 mol

CO2
+ Ca(0H)2
CaC034'
+
H2O
0,06 0,06
Ta c6:
ncophiniJng
= ^coi = 0,06 mol
Ap
dung dinh luat bao toan khoi liTdng, ta c6: m^x + mco = my +
mcoj
o my = 8,72 +
0,06.28
-
0,06.44
= 7.76 (g)
Bki 16. Cho hoi nude di qua than
n6ng
d6, thu
dugc
15,68 lit h6n hgrp khi X
(dktc) gom CO,
CO2
va
H2.
Cho toan bp X tdc d\ing h6t vdd CiiO (du) nung
'n6ng. thu dugrc h6n hgp
chat
rdn Y. H6a tan toan bp Y

bSng
dung djch
HNO3
(loSng,
du) dugrc 8,96 lit NO (san phki khu duy nhat, or dktc). Tinh ph^ trSm
ve the tich khi CO trong X.
Gidi
H2O
+ C —^ CO + H2
XXX
159
Phan
dgng va pluwng
ph&p
giai
H6a hpc
11
VP
cfl
-
P5
Xuan
Hung
2H2O
+ C
CO2
+
2H2
2y
y 2y

Ta
c6 : nx = 0,7 mol
=>
2x + 3y =
0,7
(1)
hhX {CO,
H2}
+ CuO Cu +
HNO3
-*
0,4
mol NO.
CO
+
CuO
->• CO2 + Cu
H2
+
CuO H2O + Cu
3Cu
+
8HNO3
3Cu(N03)2
+
2N0
+
4H2O
0,6
0,4

Trong phan ung
khur
oxi kirn loai boi CO,
H2
Tact:
n^po^ | =
no(i„,ngCuG)
=
ncu
= 0,6
mol =>
2x + 2y
=
0,6
(2)
Tir(l)(2)=^x
=
0,2; y
= 0,l
Vay
:
%Vco
=
—.100%= 28,57%
0,7
BAITAPTRACNGHllM
Cfiu
1. Cho 38 gam hon hdp gom
Na2C03
va

NaHCOj
tac
dung
vdi
dung dich
HCl sinh
ra
8,96 lit
CO2
(dktc). Vay khoi lifdng Na2C03 trong h6n hdp 1^:
A. 21,2
g B. 16g C. 10,6g D. 5,3 g
Cfiu
2.
Dung djch chaft nao sau day khong the chiJa trong blnh thuy tinh
A.
H2SO4
B.
HNO3
C. HF D. HCl
Cflu
3.
Nguyen to
R
tao thanh hdp chat khi vdi hidro c6 cong
thuTc
h6a hoc
la RR,.
Trong hdp chS't c6
hoa tri

cao nhat
vdi oxi thi R
chiem 27,27%
ve
khoi lUdng.
Vay
R
Ik:
A.C
B.Si
C. S
D.Ge
Cfiu 4. Dot chdy ho^n to^n 8,96 lit hon hdp gom CO,
CO2
thi can dting
het 11,2
lit khong khi. Vay %
ve
th^ tich ctia CO trong hon hdp 1^:
A. 40%
B.
50%
C.60%
D.
70%
Cfiu
5. Din tir tir 5,6 lit khi
CO2
(dktc)
v^o

200
ml
dung dich
Ba(0H)2
0,5M.
Vay khoi Itfdng ket tiia thu
diTqJc
Ih:
A.
lOg B.
ll,4g
.
C. 19,7g
D.ke'ttiiatanhe't
Cflu 6. D6't
cMy
hohn toan m6t mau graphit nang lOg trong
dd c6
4% tap chS't.
Dan tohn bO
sin
pham
vko
500
ml
dung dich
Ba(0H)2
IM.
Vfiy khd'i Itfdng
kettuathudi/dcia:

A. 39.4g
B. 19,7g C.
13.13g
D. 12,3g
160
Cflu
7. DSn V (lit) khi
CO2
(dktc) vao 200 ml dung dich
Ca(OH)2
0,1M thu diTdc
Igam ke't tua. V$y V c6 gid tri
la:
A. 0.224 B. 0,336
C.
0,224 hoSc 0,672 D. 0,048
Cfiu
8:
ThSi mot luong khi CO diT di qua ong diTng h6n hdp
2
oxit Fe304 v^ CuO
nung ndng
den khi
phdn uTng
xay ra
hoan toan
thu
dtfdc
2,32 gam hon hdp
kim loai. Khi thoat

ra
di/dc di/a vao binh
duTng
dung dich
Ca(OH)2
dtf
thay cd
5 gam ket tua tr^ng. Khoi liTdng hon hdp 2 oxit kim loai ban dau
la
A. 3,12 gam
B.
3,21
gam
C. 4 gam D.
4,2
gam
Cfiu 9: D6't chdy hoan tohn 8,96
lit
hSn hdp gom CO,
CO2
thi cin dilng h6't 11,2
lit khdng
khi.
Vay %
ve the
tich cua CO trong hSn hdp
la:
A. 40%
B.
50%

C.60%
D.
70%
Cfiu 10: Cho tfif tit dung djch chiJa
a
mol HCl vao dung dich chtfa
b
mol
Na2C03
dong thdi khuSy deu,
thu
diTdc
V
lit khi
(d
dktc)
v^
dung dich X. Khi cho dir
nifdc
voi
trong
vao
dung dich
X
thay
cd
xufl't
hi$n
k€i
tua. Bieu thtfc lien

h$
giffa V vdia,
bla:
A. V
=
22,4(a
-
b). B. V
=
1
l,2(a
-
b).
C.
V
=
11,2(a
+
b).
D. V = 22,4(a + b).
"
Trich de thi tuyen sinh
Dai hoc
khoi A
nam
2007"
Cfiu 11:
Dot
chdy ho^n to^n
m gam

FeS2
b^ng
mOt lufdng
O2
vilfa
dii, thu
diTdc
khi X. Hap thu het X v^o
1
lit dung dich chtfa
Ba(0H)2
0,15M
vk
KOH 0,1M,
thu diTdc dung djch
Y v^
21,7 gam ket tda. Cho Y v^o dung dich NaOH, thSy
xuat
hi$n
them ket tiia. Gid tri cda
m
1^
A.
23,2 B. 12,6
,
C. 18,0 D. 24,0
HUGlNG
DAN
GIAI
Cfiu

1.
Viet 2 phiTdng trinh phdn tfng:
NazCOj
+
2HC1
^
2NaCl +
CO2
+
H2O
X
X
NaHCOa
+
HCl
-»
NaCl +
CO2
+
H2O
y
y
Tacd:
106x + 84y = 38
X + y = 0,4
=>
X
= y = 0,2
Khd'i Ivrdng
Na2C03

=
0,2.106 = 21,2
g
=>
Chpn A.
161
Cfiu
2. Khong
diing
binh thuy
linh
de dUng HF
vi:
4HF
+ SiOz
-*•
SiF4 +
2H2O
=>
Chon C
Cfiu
3.
Cong thtfc hdp chat khi Ih
RH4,
nen cong thuTc cua hdp chat c6 hoa tri
cao nha't vdi oxi la
RO2
Taco: ^27^ R = 12, vay R la C Chon A.
R
+ 16.2 100

Cfiu
4. The
tich
oxi trong khong khi da dung =
20%.
11,2 = 2,24 lit
PhiTdng
trinh
chay 2CO +
02^
2CO2
The
tich
CO = 2,24.2 = 4,48 lit => %C0 = (4,48/8,96). 100% = 50%
=>
Chon B.
Cfiu
5. S6'
mol CO2
= 0,25 mol, so mol
Ba(0H)2
= 0,
Imol.
Xet ti le
nco2
=
nBa(OH)2
>
2 do do ket tua tan het.
=>

Chon D.
Cfiu
6.
Khoi
liTdng
C nguyen chat = 10.96% = 9,6g,
so mol C =
9,6/12
= 0,8 mol
PhiTdng
trinh
phan drng C +
O2
^
CO2
So mol
CO2
= 0,8 mol, so mol
Ba(0H)2
= 0,5mol.
Xdt
ti 1$ nC02:
nBa(0H)2
= 1.6 nen xky ra 2 phan
tfng
sau day
CO2
+
Ba(OH)2
-> BaCOj +

H2O
XX
X
2CO2
+
Ba(0H)2
Ba(HC03)2
2y
y y
Ta CO
h^ phOtfng
trinh:
x + 2y = 0,8 x + y = 0,5
Gidi
ra ta dtfcfc x = 0,2 y = 0,3
Khoi
liTdng
BaCOj =
0,2.197
= 39,4g. => Chon A.
Cfiu
7. S6' mol ket tua = 0,0Imol, so mol
Ba(0H)2
= 0,02 mol
Tri^ng
hffpl.
So mol
CO2
chi du ta
3

0,01 mol ket tua
CO2
+
Ca(OH)2
-> CaCOj +
H2O
0,01 mol 0,01 mol
The
tich
CO2
= 0,01.22.4 = 0,224 lit
rrU&ng
h^p
2.
So mol
CO2
dir h6a tan m6t phan ket
tiia
CO2
+
Ba(OH)2
-> BaCOj +
H2O
0,01 0,01 0,01
2CO2
+
Ba(0H)2
->
Ba(HC03)2
2y

y y
162
Ta
CO 0,01 + y = 0,02 =>y
=
0,01
Tong
so mol
CO2
= 0,01 + 0,02 = 0,03 mol
The
tich
CO2
= 0,03.22,4 = 6,72 lit
Vay
chon C.
Cfiu
8: Cdc phan urng
Fe304
+ 4C0
—3Fe
+
4CO2
CuO + CO —^ Cu +
CO2
CO2
+
Ca(0H)2
> CaCO, +
H2O

Ho
(trong
oxio = "co =
n
f^Q^
= n
Caco^
~
0,05
mol
m„xii
=
niKi
+ m,„i,r„ngoxit = 2,32 + 0,05.16 = 3,12 (g).
=>
Dap an A.
Cfiu
9: The
tich
oxi trong khong khi da dung =
20%.
11,2 = 2,24 lit
PhU'dng
trinh
chay
2CO
+
O2
^
2CO2

The
tich
CO = 2,24.2 =
4,48
lit
=>
%C0 =
(4,48/8,96).
100%.
= 50%
=:>DdpanB.
Cfiu
10: Khi cho dung dich HCl
tiif
tir vao dung djch NazCOj, xdy ra phdn xtng
theo
trinh
tiT
sau:
HCl
+ Na2C03-> NaHCOj + NaCl (I)
b
b b
HCl
+ NaHCOj -> NaCl +
CO2
+
H2O
(2)
(a - b) (a - b)

Sau phan
ifng,
cho dung djch
Ca(0H)2
diT
vSo dung dich X c6 ket
tiia,
chiirng
to sau phan
tfng
(2) NaHCOj
diT => HCl
het =>
tinh
the
tich
CO2
theo
HCl.
Theo phan tog (2) => n^o^ = (a -
b)
mol
V^QJ
= 22.4(a - b) (Ut)
=>
Dap dn A.
2Fe203
+
8SO2
(1)

0,3
Cfiu
11: Ptptf:
4FeS2
+
IIO2
0,15
Tac6
nB„oH)2
= 0,15 mol
nKOH
= 0,l mol
n
2^ = 0,15 mol
n^^.=0,4mol
Khi
cho
SO2
vao dung djch X thu diTdc 21,7 (g) -» BaS03. Cho Y
i&c
dung
vdi
dung djch NaOH thay xua't hien them ket tiia, chtfng to trong dung dich Y
c6 ion HS03'.
VI:
Ba^* +
HSO3-
+ OH"
->
BaS03 i +

H2O
han
dgng
phuong
ph^p giSi H6a hgc 11
VP
CO
-
D8
Xufln
Hung
Tac6:nBaco,
= 0,1 mol
Ptptf: SO2 + 20H- > SOi^- + H2O (2)
0,1 0,2 0,1
SO2
+ OH" >
HSO3"
(3)
0.2 0.2
Ba^*
+ SOj^- > BaS03 i
0,1 0,1
Tac6: n _= 0,4 - 0.2 = 0,2 mol
OH
Theo
ptptf
(2), (3) ta c6:
n^o^
= 0.1 + 0,2 = 0.3 mol

Theo
ptptf
(1) ta c6:
rip^s^
=
"J^soj
= 0.15 mol
=>
mpcsj
= 120.0,15 = 18(g) =>Ddp an C.
BAI
TAP
Ti;
GlAl
)ai
1.
Vie't
cic phifdng
trlnh
hoa hoc cua phan iJng bieu dien sd do chuyen h6a
sau :
a) C
> CO2
>
CO
>
FejOi
>
Fe(N03)3
>

FezOj
b)
CO2
>
NuiCOi
>
Na2Si03
>
HaSiOj
>
SiOz
c)
CO2
>
CaC03
>
Ca(HC03)2
> CO2 >
C
>
C
>
CO2
d)
Si
>
SiOz
>
K2Si03
>

K2CO3
> KCl
/
\
SiF4
CO
).C0Cl2.
)ai2.
a) Um the n^o de phan biet
khi CO2
khi
O2
b)
L^m thd' n^o d^ phan bipt muoi natri
cacbonat
va muoi natri sunfit?
Hudng
dJn :
a)
Dilng
que dom c6n than hong bung chay ->
O2.
b)
Cho tac dung vdi axit HCl sau do dan san pham qua nU'dc brom -> Na2S03.
[Jai
3. Cho
c&c
cha't:
silic,
natri silicat, natri

cacbonat,
magie silixua, axit
silixic.
Hay
lap th^nh mpt day chuyen hoa
giffa
cac cha't tren va viet cac phiTdng
trinh
hoa hoc.
Sal
4. Khi nung mot hon hdp gom cat tr^ng va than coc trong 16 di^n den
3500"C. thi thu di/dc mot hdp chat chufa khoang 70% Si va khoang 30%C.
Viet
phiTdng
trlnh
h6a hoc cua phan i?ng do, biet
ring
mot trong c^c s^n pham
cua phan i?ng la cacbon mono
oxit.
Bip so : Hdp chat
Ih
SiC.
Ski
5. Xdc dinh the
tich
hidro (dktc) thoit ra khi cho
Irfdng
dtf dung dich natri
hidroxit

t^c dung vdi mot hon hdp thu duTdc bing cAch nau chay 6g magie
vdi
4,5g
silic
dioxit.
Gia
sijT
phan
tfng
di/dc tien h^nh vdi hi^u
sua't
100%.
Bips6:
VHJ
=3,36
lit.
I
HUcJfnfi
dan: 2Mg + Si02 2MgO + Si
Si
+
2NaOH
+
H2O
> Na2Si03 + 2H2t.
Bai
6. Cho dung dich mu6'i Ca(HC03)2 tdc dung vdi cdc chS't sau :
K2CO3,
Ba(N03)2,
NaOH,

HNO3,
Ca(0H)2,
Ba(OH)2.
Viet
phi/dng
trlnh
phan urng
xay ra (ne'u c6).
Biki
7.
Bing
phiTdng
phap
hoa hoc hay nhan biet:
a) CaCOa, Na2C03,
Na2S04,
CaS04.2H20
(dilng
ntfdc v^ dung dich
HCl).
b)
Dung dich : NaOH, Na2S03, CaCb, Ca(HC03)2, Na2S04.
Bai
8. Hoan th^nh cac phtfdng
trlnh
hoa hoc sau :
1.
Mg + Si > 2.CaO + C-^^? + C0
3.
Fe304

+ CO ? +? 4. Si + F2 >?
5. Si + KOH+? >? +
H2T
6.
CO2 diT
+ Ba(0H)2 >
7. C +
KCIO3
> 8. C + S >
9. C + H2S04dH- )-?+? + H20
10.
Na2SiO3 +? +
H2O
>? + NazCOj.
Bai
9. Khi nung 97,7 gam hon hdp X gom
NH4HCO3,
NaHC03 v^ Ca(HC03)2
den
khoi
li/dng
khong doi thu difdc 32,4 gam chat r^n Y. Cho cha't r^n Y tic
I
dung het vdi dung dich HCl lay diT, thu diTdc 4,48 lit khi
(dktc).
Xic dinh
•
thanh phan % cac muoi trong hon hdp.
Dap so: %NH4HC03 = 32,37%; %NaHC03 = 34.43%; %Ca(HC03)2 = 33.2%.
Hudn};

dan :
NH4HCO3
-A
NH3
+
CO2
+
H2O
2NaHC03 NazCOj +
CO2
+
H2O
Ca(HC03)2
CaO +
2CO2
+
H2O
R^n
Y gom Na2C03 \h CaO.
hki 10. Khur a gam mot
oxit
s^t bkng khi CO d nhi$t dp cao thu diTdc 0.84g s^t
\h 0.88g
CO2.
a) Xac dinh cong thtfc
oxit
s^t da diJng.
b)
Tinh
the

tich
dung dich HCl 2M can de h6a tan het a gam
oxit
s^t ndi tren.
Daps6': a)
Fe304
b)VHci =
20ml.
165
Phan
dgnp vi phuong ph^p
giai
H6a hpc 11 VP co - P8 Xuan Hung
Bai 11. Cho 2,464 lit khi
CO2
(dktc) vao dung dich NaOH, sinh ra
1
l,44g hon
hcJp hai muo'i. Xac dinh khoi ItfcJng moi muo'i trong hon hdp thu dtfdc.
Ddp so':
niN.jCo,
= lO-^g ; mNaHco, = 0.84g.
Bai 12. Mot dung djch chufa a mol NaHCOj va b mol NazCOj.
a) Khi them (a + b) mol CaClz hoSc (a + b) mol
Ca(0H)2
vao dung dich tren
thi khoi liTdng kct tua thu dU'dc U-ong hai trUfJng hdp c6 b^ng nhau khong?
Giai thich?
b) Tinh khoi liTdng ket tiia thu difdc trong truTcfng hdp a =
0,1

mol va b = 0,2 mol
Dap so
:
a) LiTdng ket tua khong bang nhau.
h)
mc,co.,
(1)
= 20g
;
m^co,
(2)
= 30g.
Bai
13.
Nhan biet cdc chat sau :
a) Cac dung dich BaC^, NaHCOj, Na3P04,
H2SO4
va NajCOj (khong dung
thuo'c thijf n^o khac) ^
b) Cac khi:
CO2, SO2,
CO, O2 va
H2S.
c) CaCOj,
FejOj,
CuO,
Si02
(chi dung mot hoa cha't).
Bai 14. Tir nguyen lieu chinh la KCl, da voi, khong khi, niTdc va chat xuc tac c6
dii,

hay viet cac phiTdng trinh dieu che cac chat:
K2CO3, NH4HCO3
va
NH4NO3.
3ai 15. Cho hon hdp BaCOj,
(NH4)2C03
tac dung vdi dung djch HCl diTdiTdc dung
djch A va khi thoat ra. Cho A tac dung vdi dung dich
H2SO4
loang difdiTdc dung
djch B va ket tua. Cho
B
tac dung vdi dung djch NaOH dtf dtfdc dung djch C va
khi.
Viet cdc phiTdng trinh phan ufhg dang phan tijf
\h
ion thu gon.
Dap so : dd A
Bad,
^ +khi:C02
NH4CI
dd B :
NH4CI
+ ket tiia : BaS04
dd C
:
NaCl + khi: NHj.
B^i 16. Cho luong hdi nifdc qua than nong do, sau khi loai het hdi niTdc thu diTdc
hon hdp khi X gom
CO2,

CO v^
H2.
Tron hon hdp X vdi oxi
diT
vao Wnh kin
dung tich khong doi
diTdc
hon hdp khi A d nhi^t dp 0"C v^ ap suat Pi. Dot chay
ho^n to^n A roi dito blnh ve nhi?t dp 0"C tW ip suat cua khi trong binh (hon
hdp B)
Ih
P2
= 0,5?,. Neu cho NaOH r^n
vho
binh de hap thu khi
CO2.
c6n lai
mot khi duy nha't (nhi^t dp trong binh 0"C) thi dp suat trong binh la
P3
= 0,3P|.
a) Tinh th^nh phan % the tich
cic
khi trong A.
b) Can bao nhieu kg than cd chiJa 4% tap cha't trd de thu dUWc 1000m' hon
hdp X do d 136,5"C
va
2.24 atm. Biet rhng cd 90% cacbon da bi dot chay.
Dap so': a) %Vcoj =10%; %Vco= 10%;
%W^^^
=30% va %Vo^ =50%.

b) mc = 370kg.
Bai 17. Cho x mol CO2 hap thu hoan toan vao dung djch chtfa y mol
Ca(OH)2.
N6u cac kha nang cd the xay ra giai cdc b^i
loAn
tiTdng i?ng.
"^lai 18. Hoan th^nh cac phan ijrng hda hpc sau :
X + NaOH +
H2O
>
Y +
H2t
Y +
E(ichi,
+
H2O
>
Zi + NajCOj
Z T + H2O
T + HF >M + H20
T + N >Si + P
P + FejOj
>
Fe + E
E + Mg
>
N + MgO
N +
H2O
P +

H2.
t"
Daps6':X:Si Y: NajSiOj E:
CO2
Z: H2Si03
T:Si02 M:SiF4 N: cacbon (C) P: CO.
Bki
19. Dan 5,6 lit
CO2
(dktc) vao binh chtfa 200ml dung dich NaOH nong do
aM;
dung dich thu diTdc cd kha nang tac dung to'i da 100ml dung dich KOH
IM.
Tim gia tri cua a.
Dip
so : a = 2
Bki
20. Hap thu V lit CO2 (dktc) vho dung dich
Ca(0H)2
thu diTdc lOgam ket
tua. Loai bd ke't tua roi nung ndng phan dung dich c6n lai thu diTdc 5 gam
ket tua nffa. Tinh V.
Ddpso': V = 4,48 lit
167