bộ đề hóa học 9 ôn thi cho lớp 10

5450-0
NHA GIAO LfU TU PHAM SY LL/U
THI/VIEN Vm
Nhiem
vu cua cac trUdng chuyen, 16p chon clUdc nganh Giao due
giao
nhiem vu la dao tao nguon nhan
lUc,
boi dudng nhan tai de gop ph^n day
manh
sU
nghiep cong nghiep hoa va hien dai hoa dat nudc.
De
dap ilng yeu cau do, Nha sach Khang Viet xin tran
trong
gidi thieu cuoh
sach:
Bp de Hoa hoc 9 on thi vdo 10 cua Nha
giao
Uu
tii Pham Sy
Luu.
Sach
gom 3 noi dung chinh:
-
Phan 1; Cac phUdng phap giai nhanh bai tap hoa hoc.
-
Phan 2: Tuyen chon gi6i thieu de thi tuyen sinh
ciaa
mot so'trUdng chuyen.
-

Phan 3 : Ba mUdI de thi thii.
Phan
phiftfng
phap : giiip cac em c6 each giai dung, hay va nhanh gon
nhat, thich hdp v6i
tufng
dang bai tap. Moi phUdng phap
dUdc
minh hoa
bang nhieu vi du dl hieu de cac em c6 the van dung
dUdc
mot each nhanh
nhat. De dam bao yeu cau chuan kien thiic, ki nang nen cac vi du da
dUdc
can
nhfic,
Ixia
chon chu yeu tii cac de thi DH & CD cua bo GD & DT cung
vdi Idi giai
dUdc
trinh bay chi
tiet
theo
each giai tU luan phu hdp vcti kieu ra
de thi tuyen vao cac trUdng chuyen va 16p chon.
Phan
giofi thieu de thi tuyen sinh cua cac
trtfofng
chuyen : gom 30
de cua nhieu trUdng chuyen

tren
ca nUdc
trong
5 nam gdn nhat. Moi de bao
gom nhieu kien thiic
trong
tam va nhieu ddn vi kien thiic nang cao, doi hoi
tinh ta duy sang tao, sU van dung Unh
boat
cac phUdng phap giai
toan
hoa
hoc hien dai. Phan nay giup cac em trUcfc ki thi tuyen can nhdc, chuan bi
cho minh kien thiic va ki nang lam bai hieu qua de dat thanh tich tot nhat.
Phan
de thi thii: gom 30 de. Moi de c6 kien thiic bao
quat
nhieu van de
trong
tam va nang cao cua chUdng trinh hoa hoc
THCS
c6 do kho nhat
dinh,
doi hoi hoc sinh phai hieu biet, suy luan mcl rong, c6 kl nang van dung
linh
boat
cac phiidng phap giai
toan
hoa hoc. Nhieu van de dUdc lap lai ci
cac

de khac nhau nhSm giiip cac em nim dUdc kien thiic va phUdng phap
giai tot nhat.
Cuo'n
sach nay la cong cu hiiu ich khong the thieu doi v6i hoc sinh
trong
qua trinh hoc tap va on luyen mon hoa chuan bi cho ki thi tuyen vao 16p 10
chuyen, sach cung giiip cho cac bac phu huynh c6 dUdc tU lieu de c6 phUdng
hu6ng
tU
van, giiip dd cac em hoc tap dat ket qua tot
trong
ki thi.
Mac
du da rat c6
gkng
va dau tU nhieu
thbi
gian cung v6i nhiing tich luy
trong
nghe
ciia tac gia nhUng chac chan khong tranh khoi nhiing thieu sot,
chiing
toi rat mong nhan diidc nhiing dong gop cua ban doc cho noi dung
ciia
cuoh sach hoan thien hdn
trong
Ian tai ban. ;
Xin
chan thanh cam dn!
O/

TNHHMTVDVVnKhan^
Vi6l
Phan I:
PHimNG PHAP GIAI
TOAN
HOA HOC
1.
Quy
tac
thCr
ta
uu tien cua phan Crng
(/)
Phan
ifng
oxi hoa khu:
-
Phan irng
theo
chi^u:
chat
o.xi hoa
manh
licfii
oxi hoa
chat
khit
nuiuh
hc/ii
tao

thanh
chat
o.\i hoa yen hem vd
chat
khifyeh
hon. vu ,
-
Khi xay ra phan irng ciia h6n hop chat oxi hoa vori h6n hap cha't khir:
Phan
I'Oi^
mi
tien
la
chat
o.xi hoa
manh
nlid't o.xi hoa
chat
klu't
nianli
nhat
tao
thdnli
chat
o.xi hoa yen
nhat
vd
chat
khi'f
yen

nhat.
(2)
Phan
itng
cua cac chat
dien
li
trong
dung
dich
theo
thii
tit uu tien sau:
-
Phan ii-ng trung hoa:
H"^
+
OH"
-> HgO
-
Phan irng tao ket tiia hidroxit:
M"^ +
nOH"
->
M(OH)„
-
Phan irng hoa tan hidroxit lu5ng ti'nh
trong
k'lim
dir:

M(OH)„
+(4-n)0H
->[M(0H)4f
Hay :
M(OH)„
+
(4 -
n)OH
->
MO^^ + 2H2O
i'i
Vi
du 1. Cho 29,8 gam h6n hop bot gom Zn va Fe vao
600ml
dung dich
CUSO4
0,5M. Sau khi cac phan ling xay ra hoan toan, thu diroc dung djch X va 30,4
gam h6n
hofp
kirn loai. Phdn tram va khoi lucmg ciia Fe
trong
h6n hop ban
ddula
A.
56,37%
B.
64,24%
C.
43,62%
D.

37,58%
Gidi
29 8 29 8
"CUS04
= 03
moK ig- =
0,450
< (n^,, +
np,)
<
^
=
0,532
Zn + CUSO4
^
Cu +
ZnS04
x->x->
X
mol
56
(I)
Fe +
CuS04 ^Cu
+
FeS04
CUSO4
het.
(2)
y^y-> y mol

Nd'u
chi
CO
Zn phan urng, Fe kh6ng phan ling thi: n^,,
= n^u =
0,3 mol.
Kh6'i
luong kim loai sau phan
urng
giam so vai ban
diJu:
y;
=>mKL =mp,
+mc., - 29,8 - Am
=
29,8-0,3
x
(65-64)-29,5
gam
<
30,4 gam .
Neu
Fe cung tham gia phan iing het thi kim loai chi c6 Cu .
Trai
d^
ra.
Vay Zn het, Fe con du. Zn phan ung lam kh6'i luong kim loai sau phan ii:ng
giam, Fe phan ling lam khd'i luong kim loai sau phan irng tang:
He PT:
ncuS04 =(x

+
y)
=
0,3mol
fx
=
0,2
Jmz„
=0,2x65
= 13 gam
Am =
(-lx
+
8y)
=
0,6gam
^[y = 0,l
^[mp^
=
29,8-13
= 16,8 gam
>%mp^ =
—xl00%
=
29,8
56,37%
Chon
A.
//. V(
Ik•

. V;
//// v',-;.' !0
-
Phm
S/Liru
Vi
du 2. Cho 1,68 gam Fe va 0,36 gam b6t Mg tac dung v6i 375ml dung djch
CUSO4,
khua'y nhe cho de'n khi dung djch mat mau xanh. Nhan tha'y
khd'i
lugnig
kim
loai
thu duoc sau phan
ling
la 2,82 gam. N6ng d6 mol cua dung dich
CUSO4
ia
A.0,15M
B.0,10M C.0,20M D. 0,25M
Gidi
=0.03mol;
n^g =0,015mol
Phan
ling
xay ra
theo
thir
tu uu tien sau : • , !
Mg

+ CuS04->.MgS04 + Cu (I) • '
Fe
+
CUSO4
^
FeS04
+ Cu . (2)
Sau phan irng dung dich mS't mau xanh chung to rang
CUSO4
da phan ung het,
CO
the con du kim
loai
sau phan
ling.
Gia
sij Fe phan iJng het khdng con du
thl
khoi
luong Cu thu dugc se la :
mc„
= (0,03 + 0,015)
X
64 = 2,88 gam > 2,82 gam (theo gia thie't).
Fe con dir sau phan ung.
Goi X
la so mol Fe phan
ling.
Am :
do tang

khoi
luong kim
loai.
Ap
dung phuong
phap
TGKL
ta c6:
Am
=(64 - 24).0,015 + (64 - 56)x = 2,82 - (1,68+0,36) = 0,78gam
0,015 +
0,0225
0,375
>x
= 0,0225mol =>
CM(CU.SO4)
=
-—rrri
= 0,1 M Chon B
Vidu 3. Cho m, gam Al vao 100ml dung djch gom Cu(N03)2 0,3M va
AgNOj
0,3M.
Sau khi cac phan
ling
xay ra hoan toan thi thu duoc m2 gam chat rdn X.
Neu
cho m2 gam X tac dung
v6i
luong du dung dich HCl thi thu duofc
0,336

lit
khi
(a
dktc).
Gia trj
ciia
m, va m2 Idn iuot la
A.8,10va5,43. B. 1,08 va 5,43. C. 0,54 va 5,16. D. 1,08 va 5,16.
Gidi
ncu(N03)2 "
"AgNO,
=
0,1-0,3
= 0,03
mol;
n^^ = 0,015 mol.
Phan
ling
khir
AgNO^
va
Cu(NO,)2bori
Al
theo
thii
tu uu tien, sau do chat rdn
2
tac dung
v6i
HCl tao khf H2=> Al duda

khijf
HCl: n^nu^, =-n^^
Xet
toan qua
trinh
: - Chat
khir:
Al.
-
Chat oxi hoa :
AgNO,,
Cu(N03)2, HCl.
Al
+
3AgNO,-^3Ag
+
Al(N03)3
(1) 0
0,01-^0,03^ 0,03^ 0,01
2Al
+
3Cu(NO,)2
^2Al(N03)3+3Cu (2)
0,02 -> 0,03 -> 0,02 0,03
•-y
iiHiii
i-ii f uy Til luiuiig
Tici
6HC1 + 2A1
2AICI3

+ 3H2
0,03 <-
0,01
<-
0,01
<- 0,015 mol
(3)
=>
n^i = 0,01+ 0,02 + 0,01 =0,04mol =>m, =0,04x27 =
l,08gam
m2
=
m^g
+ +
niAKDu)
= 0,03(108
+
64)
+0,01 x
27 = |5,43 gam
=> Chon B.
Vi
du 4. (DHB 2009) Cho 2,24 gam b6t sdt vao 200ml dung dich chiia h6n hop
gom
AgNOj
0,1M va
Cu(N03)2
0,5M. Sau khi cac phan
ting
xay ra hoan toan,

thu
duoc dung djch X va m gam chat rdn Y.
Tinh
gia trj cua m.
A.
5,08 gam B. 4,08 gam C. 3,72 gam D. 6,24 gam
Gidi
npe =0,04
mol;
nAgwo,
= 0,02 mol;
^c^^no^,)^
=^,1 mol
Su
khijf
xay ra
theo
thii
tu uu tien:
AgNO,
tac dung het sau do Cu(N03)2 bi
khii.
Fe + 2AgN03 ^2Ag +
Fe(N03)2
0,01<-0,02-). 0,02
(1)
(2)
,01/::.:;
14
Fe + Cu(N03)2 ^Cu + Fe(N03)2

0,03 0,10
0,03^ 0,03-> 0,03 , j.|
0,00 0,07
Theo cac
PTHH
: Cu(NO02 con du 0,07 mol, Fe phan
ling
hd't.
Chon
B.
m,.„
= 0,02
X
108 + 0,03x64 = 4,08 gam
Vidu 5. Hoa tan 13,8 gam Na2C0, vao nude.
Vira
khuay
vifa
them
tirng
giot dung
djch
HCl IM cho t6i dii 180ml dung dich axit, thu duoc V lit khi.
Viet
phuong
trinh
phan ufng xay ra va tmh V
(dktc).
'
"K2CO3

=
0,1 mo! =^
HHCKcdndung) = 0,2mol >
HHCKCIC^ra)
= 0,18mol
PTHH
theo
thii
tu uu tien:
HCl
+ Na2C0, -> NaCl + NaHCO,
HCl
+ NaHCO, NaCl +
CO2
+
H2O
=
0,18-0,10 = 0,08
mol
m
nco2
= n
HCl
~"Na2C03
V
= 0,08x22,4= 1,792
lit
Vi
du 6. Them tir tir dung djch NaOH 2,5M vao 400ml dung djch X chiia HQ IM va
AlCl,

0,5M.
Ti'nh
the
tich
dung djch NaOH cin dung de thu duoc ket
tiia
16n nhat.
Gidi
Phan
ling
theo
thii
tu uu tien: ; •
NaOH
+ HCl ^ NaCl +
H2O
(1) '
Dp
dS/l^>a
hoc y on
Iht
y^o
IP-Phaa
3/ MT
3NaOH
+
AlCl,
Al(OH), + 3NaCl (2)
NaOH +Al(OH),->Na|Al(OH)4l (3)
H*

+ OH ->H:0 (1)
Ar'* + 30H ->Al(OH), (2)
"nci = 0,4x1 =0,4mol;
11^10,
=0,4x0,5
= 0,2 mol
De thu
dirge
lircnig ket tiia
nhieu
nhat:
khoiig
xiiy
ra
phau
iriig
(3).
Tir(l)
va (2) ta c6: nf^yQ^ = 3nA|Q^ + =3X0,2 + 0,4 =
1
mol
Vay the ti'ch
dung
dich NaOH can
dung
la : |V = 400 ml
Vidu
7.
Dung
dich D g6m cac

chat
NaAlO, 0.16 mol;
Na,S04
0,56 mol; NaOH
0,66 mol. Can
them
bao
nhieu
ml
dung
dich HCl 2M vao
dung
djch D de: (a)
Dugc
khoi lirctng ket tua
Idii
nhat.
(b)
Dugc
ket tiia ma sau khi
nung
den khoi
lugng
khong
d6i. thu
dugc
chat
rdn can
nang
5,1 gam.

Gidi
(a)
•NaCl + H,0
NaOH + HCl
0,66 0.66
NaAlO, + HCl + HpNaCl + AI(0H)3
'
X X X
AI(OH),
+ 3HC1 ^
AlCl,
+ 3H2O
y
3y y
D6 khoi
lugng
ket tiia thu
dugc
loii
nhA't thl .so mol HCl can
them
:
"HCl
="Na()H
+"N;.AI()2 = 0,66 + 0,16 = 0,82 (mol)
V^^ci
= ^ = 0,41 lit = |410ml
(I)
(2)
(3)

(b) n.
5,1
=
0,05 (mol)
'Ai20,-,y2
2Al(OH),->Al20,+ 3H3O (4)
0,1 0,05
THI:
Khong xay ra pluin irng (3) : x = 0,05 x 2 = 0,1 mol ; y = 0 mol
S6 mol HCl
Clin
dung:
n^ci = 0,66 + 0,1 = 0,76 mol
'
V-i = — = 0,38 lit =
1380ml
TH2: Co phiin irng (3): x=
0,16mol
;y
=0,16-0,1
=0,06
mol
So mol HCl can
dung:
= 0,66 + 0,16 +
3.0,06
=
1
mol ' '
^HCi

=-=0.5
lit =
I
500 mil
2.
Phiiong
phap
bao
toan khoi luong.
Dinh
ludt:
Trong
phdn
ung hod hoc
long
khoi
lumg
cdc sdn
phdin
bdng
long
khoi
luang
cdc
chat
tham
gia
phdn
dug.
A

+ B^C + D ,
m^ + mp = m^ + m^j
(A,
B : vira dii
hoac
con du)
m^, la khoi
lugng
ciia A, B
tham
gia
phan
ung
m^, m^: la khoi
lugng
ciia C, D tao
thanh
'
Apdung:
-
Phan
ung c6 n
chat
ma
biet
dirge
khd'i
lugng
cua (n - 1)
chat

=> khoi
lugng
chat
con lai.
-
Trong cac bai
toan
Xiiy ra
nhieu
phan
I'rng, c6 the
khong
ciin viet day dii
cac
phuong
trinh
phan
itng, chi can lap so do
phan
irng de
thay
moi
quan
he ti le
mol
giOa
cac
chat
can xac djnh va nh&ng
chat

ma de cho. Sau do lip
dung
dinh
luat
de tim ket qua.
-
Khi
CO
can
dung
dich thi kh6'i
lugng
mu6'i thu
dugc
bang
tong
khoi
lugng
cac
cation
kim loai va
anion
goc axit.
-
Tfnh khoi
lugng
dung
djch sau
phan
ihig:

'"(dd s;ui pif) - "^(dd
IriAK-
pit)"'' ''"(clut
l:m) '^(chiVl ki
lliia)
'^(chal bay hai)
Vi du I. Cho mot
luong
khf clo du tac
dung
vdri 9,2 gam kim loai
sinh
ra 23,4 gam
muoi kim loai hod trj I. Hay xac djnh kim loai hoa trj I va muoi kim loai do.
Gidi
Dat M la ki'
hieu
hoa hoc ciia kim loai hoa trj I. ' i
PTHH : 2M + Clj -> 2MCI ,
23,4-9,2
DLBTKL:9,2
+ mci,
=23,4g^
n^
=2nc,2
=2x.
9,2
71
=
0,4 mol

=>M
=
0,4
•
= 23(Na)=^Mu6i: NaCl
Vi du 2. Cho 7,8 gam h6n hgp kim loai Al va Mg tac diing vdi HCl thu
dugc
8,96
li't
H, (6 dktc). Hoi khi c6 can
dung
dich thu
dugc
bao
nhieu
gam muoi
khan.
Gidi
=(8,96:22,4)
= 0.4 mol
PTHH: Mg + 2HC1 MgCl, + H:t (I)
2AI
+ 6HCI 2A1CI, + 3H,T (2)
"HCI
=2.nH2
=0.8mol So mol goc Ch n^._ =0,8mol
=> m^i- =
0,8X
35,5 =
28,4gam

cr
Vay khoi
lugng
mu6'i
khan
thu
dugc
la:
m, ^^-
= 7,8 + 28,4 =
36,2gam
Vi
du 3. Hoa tan
hoan
toan
3,22 gam h6n hop X g6m Fe, Mg va Zn
bang
m6t luong
vCra du
dung
dich H2SO4 loang, thu diroc 1,344 1ft hidro (0 dktc) va
dung
djch
'
chiia m gam mu6'i. Tinh m?
Gidi
nH2S04
=(1-344:22,4)
= 0,06
mol

PTHH:
M +
H2SO4MSO4
+
H2
DLBTKL:
m^^g,
= nix + "1H2S04 -"IRZ
=3,22
+
98
X
0,06
-
2
X
0,06 = |8.98 gam
Vidu
4. Hoa tan 10 gam h6n hop 2 mudi
cacbonat
cua cac kim loai hoa trj
11
va
III
bang
dung
djch HCI du thu
dugc
dung
djch A va

0,672
lit khf (dktc). Hoi c6 can
dung
dich A thu
dugc
bao nhieu gam muoi
khan?
Gidi
n^„^
=
(0,672:22,4)
= 0,03mol
Goi
2 kim loai hoa tri II va
III
Ian lugt la X va Y ta c6 cac PTHH:
XCO,
+ 2HC1 ->
XCl,
+ CO, + H,0 (1) ,
Y,(C03),
+ 6HC1
2YC1.,
+ 3CO, + SH.O (2) ' '
Tir(l)
va(2)
nH20=nco2
=^'^3 mol; =
2.0,03
= 0,06 mol

^
"iHCi(ph:min,g)
=0,06x36,5
=
2,19gam
Ggi
X
la khoi lirgng mudi khan: XCK va YCl,
!
Theo
djnh luSt bao
toan
khdi lugng ta c6:
,
10 + 2,19 = x +
44.0,03
+
18.0,03
x =
10,33gam
Vidu
5. Cho 4,96 gam hdn hgp rdn gdm Ca va CaC, tac
dung
vdi
nude
dir thu
dugc
2,24 1ft hdn hgp khf A (dktc). D3n A qua dng
dung
bdt Ni

nung ndng
mot thdi
gian thu
dugc
hdn hgp khf B. Tiep tuc
dfin
hdn hgp B qua
dung
dich
nude
brom
du thl khdi lugng bhih
dung dung
dich brom
tang
m gam va cd
0,896
1ft hdn hgp
khf
D (dktc)
thoat
ra, ti khdi ciia D so vdi
H2
la 4,5. Tfnh m.
Gidi
Cac PTHH : Ca +
2H2O
-> Ca(0H)2 + Hj
CaC2
+ 2H2O ^

Ca(0H)2
+ C2H2
HC^CH
+ H2 ''^y"^ )H2C = CH2
H-C
= C-H + 2H2 )CH3-CH3
CH = CH + 2Br2
>
^v^fH -
CHBr2
Ni,,»
CH2
=
CH2
+
Br2
'
>
BrCH; -CH.Br
Tacd: n^ =^ = 0,1 mol; n^ = = 0,04 mol
22,4 ° 22,4 ,
Goi
X
va y
liin
lugt la sd mol Ca va CaC,
Theo
cac PTHH ta cd : sd mol H2 va
C2H2
liin

lugt la x va y.
>HePr:
'40x + 64y = 4,96 _ [x = 0,06
x+y
=0,1
=>
=0,06x2 +
0,04x26
= 1,16 gam
Theo
DLBTKL:
m^ = mg =
l,16gam
Va:
m =
mB-m^
=
l,16-0,04x2x4,5
= [oj8jam
y
= 0,04
Vi
du 6. (DHA
2009)
Xa
phdng
hda
hoan
toan
1,99 gam hdn hgp hai

este
bflng
dung
djch NaOH thu
dugc
2,05 gam mudi ciia mot axit cacboxylic va 0,94 gam
hdn hgp hai ancol la
ddng
dAng ke tiep
nhau.
Cdng
thuc
ciia hai
este
dd la
A.
HCOOCH, va HCOOC.H,. B. C,H.,COOCH., va
C2H5COOC2H5.
C. CHjCOOC^H, va
CH^cboCH^.
D. CH,COOCH, va CH.COOC^H,.
Gidi
DLBTKL
: m^^oH = "imuoi + "lancoi - niesie - 2,05 + 0,94 -1,99 =
1
gam
=^nN.oH=
(1:40)
=
0,025

mol
Do tao mdt mudi va hdn hgp 2 ancol nen 2
este
cd
cung
gdc axit vdi
dang
RCOOR': RCOO R' + NaOH -> RCOONa + R^OH
Mnax^N.
= ^ = 82 M, = 15 (-CH,)
^ROH
=
Trbr
=
37,6
=>
2
ancol
la CH,OH va C^H^OH
0,94
0,025
2este:
CH,COOCH, va CH.COOC^H, => Chgn D.
Vidu
7. Cho 200 gam mot loai
chii't
beo cd
chi
sd axit la 7 tac
dung

vdi mdt lugng
NaOH vua du thu
dugc
207,55
gam
hdn
hgp mudi khan. Tfnh khdi lugng NaOH
da tham gia pluin ung.
Gidi
Chi
so axit la
.\<'>'niii
KOH de tiling hoa axit heo co
tron}>
/.i;
chat
heo.
Sd mol KOH trung hda axit beo :
l^!:!^—-^T^
_
Q
Q25,,^Q|
56
PTHH:
RCOOH + KOH ^ RCOOK + H.O (I)
C,H,(0C0R)3 + 3NaOH ^
C3Hs(OH)3
+ 3R'COONa (2)
Ta cd :
nM.,oH

drujig
hoa
axil
beo) =
"KOH
(Irung hoa axit beo) =
"HitJ
^ ^'^^^
Goi
X la sd mol
triglixerit
cua chii't
beo=>
nNaOH(.huy
phA,>
trigiixci.)
= 3x
"NaOH(ph;iii
linj.)
=
iNaOHCIhiiy
phiii
Iriglixeril)
+
"Na()H(lrunp
hoa
axil
boo) ~ (^X +0,25)
DLBTKL
:

mc,,,,
+ mN,OH(phan m.g) = ^ 61 +
"Iplixerol
+ mH20
200 + (3x +
0,025)
X 40 =
207,55
+ 92x + 18 x
0,025
=> x = 0,25 mol
niNaOH
= (3 X 0,25 +
0,025)
x 40 =
|31gam
bg de H6a kx; 9 on Ihi viio lO- Pham
S/ltSu
3. Phaong phap
tang,
giam khoi luong
Nguyen
tdc:
So siinh khoi lugiig
cua
chii't
can xac
dinh
v6i
luoiig

san
pham
cua no ma gia
thiet
cho
biet,
de tir
khoi lirong tang
hay
giam
nay, ket hop
\d\n
he ti le mol
giiJa
2
chat
de
tlm
ra
iiroiig cha't
can xac
djnh
(c6
the
la so
nhom chuc,
so
mol, ).
Pham
vi su

dung:
Doi
v6i cac bai
toan
phiin
irng
xay ra
thu6c phan ling phfin
huy,
phan
u'ng
giua
kim
loai manh, khong
tan
trong nirdc diiy
kim
loai
yeu ra
khoi dung djch
muoi
phan
ung,
phan ling trung
hoa
axit
cho
biet krgng muoi
tao
thanh, Dac

biet
klii
chira biet
ro
phan
u'ng xay ra la
hoan toan
hay
khong
thl
phirong phap
nay
to ra
ra't hieu
qua.
Vi
du I.
Nhung
mot
thanh
sdt
naiig
8 gam vao
500ml dung djch
CUSO4
2M. Sau
m6t
thofi gian
UYy la sdt ra can hi
thay nang

8,8 gam. Xem the
tfch diuig djch
khong thay
doi
thi nong
do
mol/li't
cua
CUSO4
trong dung djch
sau
phan ii"ng
la
bao nhieu?
Gidi
"cu,S()4
=0,5.2
=
1,0 mol
Fe
+
CUSO4
^
FeS04
+ Cu
x mol
X
mol
PTHH
:

TGKL=>
Am = (64 -
56)x
= 8,8 -
8,0
^ x = — =
0,1 mol
8
I
111-
0,9
• "cu.so4(ci.r)
=
(1,0-0,1)
=
0,9mol
Cc„.s(),(du,
- ^ -
[l.HM
Vi
du 2. Hoa tan 20 gam h6n hop hai
muoi cacbonat
kim
loai
hoa tri 1 va 2
bang
dung djch
HCl du
thu duoc dung djch
X va

4,48
lit
khi (odktc). Ti'nh khoi luong
muoi
khan thu dirge khi
c6
can dung djch
X.
Gidi
Goi kim loai
hoa
trj
1 va 2 Ian
lirgt
la A va B.
"a)2
=(4,48:22,4)
= 0,2
mol
Cac
PTHH
: A.CO, +
2HC1
^
2ACI
+ CO,T + H.O (I)
BCO,
+
2HCI
BCK + CO,! + H^O (2)

Theo
(I) va (2) ta
nhiui
thay cur
I mol CO, bay ra tuc la c6 1 mol
muoi cacbonat
chuyd'n thanh muoi clorua
va
khoi luong tang them
11 gam (goc CO^" la 60
gam cluiyen thanh
goc
2C1
c6
khoi luong 71
gam).
Vay
CO 0,2 mol
khi
bay ra
thi khoi krgng muoi tang
la:
0,2.1
1
= 2,2 gam
Vay tdng khoi lugng muoi clorua khan thu dirge
la:
m
(muoi
;i

khan)
=20+
2,2=
22,2
gam
ci/ imnmvuwnKhang
vie/.
Vidu
3.
Nhung
mot
thanh kim loai
M hoa trj II vao 0,5
li't dung djch
CUSO4
0,2M.
Sau
mot
thcfi gian phan itng, khoi lugng thanh
M
tang
len 0,40 gam
trong
khi
nong
d6
CUSO4
con lai la
0,1M.
(a) Xac

djnh kim loai
M.
(b)
Lay m gam
kim loai
M cho vao 1
li't dung djch chira
AgNO,
va
Cu(NO,)2, nong
do m6i muoi
la
0,1M.
Sau
phan urng
ta
thu
dugc
eha't rdn
A c6
khoi lugng
15,28
gam
va
dung dich
B.
Tinh
m
gam?
Gidi

'•'}^''•
(a)
ncuso4(b<i) =0,5.0,2
=
0,1 mol;
ncuso4(d.)
=0,5.0,1 =0,05mol •>
>
• :
ncu.s()4(|«) =0,10-0,05
=
0,05 mol
PTHH:
M +
CUSO4
->
MSO4
+ Cu
Am
=
0,05.(64-M)
= 0,4 = 56 =>M la Fe
(b)
Ta
chi biet
so mol
eija
AgNO,
va so mol cua
Cu(NO,),.

-
Nhung khong biet
so mol
ciia
Fe. > '
Day HDHH
:
PTHH
theo
tM
tir iru
tidn
:
Fe^*
Cu^* Ag^^ rFe +
2AgNO,
^Fe(NO,)2+2Ag
Fe
Cu Ag |Fe +
Cu(N03)2-^Fe(N0,)2+Cu
-
Neu chi c6
AgNO,
phan irng:
Fe
thieu, Cu(N03)2 khong phan vtng,
AgNO,
vira
du
(hoac

con du).
=>
m,,.„)
=mAj.
=0,1.108
=
10,8g<
15,28
gam
-
Neu
Cu(NO,),cung
phan ung het,
Fe vua dij
hoae
eon
du.
' „
=>
m(rnn)="lAp+mCu+mFc(d,r)
= 10,8
+
0,1.64
+
mp,,,,,
=
17,2
+
mp,,,,,
>

15,28
V;)y
Fe
phan I'rng het,
CUSO4
da
tham
gia
phan ting nhung
eon
dir.
Goi
X la so
mol
CUSO4
tham
gia
phan urng.
m
rAn)
=^'"Ai.
+
'Ticu
=
10,8
+
X.64
=
15,28
=>

X
=
0,07 mol
1
"K-
=
"OK, r)
+ 2
"Ag
=
0,07
+
0,05
=
0,12mol
=> m =
0,12.56
=
|6,72
gam
Vi
du 4.
Nhung
mot
thanh
sdt va mot
thanh
kem vao
cung
mot coc

chi'ra 500ml
dung djch
CUSO4.
Sau mot
thtJi
gian
lay hai
thanh
kim
loai
ra
khoi
coc
thi
m6i
thanh
c6
them
Cu bam vao,
khoi lugng dung djch trong
coc bj
giam
mat 0,22
gam. Trong dung djch
sau
phan ung, nong
do mol
ciia
ZnS04
gap 2,5

liin
n6ng
do
mol cua
FeSOj.
Them dung dich
NaOH
du vao c6c, Igc lay ket
tua
roi
iiung
ngoai khong khi
den
khoi lugng khong d6i, thu
dugc
14,5 gam
chdt rdn.
So gam
Cu
bam
tren
m6i
thanh
kim
loai
va
n6ng
do mol cua
dung djch
CUSO4

ban dau
la
bao
nhieu?
• Gidi
••••'^
PTHH:
Fe +
CUSO4->
FCSO4
+ Cu (1)
Zn
+
CUSO4
->
ZnS04
+ Cu (2)
Goi
a la so mol cua FeS04
Trong
cuiig
mot dung
djch
nen tl le v6 nong do mol
ciia
cac chat trong dung
dich
cung chmh la ti le ve so mol.
Theo bai ra:
CM/z„s(i4

= 2.5 x
CM/CUS04
=> nznS04 = 2,5 x
n^^^^^
.
Khoi
luong thanh sat tang : (64 - 56)a = 8a gam
Khoi
luong thanh kem giam: (65 - 64)2,5a = 2,5a gam
Khoi
luong
ciia
hai thanh kim
loai
tang: 8a - 2,5a = 5,5a gam
Mii
thirc te bai cho la : 0,22g
5,5a = 0,22 ^ a = 0,04 mol
Vay
khoi
luong Cu bam tren thanh sdt la: 64.0,04 = 2,56 gam
va
khoi
luong Cu bam tren thanh kem la: 64
x
2,5
x
0,04 = 6,4 gam
Dung
djch

.sau phan
ling
1
va 2 c6 : FeS04, ZnS04 va CUSO4 (neu c6)
Ta
CO
so do phan u'ng:
2FeSO,
->2Fe(OH)2
a ->
+OT,I
.0
^Fe20,
amol->
a—> 0,5a
TIFCJO.,
= 160.0,5a = 160.0,5.0,04 = 3,2gam
11-3
CUSO4
bmol
->.Cu(OH)2
b
•
CuO
b
moio
=^0b = 14,5-3,2 =
ll,3g=>b
= 0,14125 mol
IncuS(34(b;m

L) = a + 2,5a + b = 0,28125 mol
0,28125
-CUSO4
0,5
0,5625M
Vi
du 5. Nung 6,58 gam
Cu(N0,)2
trong
blnh
kin khong chua khong khi, sau mot
thori
gian thu dugc 4,96 gam chat
rfui
va h6n hop khf X. Hap thu hoan toan X
vao
nude
de' duoc 300ml dung
djch
Y. Dung
djch
Y c6 pH bang
A.
2. B. 3. C.4." D. 1.
Gidi
tv,
2Cu(NO,)2
X
mol ->
->2CuO

+ 4N02 t+O, t
X
-> 2x
—>
0,5x
4NO2
+ O2 +
2H2O
^ 4HNO3
2x
0,5x -> 2x mol
Do
giam
khoi
luong:
Am
=
ni^(j^l
+mo,t = 2x.46+ 0,5x.32 = 6,58-4,96 = 1,62 gam
1,62
HNO,
0,1M-
108
=
0,015 mol
=>
n^^fgo^
=2x =0,03 mol
=>CHMO,
=

0.03
,'
J
MS'"'™
->H^
+NO3
0,1M
pH
= l
Chon
D.
Cl/
mmMiyDVVnKhang K/c,
Vidu 6. (f)HA
2011)
Trung hoa 3,88 gam h6n hop X g6m hai
axit
cacboxylic no,
don
chufc, mach ho bang dung
dich
NaOH, c6 can toan bo dung
djch
sau phan
iJng
thu dugc 5,2 gam muoi khan. Neu d6't chay hoan toan 3,88 gam X thl the
tich
oxi (dktc) cdn dung la
A.
4,48

lit.
B. 3,36
lit.
C. 2,24
lit.
D. 1,12
lit.
Gidi
+23-1=22
-
\ in =
RCOOH
+ NaOH RCOONa + H,0
a mol
a mol
-
n
Axil
=
a =
•
Tang 22 gam
5,2-3,88
-^COONa
22
=
0,06 mol
Axit
no, don
chiic,

mach ho
=>
2 dong dang
lien
tie'p:
C-H2-O2
-
, - 3,88 - 7
MAXU
= 14n + 32 = => n = -
0,06 3
^;.^2u^2
+
^02 nC02 + nH20
Chon
li.
4.
Phuong phap gia In
trung
binh
3.^-2
X0,06x22,4=
3,36 lit
1.
Khdi niein gid tri trung binh :
Co cac so duong A, B, a, b > 0. Trong do A < B.
Ta CO:
A.a < B.a
A.b
< B.b

(A.a
+ A.b) < (A.a + B.b) < (B.a + B.b)
Chia
tat
cii
cho (a+b), ta duoc: A <
i^^-^ii^l^^
<B <; •
(a + b)
A
= -—'• 5^ (A : la gia tri trung
binh
ciia
A va B).
(a+b)
2. Cac gid tri trung binh trong bai tap hoa hoc:
Nguyen tu khoi trung binh : nguyen tir
khoi
ciia
nguyen to c6
iihiSu
d6ng vj la
nguyen tu
khoi
trung
binh
ciia
h6n hgp cac dong vj c6
tinh
deii

ti le phan tram
so nguyen tu
ciia
m6i d6ng vj.
Cong thuc :
-_Xi.A|
+X2.A2 +- +
x„.A„
100
(vdi:X|+X2+ x„
= 100)
Neu
CO
2 dong vj:
-_x,.A,
+(100-x,).A2
/\ -~
100
Ta
CO
the thay the ti le phdn tram s6' nguyen tii
(X|,X2,X3 )bang
s6' nguyen
Bp
d:
Hda
hoc 9
on Ihi vAo lO -
Pham
S/

Ltfu
tir
(n|,n2,n, )cua
m6i dong vj :
—
_ n|.A| +
112^2
+
n^.A^
n,+112+113.
Khoi
Imng mol trung binh : 'i!
'
J
Khoi
lugiig
mol trung blnh
ciia
h6n hop la
khoi
luoiig
ciia
1 mol h6n horp c6
tinh
den % so mol
ciia
m6i chat trong h6n hop.
-
Xet h6n hop X gom 3 chat A, B, C:
Chat

A
B C
Phan
tu
khoi
:
MA
MB
Mc
mol
a
b
c
%
so mol :
XA
—
aMi +
bMn
+cMp
Mx
=
(a +
b +
c)
100
Mx
:
khoi
luong mol trung blnh

ciia
h6n hop X.
Neu
h6n hop gom i cha't ta c6 cong
thiic
tong quat
tinh
Mx :
-__^IiiiMi_Zx,Mi
100
Thucnig
gap h6n hop gdm 2 chat: A va B.
Taco:
%A
+
%B
= 100%B =
(100-%A)
-
aM^+bM,
X^M^+(100-X^)MB
(a + b) 100
-
Trong cac h6n hop
khi:
% the
tich
= % so mol.
Cdc
gid tri trung binh khdc:

S6
nguyen tu trung binh (C, H , O , N), d6 kh6ng no trung binh (k ), so
nhom chiJc trung binh, phan til
khoi
trung blnh (M
)
(a.MA+b.MB)
H6n
hop
X
A:C^HyO,N,
(amol)
B:CpHqO,N,(bmol)
M
=
C
=
H
=
0
=
N
=
(a + b)
(xa
+ pb)
(^
+ b)
(y^+qb)
(a + b)

(za + rb)
(a + b)
(ta +
sb)
(a + b)
O/ nun
MTV
DVVn
Khang
Viet
Vi
du 1. Trong tir nhidn, nguyen to dong c6 2 d6ng vj la
29Cu
va
29Cu.
nguyen
tijr
khoi
trung binh
ciia
d6iig
la 63,54. Thanh phan
philn
tram tong so nguyen
tir
ciia
d6ng vi
29Cu
la
A.

27%.
B.
50%. C. 54%.
Gidi
73%.
Goi
phan tram s6' nguyen
tirciia
dong vj
29Cu
la x
Phan
tram so nguyen
tir
ciia
dong vj
29Cu
la : (100% - x)
Chon
A.
T.c6:I=''^^^
=63,54^
X
= 27%
Vi
du 2. Trong tu nhien nguyen to brom c6 2 dong vj trong do dong vj
33
Br chiem
54,5% ve so lugng. Biet nguyen tu
khoi

ciia
brom la
79,91.
Xac djnh s6'
khoi
ciia
dong vj con lai.
Gidi
Goi
s6'
kh6'i
ciia
dong vj con lai la A.
Phan
tram so nguyen
tii
ciia
dong vj c6 so
khoi
A la: 100% - 54,5% = 45,5%
„
, -
54,5X79
+45.5XA
^ ^,
Ta
co: A = =
79,91 =>
A = 81
100

Vay
dong vj con lai c6 so
khoi
|A
= 81
Vi
du 3. Cho 6,2 gam h6n hop 2 kim
loai
kiem thuoc 2 chu ky
lien
tiep trong bang
tudn
hoan hoa hoc phan
ling
vdri
HiO du, thu dugc 2,24 lit khi (dktc) va dung
djch
A.
Ti'nh
thanh phdn % \i
khoi
lugng
tirng
kim
loai
trong h6n hgp ban ddu.
Gidi
= (2,24:22,4) =
0,1
mol

=>n,,,
=2nH2
=0,2
mol
=>M(A,B)
= ^ = 31g/mol
Gia
su: A < B
=>
A <
31
< B ^ A = 23(Na);B = 39(K)
39
23
'Na
8" 1
%K=—.
100% = 62,9%
6,2
%Na
= 100-62,9 =
37,1%
."K
=nNa =0,1 mol
Vidu 4. Cho
28,1
gam quang d6l6mit g6m MgCOj va BaCOj trong do %MgC03 la
a% vao dung djch HCl du thu dugc V
(lit)
CO, (o

dktc).
Xac djnh V
(lit).
Gidi ,j
PTHH:
2HCl
+
MC03->MCl2+C02
t+ HjO ' '
0.143 =
^snco2
=n3,eo3+nMgC03
<^ =
0,3345mol
=>0,143.22,4 = 3,2
lit
<
Vco2
^0,3345.22,4 = 7,491it
Bp
dSH6a
hoc
9
on
Uii
v6o
lO- Phm Sy
Lmi
Vay:
|3,21it

<Vco,
<7,49Iit
Vi
du 5. Hoa tan 115,3 gam h6n hap gom
MgCO,
va RCO, bang 500ml dung djch
H2SO4
loang ta thu duoc dung djch A, chat rdn B va 4,48 lit CO.
(dktc).
C6 can
dung djch A thi thu diroc 12 gam muoi khan. Mat
khac
dem nung chat rdn B
tori
khdi
luong kh6ng doi thi thu
duac
11,2 1ft CO. (dktc) va chat rdn B,.
Tinh
nong
do
mol/lit
ciia
dung dich
H2SO4
loang da dung,
khdi
luong
ciia
B, B, va nguydn

tir
khdi
ciia
R. Biet trong hdn hap ddu sd mol
ciia
RCO, ga'p 2,5 Idn sd mol cua
MgCO,.
Gidi
Cdng
thiic
trung binh
ciia
hdn hop:
MCO3
.
Tdng
sd
mol
khi COj =
^'^^^^
' ''^ = 0,7mol
22,4
M
=
11^-60
= 104,71
1.24 + 2,5.R
0,7 3,5
Mudi
khan khi cd can dung djch A la MgS04

nMgSO4=(12:120)
=
0,lmol
Chat rdn B gom mudi
cacbonat
con dir va BaS04
Theo phuang
phap
TGKL:
=
104,71 =^R = 137 =^RlaBa
V'«
Co 0,2 mol gdc
CO3'
da chuyen thanh gdc SO
2- .
'4
•
=>
Khdi
luong tang: 0,2.(96-60) = 7,2 gam
Do
CO
12g MgS04 tach ra ntn
khdi
luong chat rdn B la:
me = 115,3 + 7,2-12= 110,5gam
0\^X
rdn B, gom cac
oxit

CaO, BaO va BaS04
11
2
m„.
=110,5-—.44 =
'Bl
22,4
88,5 gam
Vidu 6. A la hdn hap gom hai chat ddng ddng
lien
tia'p cua ancol
etylic.
Cho 7 gam
hdn
hap A tac dung het vdi Na thu dugc 1,12 lit khi H,
(dktc).
Tinh
% sd
mol
va
%
khdi
lugng mdi chat trong hdn hap A.
Gidi "•
Goi
2 chat ddng ddng
lien
tiep A, B => MR = M^ + 14
CrPT
cua A, B la

C„H2n+,0H
(a mol) va
C„+,H2n+30H
(b mol)
CT
trung binh cua hdn hop
C-H2-^,0H
(a + b) mol
^u^2^.P^
+ Na -> C-H^-^ONa + -H,
0,1
mol
1,12
00
A
=
0,05
mol
(14n
+ 18)< 14n + 18 = — = 70< 14n + 32 2,71 <n<3,71=>n = 3
^
0,1
CTPT
cua A, B la
CjH^OH
va C4H,OH . Ta c6 :
(a + b) = 0,10 fa =
0,029
H6PT:
•

|60a + 74b7,0 [b = 0,071
Thanh phdn hdn hap :
oh
XJ/:
%Vc3H70H=29%
%Vc4H90H-71%
%m
%m
C3H7OH
C4H9OH
24,86%
75,14%
Vi
du 7. A va B la hai chat trong day ddng dang
ciia
etylen. Ca hai d^u la chat kh
trong
dieu kien thudng. 5 gam hdn hap A, B chid'm th^
ti'ch
bang thi
tich
ciia
4,4
gam khi CO. trong cung
ditu
kien ve nhidt dd va ap
sua't.
Xac djnh % v^ thd'
tich
ciia

mdi chd't trong hdn
hc»p.
Gidi
Cong
tht(c: CT
ciia
A, B la :
C„\{.„
(a mol) va C^H,, (b mol).
CT
trung binh
ciia
hdn hop
C-H2-
(a +
b)
mol
Gia
sir n < m va A, B deu la chat khi trong
dilu
kien thudng
=:>2<n<
X <m<4 (1)
4,4
"(hh)-
"CO2
M=
14x^
44
=

0,1 =>(a + b) = 0,l mol
-=50(g/iTiol)=^
x=
—=3,57
0,1 ' 14
Tir(l)=>2<n<
x=3,57<m<4 (2)
Tir
(2) va n, m d^u la cac sd nguydn, ta c6:
•
n nhan 2 gia
tri
la n = 2 va n = 3 CTPT
ciia
A la
C2H4
hoac
CjH^.
•
m nhan
1
gia
tri
la m = 4 CTPT
ciia
B la
C4Hg
vay
CTPT
ciia

A, B la:
C3H4vaC4H8
(THI)
hoac
:
C,H,
va
C4H8
(TH2)
Thanh
phcin
hdn hep: % ihi
tich
= % sd mol
Khh)=(a
+ b) = 0,l Ja = 0,021
b
=
0,079
THI:
%C.H^
=121%
TH2:-
m(hh)=28a + 56b = 5
[n(hh)=(a
+ b) = 0.1
[•"(hh)=42a + 56b = 5'
%C^Hg
=179%
%C^H^

=[43%
%C^Hg
=|57%
a =
0,043
b
=
0,057
Vi
du 8. Ddt chay hoan toan 2,76 gam hdn hap X gdm C^H^COOH
C^H^COOCH,,
CH,OH
thu duoc
2,688
lit C02(dktc) va 1,8 gam H,0. Mai
khac, cho 2,76 gam X phan ung vto dii vdi 30ml dung dich NaOH
IM,
thi
dugc 0,96 gam
CH,OH.
Cdng
thiic
ciia
C^H COOH la
A.C,H,COOH.
, D.
CILCOQH.
C. C^HaCOOH. D. C^H^COOH.
THU
VIEN

TIWH IINM
THU.m
17
B6
dSOda
hga^dalhivio
lO-Phm3ylHtu
Sod6:X
+ Y + Z +
HNO3
XCNOj
)3
+ YCNOj )2 +
ZNO3
+ NO2 + NO
Bao toan nguyen to nito :
"HN03(PII)
="N/HN03
"
"N/NO
+
"N/NO2
+"N/X(N03)3 "N/Y(N03)2
"N/ZNO3
Vlit(dktc)
Jons
^ V
-
+ 3x + 4x + 3x
2x

mol
3x
mol
63x1,25
22,4
( V '
y
=
l,25x
+ 10x x63
y
=
l,25x
122,4 ,
VI
du 3. Khir het m gam
Fe304
bang CO thu
diroc
h6n hop A gom FeO va Fe.
H6n
hop A tan viTa du
trong
0,3 lit dung dich H2SO4
IM
giai phong 4,48 lit khf
(dktc).
Tinh gia tri cua m.
Gidi
Sodd:

Fe304-
Fe + H2S04H
0,2 <-0,2
Fe0 +
H2S04
0,1<-0,1
CO
FeO
->A
Fe
FeS04
+ H2
<-0,2
FeS04
+ H2O
0.3 mol H2SO4
1,2 mol H2
Fe
2+
"Fe2+
""H2SO4 =0,3 mol
"Fe304
=^nFe=0,l
mol
m
=0,1x232=
23,2 gam
Vi
du 4. (DHB 2012) D6't
chay

hoan toan 20ml hoi hop
chat
hihi
co X (chi g6m C,
H,
O) cdn vira du 110ml khi O, thu duoc 160ml hdn hop Y g6m khi va hoi. Din
Y qua dung djch H2SO4 dac
(dir),
con lai 80ml khi Z. Biet cac the tich khi va hoi
do o
Cling
didu
kiSn. C6ng thiic phan tix ciia X la
A.
C4HA
B.
C4H,oO
C.CjHgO D.C4H«0
Gidi
H2SO4 dac ha'p thu hoi
nu6c,
khi con lai la CO,
I
lOmI
O2
X:C^HyO,(20ml)-
->C02
(80ml)+ H2O (80ml)
Ap dung
dinh

luat Avogadro va bao toan s6' nguydn tir, ta co:
Oxi:
20z + 2xll0 =
2x80
+ lx80=>z = l
Cacbon:
20x =
1
x80=>
x = 4
Hidro:
20y =
2x80^y
= 8
Vay
CTPT
:
C4H8O
=> Chon D.
Vidu 5. (CDAB 2010) H6n hop Z g6m hai
este
X va Y tao boi cung m6t ancol
va hai axit cacboxylic ke tiep nhau trong day dong dang (M^ < M^). Dot
chay
hoan toan m gam Z cdn
dimg
6,16 lit khi 02(dktc), thu dugc 5,6 lit khi
CO2 (dktc) va 4,5 gam H,0. C6ng thiic
este
X va gia tn ciia m tuong

ling
la
. A. (HCOO),C2H, va 6,6. B. HCOOCH, va 6,7.
C. CH3COOCH, va 6,7. D. HCOOC^Hj va 9,5.
Gidi 'iiKf'iu -
nco2 = 0,25 mol;
nH20=
0,25 mol; no2 =
0,275
mol
nco2
= "H20 =^ Z gom 2
este
no, don chiic:
^.H^.p,
S^d6 :
C,.H2„02
-»^"'-^^-^->
>{nH^o=neo2
=0,25
mol
Bao toan so nguyen
tijf
oxi:
no(esie)
+ "0(02 p/u) =
no(C02)
"OCHJO)
=>
no(es.c)

+
2no2
(p/u) =
2nco2
+ nH20=> "o(estc)
=(2nco2
+
"H2o)-2no2(p/u)
=-!^^^^=
(0,25 +
O,125)-O,275
= 0,lmol
2 , • • \
^ 0,25
=> C =
0,1
=
2,5=>Cx
< C = 2.5 < Cy ^ tX: HCOQCHJ
, m =
0,1.(14n+32)
= 0,1
.(14.2,5+32)
= 6,7 gam
Chon
B.
Vidu 6. H6n hop X g6m 2
chat
hiru
co A, B (chi co chua C, H, O, N trong phan tir),

trong
do ti le m,,: m^, = 80 : 21. Thanh phdn %
khoi
luong cua nito trong h6n
hop X la
10,966%.
Dot
chay
hoan toan 3,83 gam h6n hop X can 3,192 lit O,
(dktc),
thu diroc cac san phdm
chay
gom khi va hoi la CO,, HjO va N,. Toan b6
san phdm
chay
trdn
dem sue vao binh dung dung djch nuoc v6i trong du. Tinh
khoi
luong ket
tiia
thu duoc.
A.
13 gam. B. 20 gam. C. 15 gam. D. 10 gam.
I. i
vi.,
Gidi
m^
=
3,83X
10,97%

= 0,42 gam mo = —
x0,42
= 1,6 gam
=> mx =mc +m„ +mo +mf^ =>
"1^
=
3,83-(0,42
+1,6) = 1,81 gam
Goi
a la so mol C va b la s6' mol H => 12a + b = 1,81 gam (1)
' [CO2 (a mol)
Xet phan
ling
chay
ciia X : 3,83 gam X—"•''^^-'''""'"2 >
H2O
— mol
Bao toan kh6'i luong oxi: 32a + 8b = 1,6 +
0,1425
x 32 = 6,16 gam (2)
a = 0,13
Giaih6(l),(2): ' o m™, =0,I3x 100= 13 gam
[b = 0,25 ^
6. Phuong phap ty chon luong chdt '
Chon
A.
Co mot so bai toan ta cho
dudi
dang gia trj tong quat nhu a gam, V lit, n mol
hoac

cho ti le the tich
hoac
ti \t s6 mol cac
chat
Nhu
vay ket qua giai bai toan khong phu thu6c vao luong chtft tong quat da
Bd dSnde
hgc96nthivAo
lO-Pham S/lM
Taco:
Mz
=19x2
= 38
So
dS dircmg cheo:
Jib
sM
32 6
"C02
-
"02{dir)
10-X =
4j
,y =
40-8x
'It.
Di^u
kien cua y: 0< y =
40-8x
< 2x + 2

=>
3,8<x<5^x=4
vay =
8=>CTPT:C4Hg
Vi du 6. Dot
chay
hoan
toan a gam h6n hop X hai
hidrocacbon
A, B thu
duoc
132.a 45a
—^j-
gam
CO2
va gam . N6'u thfem vao h6n hop X m6t nira luomg A
CO
trong h6n hop X r6i d6't
chay
hoan
toan thi thu
duoc
"~~ g^rn COj va
^^^^ gam HjO. Bigt A, B khdng lam mat mau nirdc Br,. Xac dinh
CTPT
cua A va B.
Gidi
Chon
gid tri cho
thdng

so : Chon a = 41 gam.
Dot X: n
'CO2
=^ = 3'"o'
"H20
= —=
2,5mol
Dot
1
X + -A
2 j
•
"<^°2
"^^3'^^"^°'
nH20
=
3,375mol
44 18
Vay ta c6:
1
D6't - A:
nco2
=(3,75-3)-0,75
mol;
nH20
=(3,375-2,5)
=
0,875
mol.
"H20

= ^'875 mol >
nco2
= ^-^^ "^o' => A la ankaii:
C„H2n+2
2C„H2„+2
+ (3n + 1)02
—^2nC02
+ 2(n + 1)H20
"H20
_ (n + 1) _
0,875
n
'CO2
n
0,75
'n
=
6=>CTPTcuaA:C6H|4
Do'tB:
"c
"C02 _ 1.5
nco2
=(3-0,75x2)
= 1,5 mol
nH20
= (2,5
-0,875X
2) = 0,75 mol |^ " ~ U
vay B
CO

CTPT
dang :
(CH)_^
Mat
khac
B khOng lam mat mau
nu6c
brom ndn B la
benzen:
C^H^
=
1
v
mafA TNHHMTVDVVHKhang Vict
J
Phuong
phap
quy doi hdn hpp
nhieu
chdt
thanh
so
chat
it hon
• Khi quy doi hOn hop X g6m nhieu chat (tuf ba chat tro len) thanh h6n hop
hai
chat hay chi con mOt chat ta phai dam bao nghi6m
diing
djnh
luat bao toan

(s6'mo!
nguyfen tir,
khoi
luong). ' :""
• Co thi quy doi h6n hop X v^ ba't ky cap
chat
nho, tham chi quy d6i \i mot
chat. Tuy nhien ta nen chon cap chat nao don gian c6 it phan iJng nhat de don gian
viec
tinh toan.
• Khi quy doi h6n hop X ve mot chat thi chat tim duroc c6 the la gia
djnh
khOng
CO
thifc.
Vidu
1. H6n hop X gom cac oxit sat sau: FeO, Fe^O, va
Fe304
trong do so mol cua
FeO
bang
s6' mol
ciia
Fe^O,.
Cho 2,32 gam h6n hop X tac dung vira du vdi Vml
dung dich HCl IM.Tim V.
Gidi
Dung
phep
quy d6i:

X:
FeO (a)
Fe20,(a)
Fe304(b)
quyd6i
^X:
Fe304(a)
Fe304(b)J
.X:Fe304-
my
=(a + b) = —^
232
2,32
232
= 0,01 mol
PTHH:
Fe,04
+ 8HC1
FeCl2
+
2FeCl3
+
4H2O
mol
0,10^ 0,80
Z^OHCI
=0,01x8
= 0,08
mol=>VHci
IM = — = 0,08 lit=|80ml

1
Vi du 2. Cho 11,6 gam h6n hop Fe^O, va FeO c6 ti le mol 1:1 vao 3(X)ml dung djch
HCl
2M
duoc
dung dich A.
(a) Tinh nong do mol cac
chat
trong dung djch sau phan ufng (th^ tich dung djch
thay doi kh6ng dang ke).
(b) Tinh the tich dung dich NaOH 1,5M di tac dung hd't vdi dung dich A.
Gidi
(a) Quy doi h6n hop:
^ .
JFeO
(a) 1
,.y.6,
Fe,o^(a)|n.
= a = -^ = 0,05 mol
Fe203(a)J
n^ci
=0,3x2
= 0,6 mol
PTHH:
Fe304
+ 8HC1 ->
FeCl2
+
2FeCl3
+

4H2O
mol
0,05-^
0,6
'')<JJ|
0,05^ 0,4^ 0,05^ 0,10
0,00 0,2 , f
TTieoPTHH: ^ = 0,05 <
0,075
= — HC! dir.
232
1
8
Vi
du 2: Hoa tan 200 gam SO, vao m. gam dung
djch
H2SO4
49% ta diroc dung
dich
H,S04
78,4%.
Ti'nh
gia tri cua m,.
Gidi
Qui
d6i
SO,
thanh dung
djch
H2SO4:

• '^•'b
nN? r,,-
S03
+ H20
H2S04
80-^
98
C, =122,5%
C2
= 49%
^C
= 78,4%^
lOOgam
SO, ^ ^^^^ = '22,5 gam
Hay
SO, di/oc qui d6i thanh
H2SO4
122,5%
(c-c.)!
(C,-C)
m
1
78,4-49 _29,4_2
m2
122,5-78,4 44,1 3
=>
irij
=
—
X

200 = 300 gam
Vidu
3:
Phiii
them bao nhieu gam
nude
vao 200 gam dung djeh KOH 20% dd dugc
dung djeh KOH 16%.
Gidi
Dung
moi nude dugc quy doi thanh dung
djch
c6 nong do
chflit
tan: 0,0%
20 16
m
ll2()
"^KOH 20%
16
•
m
)l70
=
200x- =
50 gam
Vi
du 4. Dc
didu
che 560 gam dung

dich
CUSO4
16% can phai
lA'y
bao nhidu gam
dung
dich
CUSO4
8% va bao nhieu gam
linh
the CuSO^-SHp.
Giai
Cach 1: phuoiig
phap
duonig
eheo.
160
Quy doi:
CuS0 5H20=>%Cf.„c(>
= .100% = 64%
^
V • 250 j (b
So do dudng ciieo:
8 48
=>mr
"''CuS04 'iH2()
_ X _ 1
m
CUSO4
8%

,ir,>
ijfro'ub
<'h
xy-'
,
'
ioa
iiii'o
<i-j'i
=
—= -
::ib
vtlu
i'.n
k
560
'Cu.sO4.5H2O
~ j
X
I
=
80 gam
va
Wf
480gam
'CUSO4
m
Vi
du 5: Can bao nhieu lit
axit

n,S04
(D = 1,84) va bao nhieu lit nuoc cat dc pha
thanh 9 lit dung
dich
H.SO^ c6 D = 1,28 gam/ml? ."v - i'
^
Gidi
Sa
do duong cheo:
H2S04(D,
=1,84) 0,28
H2O (D2
= l,00) 0,56
Mat
khac :
VH,O+VH2S04=
9 1ft (**)
Tir (*) va (**) =^
VH^O
= va
"^HJSO^
=
•VH2SO4
0,28 _ 1
VH20
0.56 2
'H20
VU,O=2VH2SO^
(*)
3 lit

Dgng 2.
Bdi toan xdc djnh thanh phdn hSn h^p
Vidu
I. Nguyen tir
kh6'i
trung
binh
ciia
brom la 79,319.
Brom
c6 hai d6ng vi b^n:
35Br
va jsBr
.
Tinh
thanh phSn % sd'nguyen
tif
cua 3^Br.
Gidi
So d6 du6ng
cheo
:
»{Br(A,=81^
^
(A-A2)
'3^Bt<A2=79)
A
= 79,319
>%S61ugng: %(^sBr) =
(A,-A)

0,319
319-79 0,319
%(^{Br)"81-79,319"
1,681
-xl00%
=
15,95%
1,681
+
0,319
Vi
du 2. Mot h6n hgp gom O., O, b
didu
kien
tieu
chudn c6 ti
kh6'i
hcri
vdri
hidro la
18.
Ti'nh
thanh
phfin
% si the
tich
cua O, trong h6n hgp.
A.
15%.
B.

25%. C. 35%.
D.
45%.
Gidi
Ap
dung so d6 duomg
cheo
: (vdi M =
18 x
2 = 36 )
03=48 ^ y36-32) = 4 ]
M
= 36
02 =32 (48-36) = 12
V
03
^4^1
V02
12 3
3+1
xl00%=
25%
Vidu
3. Gin
tr6n
hai
the'
tfch
metan voi mot the
ti'ch

ddng dang X cua metan de thu
dugc h6n
hgp
khi c6 ti
khoi
hoi so
voi
hidro bing 15. Xac
dinh
CTPT cua X.
Gidi
Ap
dung so do ducmg
cheo
: (vdi
M =
15 x
2 = 30)
X
=
M,
(30-16)
= 14
M
=
ao
CH4
= 16
(Mx-30)
14

VcH4
Mx-30
2
Mx=58
Mv
=14n + 2 = 58=>n = 4=^ CTPT cua X : C4H,o
99
CO, + NaOH -> NaHCOj
(1)
[CO2
+ 2NaOH ^ Na2C03 +
H2O
(2) ^
Tinhs6-molmu6'i:
^^^^ ^^^^^ .
_^
jnC02
=
"NaHCO,
+
"NajCO^
_ "Na2C03 =
("NaOH
"
"CO2
)
"NaOH
"^"NaHCOj
'^^^Na2COi
"NaHC03

-(2"C02
"NaOH
)
•
"NaHCOj
=(2nc02
""NaOn)
(*)
nNa2C03
=("NaOH
" "cOj )
-
Gia thi^'t iru tidn tao
thanh
mu6'i axit va sau do
kifem
du
chuy^n
mu6'i
axii
thanh
muoi
trung
hoa :
|C02 + NaOH -> NaHCOj (l)
NaHCO, + NaOH ^
Na2C03
+
H2O
(3)

Tinh
Sd'mol
mu6'i :
nNa2C03
=nNaOH(3)
=("NaOH
-"CO2)
(**)
^
(**)
Bao toan
cacbon:
n^aHcOj
="co2
-"Na2C03
2"C02
"NaOH
-
Bai toan xay ra
trong
tinh
huong sue khi CO, va
dung
djch
kidm
nen uu
tidn
tao
ra
mu6'i

trung
hoa va sau do la
khi
CO, du
chuyen
mu6'i
trung
hoa
thanh
mu6'i
axit.
CO2
+ 2NaOH ^ Na2C03 +
H2O
(2)
CO2
+
H2O
+
Na2C03
->•
2NaHC03 (4)
Tinh
sd'mol mu6i:
(4)=>"NaHC03
=2nco2(d.r)
=2(nc02
-O-^nNaOH
) =
(2"C02

""NaOn)
(*)
Bao toan
cacbon:
"Na2C03
="C02
~"NaHC03
=
"CO2
~(2"c»2 ~
"NaOH
)
("NaOH
~
"CO2
)
Nhdn
xet: Sir tao
thanh
san ph^m chi phu
thu6c
s6' mol
chat
tham gia
phan
ihig
khdng phu
thuoc
qua trinh
phan

ling.
Ne'u: n
CO2
<
"NaOH
<
2nco2
ta lu6n c6 :
"NaHC03
-(2"C02
""NaOn)
(*)
va
Na2C03
NaOH
"CO2
(**)
1.
BAI
TOAN
THUAN
Biet
so mol CO2 va so mol
kiem.
Tim so mol
muoi.
Sir dung cap
phan
ung (1) va (2) de xet
gi6i

han ti le mol;
J
_
"NaOH
_ ^
"CO2
^
Phan
ling
San phdm
T<
1
(1)
CO2
dii
hoac
du
"NaHC03
-
"NaOH
1<T<2
(l)
+ (2)
CO2
va NaOH dfeu het.
hay (l) + (3)du NaHC03
'Na2COj
-
"NaOH
"

"CO2
"NaHC03
-2nc02
~"NaOH
T>2
(2)
NaOH du
"Na2C03
-
"CO2
VJ
du 1. Nung 13,4 gam h6n hop 2 muoi
cacbonat
ciia 2 kim loai hoa trj 2, thu
duoc
6,8 gam
chat
rdn va khi X. Lucmg khi X sinh ra cho ha'p thu vao 75ml dung
djch
NaOH IM, kh6'i iuong mud'i khan thu dirac sau
phan
limg
la (cho H = 1,
C=
12, 0= 16,Na = 23) ,,
A.
5,8 gam. B. 6,5 gam. C. 4,2 gam. D. 6,3 gam.
Gidi
MO
+ CO, '

6,6
MCO3
DLBTKL:
mco2=
13,4-6,8
= 6,6 gam
=>nco2=
^
nNaOH=
I-0,075
=
0,075
mol
_^
"NaOH
=
0,15 mol
'CO2
=
- => Tao mu6'i NaHCO, va CO, dU.
0,15 2 • '
PTHH
:
CO2
+ NaOH -> NaHC03 =>
"NaHC03
="NaOH
=0,075
mol
"iNaHC03

=
0,075
X
84 = 6,3 gam
i:>Chon I).
Vidu 2. Hap thu het
2,688
lit CO, (dktc)
bang
200ml dd NaOH IM. Sau th6i gian
phan
ung, c6 can dung dich thu
duoc
m gam
chat
rdn. Tim m.
Gidi i»
2 688
nco2=
^^ = 0,12 mol <
n^aOH
=0,2x1 = 0,2 mol
Tao
thanh
h6n hop 2 muoi NaHCO, va Na,CO,.
CO2
+ 2NaOH->
Na2C03
+
H2O

CO2
+ NaOH ^ NaHC03
TircacPTHH,tac6: ,
"Na2C03
="NaOM-"C02
="'20-0,12
= 0,08 mo!
"NaHC03
=
2"c()2
"
"NaOH
= 2
X
0,12 -0,20 = 0,04 mol
=>m=
0,08x106
+ 0,04x84= 11,84 gam
Vr
J« 3. (CJ)AB 2012) Ha'p thu
hoan
toan
0,336
i.'t khf CO, (dktc) vao 200ml dung
dich
gom NaOH 0,1M va KOH 0,IM thu dirge dung djch X. Co can toan bo
dung djch X thu
duoc
bao nhidu gam
chat

riin
khan?
A.
2,44 gam B. 2,22 gam C. 2,31 gam D. 2,58 gam.
33
Gidi
Quy
d6i dung
djch
h6n hop
NaOH
0,1M va
KOH
0,1M thanh
1
chat la
MOH
0,2M.
"MOH
=
0-2.0,2
=
0,04 mol
va
nco^
=
0,336
:
22,4
=

0,015 mol
^
JlMOH
^ 2:2^ = - > 2 =^ Tao M,CO, va MOH con du. • '
nco2
0,015 3 • -
PTHH:
2MOH
+
CO, M,CO,
+
H,0
=^
"MJCO,
"=nco2
=0,015
mol=>nH20
=0,015 mol
DLBTKL
: m^o^ +
m^^oH
=
minuoi
+
mMOH(du)
+
mH20
.>::i
: "Van
Vdri

HMOH
=
iiNaOH
+
HKOH
=
0,02
+
0,02
=
0,04mol
=>
m,^„
0,015X 44 + 0,02(40 + 56) -0,015 x 18
=
2,31 gam => Chon C.
2.
BAI
TOAN
NGHICH
Biet so mol mud'i. Tim
so
mol
CO2.
Vidu.
Tinh
the
tfch
CO,
a

dktc cdn cho vao
100ml
dung
dich
KOH 2M
de
thu diroc
2
mudi.
Gidi
KOH
+
CO,
^KHCO,
(1)
2KOH
+
CO,KjCOj
+
H,0
(2)
"KOH
=
0,2 mol
ttH
De thu duoc
2
mudi
:
1<T

=
iljm
<
2
=>
tai <
< n,
"CO2
0,1
<
Hcoj
<
0,2
=>
2,24
lit
<
Vco^
<
4,48
lit
2
- -'002 ^
"KOH
Dgng
2. Todn CO, tdc
di/ng
vdi
dung
djch

CalOH),
hay
BoIOH),
Cac
phan i/ng c6 the xay ra
:
CO2
+ Ca(OH)2 -> CaCO,
+
H2O
(1)
'
2CO2 +
Ca(OH)2
->
CaCHCO,
)^
(2)
CO2
+
Ca(0H)2
->
CaCO, + H2O
(1)
[CO2 + CaCOj
+
H2O
^
Ca(HC03
)2

(3)
'
Tinh
so mol itiuoi:
TCfccuFrHH(l)va(2)
"Ca(OH)2(l)
="CaC03' "Ca(OH)2(2)
=
"Ca(HC03)2
'
"C02(l)
=
"CaCO,
=
"C02(2)
= 2nca(HC03)2
Hoac
:
3^
"CO2
-"CaC03
+2nQ,(HC03)2
nCa(OH)2
="CaC03 +"Ca(HC03)2
nCa(HC03)2
=
"CO2 -"Ca(OH)2
nCaC03
="Ca(OH)2 ''"Ca(HC03)2
=2nca(OH)2

~
"CO2
Vay :
nca(HC03)2 =("C02 -"Ca(OH)2)
(*)
Va
ncaco3
=
(2nca(OH)2
-
"CO2
)
(**)
HoactifcacPTHH(l)va(3)
CO2
+
Ca(0H)2
CaC03
+
H2O
CO2
+
CaC03
+
H2O Ca(HC03)2
(1)
(3)
Tiif(l)va(3)
taco:
•

•"C02(l)
~
"Ca(OH)2
*"C02(3) -"CO2 ~"C02(1) -"CO2 ~"Ca(OH)2
"Ca(HC03)2
=
"C02(3) =("C02 -"Ca(0H)2)
(*)
Bao
toan
sd
nguydn
tijT
cacbon
:
ncaC03
=
"002
- 2nca(HC03)2
=>"CaC03 ="C02
~2(nco2
"
nCa(0H)2
) ~
(2"Ca(OH)2
'^COj)
(**)
vay:
.
va

"CaCO.ii
-(2"Ca(()H)2
""cOj)
"Ca(HC()3)2
-("C()2 "Ca(OH)2
)
1.
BAI
TOAN
THUAN
Biet so mol
CO2
vd so
mol kiem. Tim
so
mol mud'i.
Six
dung cap phan
ling
(1) va (2) d6' xet
gidi
han ti 16 mol;
Bien
luin
theo cac phan umg (1) va (2): Dat
T= ""^^ =-
"Ca(OH)2
^
Gia
tri

ciia
T
T<
I
Phan u'ng xay
ra
(1)
Ca(0H)2
dii
hoac
du
San
pham
"CO2
~
"CaC03
1
<T<2
T>2
(l)
+
(2)
C02va
Ca(0H)2 d^u he't.
hay (1)
+
(3)duCaCO.,
(2)
CO2
du

nCa(HC03)2
= "COj
-"Ca(OH)2
nCaC03
=2nca(OH)2
-"CO2
nCaC03
=
0
Tuong
tir khi thay
Ca(OH)2
bang
Ba(OH)2
•
35
Cac PTHH
Hoac
:
Vidu
1. Hap thu
hoan
toan
V h't khf
C02(DKTC)
vao 600ml
dung
djch Ca(OH)2
0,5M
thu

duoc
10 gam ket tua. Tinh gia tri cua V.
Gidi
CO2
+
Ca(OH)2
->
CaC03
+
2CO2
+
Ca(OH)2
^
Ca(HC03)2
CO2
+
Ca(OH)2
->
CaCOj
i
+H2O
CO2
+
CaC03
+
H2O
^
Ca(HC03)2
"caC03
~

"ca(OH)2
~ ^'-^^
""^^
C6 2 trucrtig hop :
THl:
chi xay ra
phan
ling
(1)
Ca(OH)2
du
^ "CO2
=
"caCOj
= 0,10 mol => |V= 2,24 lit
TH2:
xay ra 2
phan
ling
(1,
2)
hoac
(1,
3):
^ "CaC03
=2nca(OH)2
"CO2
(1)
.:oH,.
•A

'
(2)
',r,«,,.
(1) a
(3)
.i4V
Cac PTHH
Hoac
:
=>
nco2
=2nca(OH)2
-ncaC03
-0.3x2-0,1 =0.5 mol^|V= 11.2 lit|
Vi
du 2. Ha'p thu
hoan
toan
x mol
CO2
vao
dung
dich chira y mol
Ba(0H)2
thu
duoc
5,91 gam k6't tua. Neu ha'p thu
hoan
toan
2x mol

C02vao
dung
dich chiia y mol
Ba(0H)2
thi luang kd't
tiia
thu
duoc
la 7,88 gam. Xac dinh cac gia trj cua x va y.
Gidi
'CO2
+
Ba(0H)2
BaC03 +
H2O
(1)
2CO2
+
Ba(OH)2
-> Ba(HC03
)2
(2)
CO2
+
Ba(OH)2
->
BaC03
i
+H2O
(1)

CO2
+
BaC03
+
H2O
->
Ba(HC03
)2
(3)
X€i
thi nghidm 2:
Ne'u
Ba(0H)2
du thi luomg
BaC03
ga'p d6i luang
BaC03
cua
TN1.
Trai v6i d^ ra
la 7,88 gam. Vay
Ba(0H)2
thieu va m6t
phdn
BaCOjda tan trong COj
=^
"caco.,
=
2nB,(OH)2
" "coa => 2y - 2x = 0,04 (1).

Xet thi nghiem 1:
Nd'u
Ba(OH)2
cung
thia'u nhu TN 1 thi trong TN2 khd'i luomg BaC03< 5,91
gam. Trai v6i de ra la 7,88 gam. Vay
Ba(0H)2
^^"g nghiem 1.
^"&1CO3 - "CO2 ~ ^ ~ ^'^-^ Tir (1) => y = 0,05 mol
VrrfM3.
(CDAB2010)Ha'pthu
hoan
toan
3,36 lit khf
C02(dktc)
vao 125 ml
dung
djch Ba(OH), IM, thu
duoc
dung
dich X. Coi the tich
dung
dich khdng
thay
d6i,
n6ng
do mol cua
chat
tan trong
dung

djch X la
A.0,6M.
B.0,2M. C.0,1M. D. 0,4M.
36
Cty
mnn
MTV D
Wn Khan^
Vict
Gidi
n^y^ =(3,36:22,4) = 0,15 mol;
nB,(OH)2
=
0,125x
1
= 0,125 mol
,
. = 1,2 < 2 ^ Co 2 muoi: BaCO, va BaCHCOj), '''^ ' •
'
0,125 • ' ' '
-y^^^ir
Tif
cac PTHH, ta c6:
nBa(HCO,)2
-"C02
-"Ba,OH)2
=0.150-0.125
=
0.025
mol •

nBaC03
=2nB,(OH)2
-"C02
=2x0,125-0,150
= 0,10 mol :£
^H->
,
Cha^ tan trong
dung
dich la BaCHCOj),: ,
^,,(1
,|))fj
;
/ i
0,025
. '
Qj^jj
N6ngd6:
C^^^,^^^^^^^
^5^^
= "'^^ =^
Chon
B.
1.
BAI TOAN NGHICH
Biet
so mol
Ca(OH)2
vd so mol ket tua. Tim so mol CO2.
De

giai
quyet bai toan ta so
sanh
so mol
Ca(0H)2
vdi so mol CaCOj.
-
Neu ncaco3 -
"ca(OH)2
• ''^y
""^
P^^"'
if'ig
('),
CO2
va
kiem
tac dung
viia
du
vai
nhau
=> n^o^
=ncaC03
=nca(OH)2
^
-
Neu
Hcaco,
<

nca(Ofi)2
• 2 truong hop xay ra
TH1:
CO2
thieu,
chi xay
ra
phan
ung (1),
Ca(0H)2
du
i=>
n^Q^
= ric^coj
.
TH2: Ca(0H)2 thieu, Ca(0H)2 tao ket
tiia
hoan toan theo
phan
ling (1),
luong
CO2
con lai hoa tan mot
phan
CaCO,
theo
phan
ung (3).
"C02
=2n^^^.,((j„,^-n( ,,co3

=»j^j^ ncaC03
.
(Giiii
tuong tu
khi
thay Ca(OH)2 bfmg BaCOH)^).
Vidu
1. Thoi V lit (dktc) khf SO, vao 200ml
dung
djch Ba(OH), 1,25M thu
duoc
32,55
gam ket
tiia.
Ti'nh V.
Gidi
;
,,;•„,
,
Ba(OH), +
SO3
>
BaSOj
+ H.O (1) ,, / , :
Ba(0H)3
+
2SO3
—> Ba(HSO,): (2) ' •
^
^ryx.

"Ba(OH)2
= V =
0.2.1.25
= 0,25 mol >
n„^,S(,3
= fj^ =
' ^,
TH
1:
nso2
<
nBa(OH)2
^ ™ Ba(OH): du.
=^
"so2=
nBaS03
=0,15(mol) ^|V= 0.15. 22,4 = 3.36_lft|
^^^f^^
TH2: nBa(OH)2
<"co2
<2nB,(OH)2=>^°2phanu-ng(l)va(2).
^
"BaC03
=2nBa(OH)2
~"S02
>f
D6
dSnda
hoc
9

6nUuvio lO-
Pham
S/LiTu
nso2
=2iiBa(OH)2
-"ft-co.,
=2x0,25-0,15=0,35
mol
. |V
= 0,35
•
22.4 = 7,84 lit
Vidu
2. Hap thu
hoan
toan
2,688
lit khf CO, (ddktc) vao 2,5 1ft dung djch Ba(OH)-,
nong do a
mol/1,
thu
duoc
15,76 gam k^'t
tiia.
Gia trj cua a la (cho C = 12, O =
16, Ba= 137)
A.
0,032.
B.
0,048.

C. 0,06. D. 0,04.
(jiai
nco, = (2,688:22,4) = 0,12 mol >
nR^co,
=(15,76:197) = 0,08 mol
Vi
khf CO, bj hap thu
hoan
toan
nen da xay ra cac
phan
irng :
JCO2+Ba(0H)2->BaCO,
i+H,0 (1)
CO2
+ BaCO, +
H2O
-> Ba(HCO,), (3)
_\,
\
^ "ftiCO,
-
2nBa(OH)2
~ "CO2
"Ba(OH)2
" 2 \2 /
i
2'
na.(OH)2=:7(0.08
+ 0,l2) = 0,10mol

0,10
=>
a =
•
2,5
=
0,04M=>Ch()n 1).
Dgng
3. To6n CO, tdc dyng vdi dung djch NaOH vd Ba(OH)j
Xet
bai
toan
theo
2
buoc:
-
Bifoc 1: xet sir tao
thanh
s6' mol cac ion CO^' va HCO^ dira vao ty le so
mol
CO, va OH .
-
Budc 2: tinh so mol kd't
tiia
CaCOj
theo
s6' mol CO^' va s6' mol mu6'i tan
theo
so mol HCOJ.
"ceo,

=(2"c;,(OH)2
""^-02) =
("OH-
-"CO2,
Tircnig
tir khi lhay Ca(OH),bAng
Ba(0H)2va
NaOH
bang
KOH.
Vi
du 1. (f)HB 2012) Sue 4,48 lit khf CO, (dktc) vao 1 lit dung dich h6n hop
Ba(OH),
0,12M va NaOH 0,06M. Sau khi cac
phan
ling
xay ra
hoan
toan
thu
dugc
m gam kct
tiia.
Gia trj cua m la
A.
19,70 B.
23,64
C. 7,88 D. 13,79
Gidi
Quy d6i Ba(OH), 0,12M

thanh
NaOH 0,24M => Ta c6 NaOH 0,30M
nco, =0.2] . nN,oH 0,3 ^ ^, ^ ,. [NaHCO,
2
lz::>l<-Jli!i2H
=_L_ = |,5<2=::.C6 2mu6i
<^
'NaOl
,=0,3
"002 0,2
Na2CO,
Tircac PTHHtatinh diroc:
Ct/
TNHHMTVDVVn
Khans
Viet
nNa2CO., =nNaOH -"coj =0,3-0,2=0,1 mol
nNaHC03
=2nco2
-"NaOH
=0,4-0,3
=
0,1
mol
Do lofp 9
chua
hoc su
dien
li ntn ta dung
phuong

phan
quy doi h6n hop NaOH
va Ba(OH),
thanh
chat
tirong
duong
la NaOH.
Thuc
te trong dung djch c6: ion CO3"
(0,1
mol); ion HCO3 (0,1 mol);
ion
Na^ (0,06 mol) va ion Ba-* (0,12 mol).
Ion
Ba"* con du sau
phan
ling
tao ket
tiia:
'r;
"B;,CO,
="CO2-
Ba^^ + cor >BaCO, i
0,1
<-0,l^
0,1
n
2+ =0,02 mol
Dung dich thu diroc c6 2

chat
tan:
-
NaHCO, : 0,06 mol
•
Ba(HC03)2 : 0,02 mol
=>
m =
0,1 X
197 = 19,7 gam
r:^.
Chon A.
Vidu
2. Cho
0,448
1ft khf CO,
(0
dktc) htfp thu het vao 100 ml dung djch chi'ra h6n
hop NaOH 0,06M va Ba(OH), 0,12M, thu
duoc
m gam ket
tiia.
Ti'nh gia tri ciia m.
A.
3,940.
B.
1,182.
C.
2,364.
D.

1,970.
Gidi
Hoan
toan
tuong tu vf du 1. Ta c6 :
nco2 =0,02 mol; n^2+ =nBa(OH)2
=0,012
mol
"OH-
="NaOH +2nBa(OH)2 =
0,1(0,6
+ 2
X
0,12) = 0,03 mol
1<
OH"
^C02
0,03
0,02
=
l,5<2=>C6 2muoi
HCO3
cot
2-
" o^-
=
2nco,
- n _ = 0,04 -0,03 = 0,01 mol
HCO3
OH

n
2-
=0,03-0,02
= 0,010mol < n 2^
=0,012
mol
Dung dich
CO
:
0,006
mol NaHCO, va
0,002
mol Ba(HC03),
=>
m =
0,01.197
= 1,970 gam => Chon I).
^'
du 3. (»HA
2011)
Ha'p thu
hoan
toan
0,672
1ft khf CO, (dktc) vao 1 1ft dung
djch
g6m NaOH 0,025M va Ca(OH), 0,0125M, thu
duoc
x gam ket
tiia.

Gia tri
ciia
X
la ' , ^.
A.
2,00. B.0,75. C. 1,00. D. 1,25.
Gidi
Quy doi Ca(OH), 0,0125M
thanh
NaOH 0,025M ta c6 NaOH 0,05M
nco2
=0,03
nN„oH=0.05
I
^
"NaOH
0,05
"C02 0,03
<
2
=>
Co 2 mu6'i
NaHCO,
NajCO,
Tir
cac PTHH ta tfnh
duoc:
iiNajCO, = "NaOH - "coj = ^.05 - 0,03 = 0,02 mol
nN,HCO.,
=2nco2-"NaOH

=0'06-0,05-0,01
mol •
.,v>i,a'<
Thuc
te'troiig dung djch c6 : ion
CO3"(0,02
mol); ion
HCO3
(0,01 mol); '
ion
Na*
(0,025
mol) va ion Ca""
(0,0125
mol).
Ion
CO^~ con du sau
phan
u'ng tao ket tua :
Ca-^ + COJ- >CaCO, i fx-0,0125x 100= 1,25 gam
0,0125-^
0,012.V
0,0125
Chon 1)
LOA/2.
DUHG
DjCH
MU(5l
NHOM
(HOAC

MU6|
KEM)
TAC
DUNG
VC^JI
DUNG
DjCI
KI^M
MANH.
Dpng
7.
MuS'i
nh5m
tdc
di/ng
vdi
dung
djch
ki§m
mqnh
=>
n
NaOH (p/ir) • "
•
Kiam
manh
NaOH , KOH thi
AKOH),
tan trong ki6m dir:
nrruu

[3NaOH +
AlCl,-^Al(0H),
+ 3NaCl (1)
PTHH
'. <
NaOH +
Al(OH),-)-Na[AI(OH)4l
(2)
hay : NaOH +
AKOH),
NaAlOj +
2H2O
(2)
PTHH
dang
ion thu gon:
Al'"
+ 30H ->
Al(OH),
Al(OH)3
+ OH ^
AIO2
+2
H2O
"NaOH(l) -•^"A1CI3
"NaOH(2)
- "NaOH ""NaOHd) = "WaOH
-^n^jQ^
^
"NaAIOi ="Na()H(2) =("NaOH

^"AICI^)
(*) H'\
Bao loan so
nguyen
tij
Al:
"AI(()H),i
="AICI,
-"NaAI()2
^("^"AICI,
-"NOOH)
f3NaOH
+
AlCl,->
Al(OH),+3NaCl (1)
Hoac:
<
4NaOH +
AlCl,->Na[Al(OH)4]
+ 3NaCl (3)
Hay : 4NaOH +
AICI3
^ NaAlO^ + 3NaCl +
2H2O
(3 )
^
"NaOH(l)
-^"A1C13
' "Na()H(2)
^'^"AICI^

' "NaAI02 =
"AICI3(2)
=>HePT:
^"
•" —
"NaAI02
-("NaOH
-3n^lQ^ j (*)
»AI(OH)3
=(4nAICl3
"""NaOn) (**)
Vay neu c6 :
3nAici,
<nNaOH <4nAia3 ^''^
^"A,-^+
<"O„-
^'^"AI-'*
"A1(OH)3
+"NaAI02 " "AICI3
3"AI<OH)3
"*"^"NaA102 = "NaOH
Oy
mm
MTVDVVn
Khang
Viet
Thi
CO
2 san ph^m
Al(OH),

va
NaA102
vdi cac lien he so mol voi cac
chat
ban
(jau
tham gia
phan
ling
nhu sau :
nNaAi02 -("NaOH-3"Aici3) ^ay n^^^ =(n
'AIO2
\"OH
""'A|-'+)
_
-3n
(*)
;^AI(OH)3i=('^"AlCl3-"NaOH)|
^ay
"Al(OH)3i
"
(^"AI-^^
~
"OH"
)
(**)
N6'u
mudi tham gia
phan
u'ng la

Al2(S04),:
cong
thuc
dang
phan
tu diroc viet
lai
nhu sau :
"AI(0H)3i
~^"Al2(S04b ~"NaOH
"NaA102 - "NaOH -''^"AI2(S04)3
(**)
(*)
•
Trong dung dich gdm h6n hop cac ion (axit),
M""^
(tao hidroxit k6't tua) va
Ar^^chap
nhan
c6 3 nhom
phan
ling
xay ra
theo
thu tu nghiem
ngat
uu tien
v^
thcri gian la:
Trung hoa - tao cac ket tua

hidroxit
- hda tan
Al(OH),.
Xn.=n.+n
, +n
OH
H* OH (taocac>lhidroxil) OH (hoa tan iAI(()n)3)
•
Ki^m
yd'u: dung djch NH3 thi
AI(0H)3
khong tan trong ki^m du:
3NH3 + 3H2O +
AICI3
A1(0H)3
+ 3NH4CI
Dang ion thu gon:
Al^"
+ 3 NH3 +3 H2O ^
AI(OH)3+3
NH4 ; iiifv"
A1(0H)3
+ 3 NH3 +3 H2O X
BAI
TOAN
THUAN
"NaOH hay n
(a mol)
OH"
••AICI3

hay
•'^i.uC'
"^"AI(OH)^i
"NaOH
K"0H-)
Uptyle:
T = - = -
"AICI3
(hay n^^3^ j b
Neu0<T<3hay0<a<3b:
=>
NaOH
(hoac
OH') thid'u,
AICI3
(hay
Al""")
du
(hoac
ca 2 t/d vira du)
hay
d
* -
"AKOH),
-^"NaOH
"AKOH),
~ 3 OH
Neu 3 < T < 4 hay 3b < a < 4b => OH' hoa tan m6t phdn
Al(OH)3
Tinh

luong
AKOH).,
va NaAlO,
theo
cac
cong
thurc
(*) va (**)
liSdSH6ohpc9Sn
Uil vAo 10
-
Pham
S/ UT
• Neu T = 4 hay a =
=
4b
n
= 0
Al(OH)^
va
"Na()H(itnhat)
-"^nAlClj,
hay
n
= 4n ,.
OH
(itnhiit) A|-'+
Hoac
:
D.

a : b > 1 : 4.
(1)
(3)
\idu 1. Tron dung djch
chira
a mol
AlCl,
\6\g djch chiia b mol NaOH. De thu
duoc
ket tua thi cdn c6
ti'
le
A.
a : b = 1 : 4. B. a : b < 1 : 4. C. a : b = 1 : 5.
Gidi
3NaOH
+
AICI3
-> Al(OH)3 +
3NaCl
4NaOH
+
AICI3
^ Na[ A1(0H)4 ] +
3NaCl
' Hay:
4NaOH
+
AICI3NaAI02+3NaCl
+

2H2O
(3")
Ket tua cue dai khi:
n^aOH
=
3nA|ci^
=> b = 3a.
: K€i tua tan bet khi:
n^aOH
-
'*"AICI3
b = 4a.
1
CO
ket tua Al(OH)3 :
Q £1 1
nNaOH<'*nAici, =>
b<4a=>l<4-=>->-=>ChonD.
b
b 4
Vidu
2. Nho tir tir 0,25 lit dung djch NaOH 1,04M vao dung djch gom
0,024
mol
j
FeCl,;
0,016 mol
Al2(S04X,
va 0,04 mol H2SO4 thu diroc m gam ket tua. Gia tri
cija

m la
A.
4,128 B.
2,568
C. 1,560
D.
5,064
Gidi
CacPTHH:
2NaOH
+
H2SO4->
Na^SOj +
2H2O
(1)
3NaOH
+ FeCl3 ->
3NaCI
+ Fe(0H)3 ^ - (2)
6NaOH
+ AI2 (SO4 )3 ->
3Na2S04
+
2A1(0H)3
(3)
8NaOH
+
AI2(804)3
->
2NaA102

+
3Na2S04
+
4H2O
(4)
"NaOH
= 0'26
(mol),
nFeci3
=
0,024
(mol)
Ket tua
A1(0H)3
tao
thanh
sau khi
trung
hoa va ket tua het Fe'^
"NaOH
(.1+4) = 0,26
-(3X0,024
+ 2
x0,04)
= 0,108 mol
6<
J^aOH
-0''0« 6.75<8«3<^
= ^ =
3.375<4

nAl2(S04)3
0,016
Al
0,032
Vay : n
n
Al(OH)3
""^"AI-^^ ""OH-
-^"Al2(.S04)3
-nNaOH(3+4)
AI(OH)
=
8x0,016-0,108
= 0,02 mol
"1 =
"1AI(OH)3
+ mFe(OH)3 =
0'02-78
+
0,024.107
= 4,128 gam
Chon
A.
BAI TOAN NGHICH
nAici3
hay n^|3^
(b mol)
"NaOH
hay
n^^^^,

(a mol)?
[n^„„^^^^
(k mol)
• Neu n
• Neu n
AI(OH)
'.1
A|-'+
n
=3.n
NaOH
AI(()H),4
hay
n
= 3.n ,^
OH-
Al^+
AI(OH),
<
n 3. ta xet 2 trucnig hop :
A!"
Aia,(hay
Al"^^) dir n^aOH
hay n
)
= 3n,
•If-*.
'AI(0H)34.
. AI(OH)3 tan mot phiin trong NaOH (hay OH") : '
Tinh

lugng Al(OH), va NaAlO,
theo
cac
c6ng
thiic (*) vh (**)
• Ne'u dung djch ban dau g6m
AlCI,
(hay
Ar^"")
va HCI (hay ) thi: >
AI(OH),
tao thanh sau khi da trung hoa het axitH*. Do do :
("NaOH ~ "HC|)-(4nAICl3
""AKOIDII)
hay
^)H"
V^ =
('*-"A ^ "AI(OH),
Vidu
1. Cho
200ml
dung djch
AlCI,
1,5M tac dung vdi V lit dung djch NaOH
0,5M, lucfng ket tua thu
duoc
la 15,6 gam. Gia tri Idn
nha't
ciJa
V la (cho H = 1,

0= 16, Al = 27)
A.
1,2. B. 1,8.
C.2,4.
D.2
Gidi
'^'^=0,2(mol)<nA,ci3 =
0,2.1,5=0.3(mol)
n
AI(OH)
'3
78
«A1(0H)3
tan m6t phdn trong NaOH .
PTHH:
3NaOH +
AICI3
-> A1(0H)3 + 3NaCI (I)
4NaOH +
AICI3
^ Na[AI(OH)4 ] + 3NaCl (2)
^'^'^(2)^%<OH)3='^"A.Cl3-nNaOH
"NaOH
=4nAici3
~"AI(OH)3
= 4 X0,3 -0,2 = 1 mol
=>V=ii^=21it=^ChonD.
0,5
^' du 2. Cho V lit dung dich NaOH 2M vao dung djch
chiJa

0,1 mol Alj (SO4 )j va
0,1 mol
H2SO4
den khi phan ling
hoan
toan, thu
duoc
7,8 gam ket tua. Gia trj
I6n
nha't
cua V de thu
duoc
luong ket tiia tren la
A.
0,45.
B.0,35.'
C. 0,25. D. 0,05.
Gidi
• ^
"NaOH = 2V (mol) , , , V " •
"AI(OH)3
"
^"A'2<-^4>3
=0'2 mol
Lircmg
NaOH
da
dung
nhieu
nha't

=>
AKOH),
da
tan m6t
pMn.
Vay
CO
cac
phan
irng
:
5, ,ji>^
2NaOH
+
H2SO4->
Na2S04
+
H2O
(1) '
6NaOH
+
AI2(804)3
3Na2S04
+
2A1(0H)3
(2)
8NaOH
+
Al2(S04)3
^2NaA102

+4Na2S04
+4H2O
(3)
Ta
CO
:
(n^aOH
-
2nH2S04)
=
(4nAici3
-
"AKOH).,
)
=>
("NaOH
-
2nH2SO4)
=
(8nAi2(S04-
"AKOH).,
) O
2V
-
2
X
0,1
=
8
X

0,1
-
0,1
2V-0,9mol
0
9
V-
—
= 0,45 1ft
Chon
A.
Vidu
3.
Them
m gam
kali vao 300ml
dung
djch
chura
Ba(OH), 0,1M
va
NaOH 0,iM
thu
duoc
dung
djch
X. Cho tir tu
dung
dich
X vao

200ml
dung
dich AU(SO^),
0,1M thu
duoc
ket tua
Y.
De
thu
duoc
luofiig
ket
tiia
Y
Ion
nha't
thi
gia
tri ciia m
la
(ChoH=
1;0=
16;Na
=
23; Al
=
27;S
=
32;K
=

39;Ba=
137)
A.
1,59. B. 1,17. C. 1,95. D. 1,71.
Gidi
"ft,2+
"
"'"^
"Ai2(S04)3
=0,02
mol;
nB^,(OH)2
=
"NaOH
=0,03
mol.
n^^^.
= +
2n3,(OH,2
+
n,,oH
=
39 +
0-06
+
0,03 =
+
0,09
Ket tiia gom
BaS04

va
A1(0H)3
.
Kd't
tua
lorn
nha't
=> Al(OH)3 khong bi
hoa
tan.
^
CacPTHH: 2K
+
2H2O
^
2K0H
+
H2
,^
3Ba(OH)2+Al2(S04)3->3BaS04+2Al(OH)3
6NaOH
+
Al2(S04
)3->
3Na2S04
+
2A1(0H)3
'
'
6KOH + Al2(S04)3->3K2S04

+
2Al(OH)3
n
=6nAu,so,i.
^ —
+ 0,09 =
6x0,02
= 0,12^
m =
l,17gam
Chon
B.
Vi
du 4.
(J)HA
2012) Cho
500ml
dung
djch Ba(OH), 0,1M
vao
Vml
dung
dich
Al2(SO4),0,lM,
sau
khi
cac
phiin ung
ket
thiic thu

duoc
12,045
gam
k6't tiia. Gia
trj
ciia
V la
A.
300. B. 75. C. 200.
D. 150.
^ .
^^^^
Gidi
. . ,) ,
3Ba(OH)2+AI2(504)3-»3BaS04+2A1(0H)3
(0
0,3V
<-0,IV->
0,3V
^ 0,2V
Ba(OH)2
+
2Al(0H)3-*Ba(A102)2+4H20
(2)
(0,05
-
0,3V)
(0,1
-0,6V)
-> (0,0.5

-0,3V)
NS'u chi
CO
ket tua
BaS04,
AKOH),
tan het
theo
phan
ling (2).
Taco:
mBa.so4
=
0,05x233
= 11,65 gam
<12,045
gam. Trai
de ra.
vay :
CO
2
phan
li-ng (I)
va
(2): kd't
tua c6
BaS04
va
Al(OH)3.
m|=niBaS04

+"1AI(OH)3(1)
~'"A1(OH)3(2)
^
233
X
0,3V+ 78(0,2V-(0,1-0,6V))
=
12,045
|V^,0,15 1ft
=
150ml|
Chon
D.
Dgng
2.
Mu5i k§m tdc dyng vdi dung djch ki^m mqnh
Phuffng
phdp
gidi
hodn
loan
tuang
tu
mud'i
nhom
tdc
dung
v&i
dung
dich

kiem
manh.
PTHH:
ZnCl2
+
2NaOH
->
Zn(0H)2
+
2NaCl
(1)
ZnCl2
+
4NaOH
Na2Zn02
+
2NaCl
+
2H2O
(2)
"ZnCl2(l)
=
"Zn(OH)2
•
"ZnCl2(2)
=
"Na2Zn02
"NaOHd)
=2nzn(OH)2'
"NaOH(2)

=
4nNa2Zn02
"Zii(OH)2
+"Na2Zn02
"
"ZnCl2
2nZn(OH)2
+4"Na2Zn02
="NaOH
HoSc
:
nzn(OH)2
=(2nzna2
~0,5nNaOH)
(**)
"Na2Zn02
=
(O-StlNaOH
"
"ZnClj
)
ZnCl2
+
2NaOH
-> Zn(OH)2 +
2NaCl
(1)
Zn(OH)2
+
2NaOH

^
Na2Zn02
+
2H2O
(3) . .
"NaOH(I)
=
2n2nc,2 •
"NaOH(2)
=
2nNa2Zn02
2nNa2Zn02
+
^n^^dj
=
nN^OH
=>nNa2Zn02
=(0.5nNaOH
-"21102)
Bao toin s6'
nguyen
tu kem :
nz„(OH)2
=
"Z"Ci2
"
"Na2Zn02
=>
nZn(OH)2
""ZnCl2

"(O.SnNaOH
"
"Z11CI2
) =
(2nZnCl2
-O.SnfjaOH)
(**)
Ket luan:
•
ZnCl2(hoac Zn^"") du
(hoac
vira
dii),
NaOH thie'u
•"Zn(OH)2
=0.5nNaOH
Ndu:
i^NaOH
<2:
'ZnCl2
-
Ne'u:
•^ii^>4=>-
"ZnCl2
>n
Na2Zn02
=
0,0
.ZnCl2(hoac
Zn^"") thie'u, NaOH vira du

(hoac
dir)
•nzn(OH)2
=0,0
'"Na2Zii02
-
"Z11CI2
45
Ne'u : 2 < < 4 => Co 2 san phdm : Zn(OH), va Na^ZnO,, chit tham gia
"ZnCl2
NaOH
va ZnCU d^u he't.
"Zn(OH)2
-2ny„Q2
~^>5'lNa
NaOH
"Na2Zn02 " ^'^"NaOH ~"ZnCl2
(**)
(*)
hay
nz.i(OH)2
=2"z„2+
-0'5n
OH"
n
1 - 0,5n
V.nO^'
OH
Zn
2+

•
NaOH thieu :
Vi
du I. Tmh the
tich
dung djch NaOH IM can cho vao 200ml dung djch
ZnCl,
2M
dc
dugc
29,7 gam ket tua.
.
,
iinmn
mmn
»:»5i!>
Gidi
J
"/,nci2 >
"zn(OM)2
"^'"^ => 2 tru6ng hofp :
"zn(OH)2
~0'5"Na()H
"NaOH = 2nz,„()„)2 = 2
X
0,3 = 0,6 V = 0,6 lit
•NaOH
hoa tan I
phfin
hidroxit luong

tinh
:
ny^^^Q^^^^
-
^^yA\C\i
""^-^"NaOH
"NaOH ='^"/Ma2 -2'izn(OH)2 = 1,6 -0,6 = 1 mol ^ V = 1 lit
Vidti
2.
(J)HA
2009) Hoa tan het m gam ZnS04 vao
nuoc duoc
dung djch X. Cho
110 ml dung dich KOH 2M vao X
duoc
a gam ket tua. Mat
khac
neu cho 140ml
dung dich KOH 2M vao X thl cung
duoc
a gam ket tua. Gia tri m la :
A.
20,125
B.
12,375
C. 22,54 D. 17,71
Gidi
ZnS04+2KaOH^Zn(OH)2+K2SO4
(1)
Zn(OH)2

+ 2KOH->K2Zn02 + 2H20 (2)
-
Trong thf nghiem 1: neu ZnS04 het, KOH con du thi trong thf nghiem 2 voi
luong
KOH nhi^u han luong ket
tija
se be hon a gam. Vay trong thf nghiem 1,
KOH
thieu va ZnS04 du.
nzn(0H)2
=0,5nKOH=^
^ = 0,5x0,22 = 0,11
'
* Trong thi nghiem 2: Neu KOH cung thieu va ZnCU tiep tuc du nhu trong tin
nghiem (1) thl luong ket
tiia
da nhi^u hon cua thi nghiem (1). Vay trong thf nghiem
(2),
ZnS04 thieu va KOH da hoa tan mot phdn Zn(OH),.
"zn(OH)2
=2ny„j^j^
~0,5"KOH
:^
A .
0,11 =
2n^„s,)^
-
0,5
X
0,14

X
2 =^
n^^so,
= 0,125 mo!
=>m
= 161 xO,125-20,125g=>
Chun
A.
PTHH
:
46
L0AI3:UmG
mCH
AXITTAC
DUNG
raiC/Vdl
DUNG
DjCH
MU6l
ALUMINAT
(1)
V6i cac axit manh,
AKOH)^
tan trong luong axit du.
,
PTHH: NaA102+HCl +
H20->
Al(OH)3
+ NaCl (1)
Ai(OH),+3HC1->A1C1,+

3H2O
(2)
Dang ion thu gon :
AIO2
+ H^ + H,0 ^
Al(OH),
^
Al(OH),
+ 3H*
Al'"
+ 3H.O
Hoac:
NaAlOj
+ HCl +
H2O->
Al(OH),
+ NaCl (1) ' ,i.
NaAI02+4HCl-^AlCl, + NaCl + 2H20 (3)
Dang ion thu gon: MO'j + H* + H,0 ^
A1(0H)3
AIO2 +
4H* ^A1''
+ 2H,0
•
Tinh
sd'mol san phdm:
^
/l^ ^ /^^
(""^1(1)
="NUAI02

-
Theo(l) va(2) : i
"HCI(2) ="HCI
~"HCI(I)
="Ha ~"NaAI02
"A1CI_,
-:^^HC\(2)
='^("HCI
~"NaAI()2
Bao toan so' nguyen tu nhom :
n^noH),
= nN,Ai02 "
IAICI.,
^ "AI(OH)3
="NaAI02
~'^(""^"'
~ "NaAI()2 ) = "j("NaAI02
~"HC|)(*)

Theo(l) va(3) :
"AI(OH)3
+"AICI3
-"NaAl()2
"AI(OH)3
+4nAic,^
="HCI
"AI(()H)I
-^(^"NaAlOo
"HCI)
(*)

) (**)
AICI3
-•J\"HCI
"NaAI02
Vay
trong trudng hop tao 2 san pham ta c6 :
hay
"AI(OH)3
-r(4nNaAi02
~"HCI)
"AI(OH)3
- 2
4n
-n ,
AIO2 H+
(*)
"AICI3
(IHCI
-"NaA102)
hay
Al
n
^ -n
\
H"^ AIO2
(**) "i-:
(2)
Khi CO, tac dung voi dung djch
NaAlO,
tao

Al(OH)3
nhung
Ai(OH)3
khong tan trong luong CO, du.
NaA102 + CO2 +
2H2O
Al(OH)3
+ NaHC03
47
Bid't
HHCI
(hay ) va n^aAiOz (hay
n^|^_
)
tim
n
AI(OH),
Up
ty Id
"NaA102
hay
n
A102-;
Neu
0 < T < 1 hay a < b =^ HCI (hay )
thieu,
NaAlO, (hay AlOj) du
"AI(OH),
hay
n ,

U.'' i
- Neu T = 1 hay a = b ^ HCI (hay ) va NaAlO, (hay AIO2 )vira dii
Al(OH)
- Neu 1 < T < 4 hay b < a < 4b hoa tan m6t
phdn
AKOH),:
Tinh
lirqmg
saii
philm
theo cac lien h6 (*) va (**)
- NeuT = 4haya = 4b=^
n
=0
va
AI(OH),
n
H
(itnhai)
=
4.n
AIO2
BAI
TOAN NGHICH
Biet
n
'NnA,02
(hay
n^_^_
) va

n^_^^^^^^
tlm nnc, (hay n^, )
Ne'u n =nN-,Ain.
AI(OH)^ NaAI02
Neu
n <
nN.,.|n,
Al(OH), NaAI02
hay
n
AIO2
'HCI
AI(OH)
'3
hay
n
AIO2
)
ta
xet 2
trircnig
hop :
• NaAlO. (hayAlO, )du=> n ^=n
H+ AI(OH)-,
• AI(OH), tan mot
phiin
trong
HCI (hay ): tir
lien
he (*) ta c6 :

hay
n
+
=
4n
_ - 3n
AIO2
AKOH)
• Neu
dung
djch
ban
dfiu
c6 NaAlO. (hay AIO2) va NaOH (hay OH')
thi:
Ket
tiia
AI(OH), chi duoc tao
thanh
sau khi da
trung
hoa het OH".
"AI(OH)3
-•r(4nN.,Ai02
-("HCI
-"NaOH),
48
Cty
TNHHMn
/'I'.v

x/,r.^.,
("HCI
~ "NaOH) -
4nNaA102
- 3>\,^oH)
•ij
Hay
(V-"0H-)
=
4n
_ - 3n
AIO2
Al(OH), y
Vidu
1- Mot
dung
djch
A c6
chiia
NaOH va 0,3 mol
NaA102
-
Cho 1 mol HCI vao
A
thu
duc«c
15,6 gam ket
tua.
Khoi
luong

NaOH c6
trong
dung
dich
A la
A. 32 gam hoac 16 gam B. 32 gam hoac 28 gam
C. 32 gam hoac 8 gam D. 32 gam hoac 14 gam
Gidi
CacPTHH: NaOH + HCI ^ NaCl + HjO (1)
NaAI02 + HCI+ H20^ AKOH),+ NaCl (2)
Al(OH)3+3HCl->AlCl,+3H2O (3)
"AKOH),
= '7^^ = (^'2 mol <0,3 mol =
nMaAi02
•
hai
truong
hop:
+ THI axit thieu: chi CO
phan
irng
(1), (2) xay ra:
("HCI -
"NaOH
•
(I - "NaOH) = (^>2 n^.^H = 0,8 => m = 32gam
AKOH),
+ TH2 axit dir:
phan
irng

(3) c6
xiiy
ra
nhung
khong
hoan
toan.
HCI
- n
NaOH
4"NaAI02-3"^,^,3„,^
<=>('-nNaOH) = 4xO,3-3xO,2
=> n|vj|()j, =0,4 => m = I6gam =>
Chon
A.
Vi du 2. (WHA 2012) Hoa tan hoan
toan
m gam h6n hop gom Na,0 va AKO, vao
nucfc
thu
duoc
dung
dich
X
trong
suot.
Them
tir tir
dung
dich

HCI IM vao X, khi
het
lOOmI thi
biit
dau
xuA't
hien
ket
tiia;
khi het
300ml
hoac
700ml
thi deu thu
duoc
a gam ket
tiia.
Gia
tri
ciia
a va m Ian
luot
la
A. 15,6 va
27,7.
B. 23,4 va
35,9.
C. 23,4 va 56,3.
Gidi
PTHH : Na,0 + 2H,0 -)• 2NaOH

2NaOH
+ AUO, 2NaAIO, + H^O
NaOH + HCI -> NaCI + H2O
^^^^ NaAI02 + HCI + H2O ^ AI(OH), + NaCI
AI(OH),+3HCI->AICl3 + 3H20
Dung
djch
X g6m NaAlO, va NaOH dir
I Them
100ml
HCI (0,1 mol) bat ddu c6 ket
tua:
phan
ling
(3) hoan
toan
=^nNaOHau="Hci
= 0,l mol \
Them
300ml
HCI (0,3 mol)
thu
duoc a gam ket
tua:
HCI
thieu,
phan
(^"g (4) xay
ra
va NaAlO, con du.

D.
15,6 va 55,4.
(1)
(2)
(3)
(4)
(5)
-
HHCI
"""NaOH
=
(0,3-0,1)
=
0,2
a =
78X0,2
=
[l5,6 gam
AI(OH)3
Th6m 700ml
(0,7 mol)
cQng
thu
duoc
a gam ket
tua:
AKOH),
tan mot
phai
trong HCl.

("HCI ~
"NOOH)
-
(0,7-
0,1) = (4nN,A,02
-3x0,2).:^
nN,Ai02
=0-3 mol
Bao loan s6'nguy^n tir
Na va
Al: n^j^o^
=-nNaAi02
=0.15mol
"NazO = |(nNaOH + nNaAI02 ) = jc"'' + ^,3) =
0,2 mol
m
=
62 n
N„20
+ '
02 n
Ai=
62
X
0,20
+
102
X
0,15
=

|27,7
gam
Chon
A.
LOAI
4 : DUNG DjCH AXIT TAC DUNG TCT TC/
V6|
DUNG DjCH
MU(5l
CACBONAl
(HOAC
MU6l
SUNFIT)
•
So
sanh lire axit: H,CO, la axit manh hon
HCO3
=>
NajCO,
CO
tinh
bazo
manh hon
NaHCO,
{I)
Trinh
tu uu
tien
cua
phdn itiig giua muoi cacbonat

vd
axit
khi
them
tCc
tit
axit
vao
dung
dich
muoi
Id:
Na2C03
>
NaHCO,
HCl.tr.C I
PTHH:
(I)
• " "" (2)
HCl
+
Na2C03->
NaHCO,
+
NaCl
•. (!)•
HCl
+
NaHCO,
^

NaCI
+ CO2 + H2O (2)
,^
[HCl+
Na2C03->
NaHCO,+
NaCl
(1)
Hay la:
<
[2HCI
+
NajCO,
->
NaCl
+ H2O + CO2 (3)
T=-
n
'HCI _ a
Na2C()3
^
Phan ung
San phdm
T<
1
(1)
Na.COj
du
hoac
du

nco2
=0
"NaHCO,
- "HCI
hay
n
=n ,
HCO3 H"*
1
<T<2
-(l)
+ (3):
NajCo,
(hay
CO^-)
va HCl
(hay
H*)
dieu het.
-
Hoac
(I) + (2):
du
NaHCO,
(hay
"CO2 ~ "HCI
"Na2C03
hay:
"C02="H "„„2-
"NaHCO,

~"Na2C03
~ "CO2
HCO^.
hay:
HCOJ
Co\~
T>2
(2) HCI (hay
)
du
"c02-"co2-
:=> Vay
trong bai toan nay nd'u
c6
khi
CO2
thoat
ra
thi trong dung djch chi
c6
the
condirHCO^.
>i
(2) Sgugfc
Iqi neu ta cho hon hap 2
chat
NujCO}
(hay ion CO^') vd
NaHCOj
(hay

ion
IICO
3 ) tit tit vdo
dung
dich
axit
thi cd 2
phdn icng tren
xdy ra
dong thai.
Vidu
/. (1) Cho
rat tir tir dung dich
A
chiia
a
mol HCl
vao
dung dich
B
chiia
b
mol
NajCO,
(a < 2b)
thu
dugc
dung dich
C va V
lit khi (dktc).

(2)
Neu cho
dung dich
B vao
dung djch
A ta
thu
duoc
dung djch
D va V,
Ii't khf.
Cac phan
lirng
xay ra
hoan toan; the'
ti'ch
khf
la'y o
dktc. Ti'nh
V va V,
theo
a, b.
Gidi
HCl
+
Na2C03
^
NaHCO,
+
NaCl

(1)
NaHC03
+
HCl
NaCl
+
COj
+ H2O (2)
Vi
sau phan ung thu
duoc
khi nen phan ling
(2) da xay
ra.
Suy
ra
Na2C0,
da
phan ung het or
(1)
S6' mol HCI
con
lai sau phan u'ng( 1)
la : (a-b)
mol
"Naiico3 = b
(mol)
>
"HCI (du)
= (^1" b) (mol)

=
(a-b):
(a)
PTHH:
"CO2 ~ "HCI (dir)
V
=
22,4(a-b)lit
(b)
Na2C03
+
2HCi->2NaCl
+
H20
+
C02
Gia sir
Na.CO,
phan urng het thi
so
mol HCl can dung lii
2b
mol.
Ma
theo
d^ bai
thi
a<2b
vay
HCl phan ung het,

Na.CO,
con du.
n^o^
=0,5a
(mol)=>
V, =
1 l,2a(lit)
du
2. Cho
tir tu dung djch chiia
a
mol
HCl vao
dung djch chiJa
b
mol
Na^CO,
dong
thdi
khuay deu, thu
dugc
V lit
khf
(d
dktc)
va
dung djch
X. Khi cho du
nuoc
voi

trong
vao
dung djch
X
thay
c6
xuQ't hien ket
tiia.
Bieu thuc lien
he
giua
V vol
a, b la " •
A. V =
22,4(a
- b). B. V =
1
l,2(a
-
b).
' '''
V
=
11,2(a
+ b). D. V =
22,4(a
+ b).
Gidi
can toan tuong tu
vf du I

(dung djch
X tao ket
tua \6\c
voi
trong du
co
ghia
la a <
2b).
Bo dS'
Hda
hoc 9 on Uii vAo lO
-
Pham
Sf
Im
HCl
+
NajCO, NaHCO,
+
NaCl
(!)
1 PTHH
:
JNaHCO^+ HCl^NaCl
+
C02+
H2O
(2)
j

[2NaHCO,
+
Ca(OH)2
->
CaCO,
+
Na2CO,
+ H2O
(3)
' Do da thu duoc khi CO,
va
sau
do
dung
dich
thu duoc tao ket
tiia
vdi nu6c vol
trong
nen
:
b<a<2b.
5
iiwinfid
i^uo!;
v-ij/<
"NaHCO.,
=
b
("lol)

>
nHci
(du)
= "
b) (mol)
nco2 =
"HCI (du)
=
(a
-
b) =>
|
V
=
22,4(a
-
b)lft
Chon
A.
Vidu
3. Nho tir tir
tirng
giot den het
30ml
dung
djch
HCl IM vao 100ml dung
dich
I
chu-a

NajCOj 0,2M va NaHCO., 0,2M, sau phan
ling
thu
diroc so mol
CO,
la
'
A.
0,020. B. 0,030. C. 0,015. D. 0,010.
Gidi
Na2C03
(0,02mol)
]
Hciiinif
[NaHCO,
'
NaHCO, (0,02 mol)
J
0,02
mol
uk-dung
[(0,04mol)J
HCl
lir tir
CO2
(0,01
mol)
PTHH:
0,01
nn)l l;\

dung
HCl
+
Na2C0,
^
NaHCO,
+
NaCl
(1)
NaHCO,
+
HCl
->
NaCl
+
COj
+
H2O
(2)
"CO2
="HCI
-"Ninco,
=0,03-0,02 =
0,01
mol => Chon I).
LOA/5: KIM
LOAI
SAT TAC
DUNG
V6l

DUNG
DICH
AXIT
SUNFURIC
Fe
tac dung voi
H2SO4
c6 cac trirong h(.rp sau day xay ra
:
H,S04
dac nguoi
:
thu dong hoa hoc
siit
(khong phan u"ng)
H,S04
loang
: Fe +
H,S04
^
FeS04
+
H21
(1)
H,S04 dac va nong
:
2Fe
+
6H,S04
->

Fe2
(SO4),
+ 3
SO2
+
6
H2O
(2)
Neu Fe con du:
Fe +
Fe2
(SO4),
-> 3FeS04
(2)
+
(2)
^
Fe
+
2H,S04
->
FeS04
+
SO2
+
2 H2O
Lien
he
nn^s(.)4
vc'rf

np^. va .san pham khi c6 2 muoi
2Fe
+
6H2SO4
->
Fe2(504)3
+
3SO2
+
3H2O
Fe
+
2H2SO4
->
FeS04
+
SO2
+
2H2O
(2')
(3)
Neu theo
(2
va 3):
"Fe -2npe2(S04),
+"FeS04
"H2SO4
=6npg2(so4),
+2npgso^
_

1
"Fe2(S04)3
" 2
"H2SO4
"FeS04 "^^Hpg -nH2S04
Fe
Neu theo
(2
va
2) :
2Fe
+
6H2SO4
->
Fe2(S04),
+
SSOj
+
3H2O
Fe
+
Fe2(S04),
->3FeS04
=>
Hp,.
= 3n,
FeS04 ^"FeCdu)
-3
1
"Fe

-
2""2'»4
= 3nFc
-"H2SO4
Bao loan s6' nguyen tu sat:
nFe2(S04)3
= 2("'''=
""''^''^4)^
2
"^^ ~
(^"f'<=
~
""2-^4
)
'^2""2'504
• Bang tong hop
:
-
Up,,
b
'Fc
T=l
o a=b
T> 3
o a>
3b
T
< 2
o
a

<
2b
2<T<3
Phan
ij"ng
va
san
philm
(1).
FeS04+
H2t
(2). Fe2
(504)3+
502?
"FC2(.S()4).,
- 2'"''''
(2). Fe504
+
5O21
"FC.S()4
-
"FC
(2) va
(2)
hoac
(2)
va (3).
Fe2(S04)3,FeS04, S02t
_ 1
"FC2(.SC:)4)3

-
2""2S04
"F<^
"Fe.S04 -^Hpe
"H2.SO4
Dieu
ki6n
H,S04
loang
H2SO4
dac, nong, du
H,504
dac,
nong,
thieu
H25O4
dac, nong.
Fe
va
H:504
deu
het.
Vi du 1. Cho 6,72 gam
Fe
vao dung
dich
chua 0,3 mol H,504 dac, nong (gia
thict
SO,
la

san phdm
khir
duy nha't) thu duoc dung
djch
B.
Tinh
so
mol
muoi
sdt thu
duoc sau phan
ling.
A. 0,04 mol Fe2
(SO4 )3
va 0,04 mol Fe504 B. 0,06 mol Fe2
(5O4
)3
C. 0,03 mol Fe2
(SO4 )3
va 0,06 mol
FeS04
D. 0,12 mol
FeS04
Gidi
"Fe
=
0,12 mol. nH2S04
=
0.3 mol.
<|V'J ' ' '

2<X
=
.!!M2SO4_^030^
2
=>c6
2mu6'i
Fe2(S04), va Fe504
npe
0,12 ' ' •
J2Fe
+
6H25O4
Fe2(504)3
+
35O2
+
3H2O
,
Fe
+
2H2SO4
^
FeS04
+
SO2
+
2H2O
nFe2(S04)3
=2"H2S04 " "Pe ^
|nFe2(S04).,

= 0,12 - 0,03
=
0,030101
nFc-so4
=3nFe
-nH2S04
[nFc.so4
=3.0,12-0,30 = 0,06mol
Vi
du 2. Mot dung djch co chu-a b mol
H,S04
hoa tan het a mol Fe thu dirge khi A
va 42,8 gam mu6'i khan. Nung lugng muoi khan do a nhiet do cao trong diau
kien
khong eo khdng khi den
khdi
lugng khong doi thu
duge
h6n hop khi B.
(a)
Tinh
gia tri eiia a, b (biet ty s6' - =
—).
' "
b
6
(b)
Tinh
ty
khoi

eiia h6n hgp khi B so \6\g khi.
Gidi
(a) Tlieo de ra : i
a 2,5
'Fc
•
H2SO4
dae va nong, khi A la SO^
•Tao muoi Fe,(S04), va FeSOj
PTHH:
2Fe + 6H2SO4 Fe2(S04), +
3SO2
+
3H2O
Fe +
2H2SO4
-4
FeS04
+ SO2 +
2H2O
Theo eac
PTHH
ta eo :
1
b-a
"Fe2(S04)3
'
^"^HiSL^j,
-»F
"FC.SO4

=3np^ -n,^2S()^
=
^x2,4a-a=0,2a(mol)
(b)
"Fc2(.S04)3
nFe,so4
= (3a - b) = 3a - 2,4a = 0,6a (mol)
>
400X
0,2a + 152
X0,6a
= 42,8 gam ^ a = 0,25 mol
>
b = 2,4x0,25 = 0,6 mol
n
Fc2
(S()4
„
=
0,2a = 0,2
X
0,25 = 0,05
mol
"FC.SO4
=
0,6a
=
0,6
X
0,25 = 0,15mol

PTHH:
2Fe2(S04)3—!^2Fe203+6S02+302 ,
4FeS04
—>2Fe20, +
4SO2
+ O2
Thu
dugc h6n hgp khf:
nso2 "^"f-V2(.s(i4),
+nFc.sc)4
=3x0,05
+
0,15
=
0,3 mol
"02 =>'5nFe2,s()^,^ +0,25np,so4 =1,5x0,05 + 0,25x0,15 =
0.1125
mol
Ti
kho-i
doi voi khong khf: d,^ = ">3x64 + 0,l 125x32 ^
(0,3 + 0,1125)
X
29 '
54
^0.
Phuong
phdpdothj
Cac bai toan tao thanh ket
tiia

khi sue khf CO. vao dung djch nir6c voi trong,
dung dich
ki6m
vao dung djch muoi nhom hay them axit vao dung djch mu6'i
1
niinat
CO
the
giai
bang phuong
phap
sir dung
gidi
han ti le mol. Ngoai phirong
^[llip
tren thl cae bai toan nay cung co the
giai
bang phuong
phap
do thi.
^ Dgng
1. Khf CO, tdc di/ng vdi dung djch Ca(OH)2 hay BaCOH),
PTHH:
CO2 + Ca(0H)2 -> CaCO, + H2O
CO2 + CaCO, + H2O -> Ca(HC0,)2
(1)
(2)
Do
thi:
nco2

=
X
nCa(OH)2=
^
nCaC03 = y
0 xi a X2 2a ^ "^^^
Tr6n
true y chgn diem y = a, trdn true x chon 2 diem x = a x = 2a. Tir
diem
a eiia true y va a eiia true x, ke vuong goe ehiing giao nhau tai di^m P. Tir P
noi
vdti
tga do O va 2a ta dugc tam giae.
Vdri
mot gia tri
ciia
s6' mol
kfi't
tiia
tren true y, ke ducmg song song vdi true x
cat tam
giae
tai 2 diem Q va R, tif Q va R ke vu6ng goe v6i true x ta co cac gia
trj
X,
va
Xi
ciia
so mol CO2.
fy

= x vai 0<x <a
Ta
eo
:
y = (2a - x) vdi a < x < 2a i j
y
= 0 vdi x > 2a
Tir
do thj suy ra :
Oiig
vdi m6t gia trj eiia x
(O
< x < 2a) thi luon cd
1
gia trj eiia y
Neu
cd mot gia trj eiia y
(O
< y < a) thi lu6n cd 2 gia trj eiia x
du. Ha'p thu hoan toan
2,688
Ift khf CO, (dkte) vao 2,5 1ft dung djch Ba(OH),
nong do a
mol/Ift
thu dirge 15,76 gam ket
tiia.
Gia trj
ciia
a la
A.

0,032
B.0,06 C. 0,04 D.
0,048
Gidi
nco2 =(2,688:22,4) = 0,12 mol;
"BaC03
=('5,76:197) = 0,08 mol
nsaCOa
'nco2
=0,12
"Ba(OH)2=
2,5a
"BaCO.,
= 0'08
0,08 2,5a 0.12
5a ^
Theo
d6 thj
urng
vdi
gia
trj
y =
0^,,^)^
=
0,08 ta
c6 2 gia
tn cua s6' mol khi
CC
la

X, =y =
0,08<0,12
va Xj =
2a
- y o
0,12
=
5a-0,08
a -
0,04M
Vay
: a =
0,04M
=>
Chon
C.
^7(7/7^
Mud'i nh6m tdc di/ng vdi dung djch
ki§m
mqnh
PTHH:
<
3NaOH
+
AlCl,
-
OH"^
+
AKOH),
-

->Al(0H),+3NaCI
(1)
->A10-+2H20
(2)
Do
thi:
' "AI(OH),
•
B
7
O'!;••••,)
i- a
P
7
O'!;••••,)
i- a
0^^-^^
:\
^•-"""^
1 ! ! \ n
:/::
0
-
"OH
3a
•''2 4a
Trcn
true
y
ehon diem

y =
a,
tren
true
x
ehon
2
diem
'^AI(OH)7
^
'OH
=
x
diem
a
eiia
triie
y va 3a
eiia
true
x, ke
vuong
goe
cliilng
giao
nhau
tai diem
P. Tit
P noi vdi toa
do O va 4a

ta duoe tam giac.
Vdfi
mot
giii
tri eiia
so
mol
kct
tiia
tren
true
y, ke
duong song song vdi
true
x
eit
tam giac tai
2
diem
Q va R,
tir
Q va R ke
vuong
goe
v6\e
x ta c6 cae gia
trj
X| va X,
cua
so

mol
OH .
Ta c6:
y
-
—
khi
0< X <3a
y
=
(4a-x)
khi 3a<x<4a
y
= 0
khi x>4a
Tir
d6
thj
suy ra:
• Uiig
voi
mot
gia
trj cua
x (O < x < 4a)
thi
luon
c6 1 gia
tri ciia
y

Nd'u
CO
mot
gia
trj eiia
y (O < y < a)
thi
luon
c6 2 gia
tn eua
x
56
yidu-
Cho 200ml dung
dich
AlCl, 1,5M
tac
dung v6i
V
lit dung
djch
NaOH 0,5M,
lucmg
ket tua
thu
duoc
la
15,6 gam.
Gia
tri

lorn
nha't
ciia
V la (cho H =
1,
O =
16,
Al = 27)
A.
1,2. B. 1,8.
C.2,4.
D.2
••'•.•r\:
Gidi
,
= lM=o,2(mol)<nA,ch
=0,2.1,5=0,3(mol)
>
"AKOH),
78 ^
ni,
Al(OH), tan m6t
phdn
trong NaOH
.
ifli
3NaOH
+
AlCl3->Al(OH)3+3NaCl
(1) ^i,: .

NaOH
+
Al(OH)3
->
NaAlO^
+
2H2O
(2)
j()fi;
;uv
PTHH
:^
D6
thi
:
"AICI,
=^'3
"AKOHXr^'^
0,9^2
"NaOH
Tir
d6
thi
vdi
n.
'Ai(OH)3
=y=o,2
.X,
=3y =
0,6mol=^V, =(0,6:0,5)

= 1,2 lit
.
X2 - (4a - y) = 4
X0,3-0,2
= 1,0
mol
V2 =
(1,0:0,5)
-
2,0 li't
V
CO gia
trj tri Idn
nha't
vSy : V =
2,0 lit
=>
Chon
D
Dgng
3.
Mud'i aluminat tdc di/ng vdi dung djch axit mqnh
• PTHH:
NaA102+
HCl
+
H20-^Ai(OH)3
+
NaCI
(1)

A1(0H)3
+
3HC1
->
AICI3
+
3H2O
(2)
D6
thi:
I
"A1(0II)3
p
"AI(OH)3
y
-
"AKV
=^
/\
Xi
a
^2
4a
Tren
true
y
chon diem
y = a,
tren
true

x
chon
2
diem
x = a va x =
4a
. Tir
•^Ifm
a
ciia
true
y va a
ciia
true
x, ke
vuong
goc
chung giao
nhau
tai diemjVT&T
57