chuyên đề ôn thi môn Hóa THPT

ca'm
nang
6n
luyjn
thi dgi hgc 18
chuyfin H(Sa hpc
-
NguySn
Van H3i
Lot
gidi:
Cach
1:
Khi hoa
tan vao
dung dich
H2SO4,
Fe(N03)2
se
phan li thanh
cac
ion.
Do
vay,
truoc
het cac em can tinh so mol cac ion nhu sau:
O'
•
Phuong trinh ion rut gon:
'
^''^''/'i

•••^'J^/•^ :>
<>^:r'^
^i^mrp
3Cu
+ 8H* + 2NO^ •
>
3Cu2-
+
'^NO
+
4H2O
Mol:
a3 <- 0,8 0,2 -> 0,3 0,2 cO/; ,
Cac
em luu
y,
Fe^* cung bi oxi hoa thanh Fe^* va giai phong khi
NO:
3Fe2*
+ 4H* +
NO;
> 3¥e^ + NO +
2H2O
Mol;
0,6 ^ 0,8 ^ 0,2 0,6 0,2
^
^
_
->
V

= (0,2 + 0,2).22,4
=
8,96
lit
->
Dap an B.
rr,fj{),p«
y 4-
Cach
2: Cac
chat
khu: Cu
- 2e ->•
Cu^^;
Fe^^
- le -> Fe^*.
Chat oxi hoa: N*-^
+ 3e -> NO
Bao
toan
electron:
2ncu
+
lnp^2+
=
Sn^o
2.a3+a6
, vr ^
riNO=
z =0,4 mol

^
V
= 0,4.22,4 = 8,96
lit
^
Dap an B.
Vi
d^ 13:
Cho
3,84
gam Cu vao
200ml
dung djch
gom
NaNOs 0,2M
va
H2SO4
0,5M,
tao
thanh Vml khi NO (san pham khu duy nhat,
0
dktc)
va
dung djch
-I
X.
Cho Vml dung dich
NaOH
2M vao
X de

thu
dugc
lugng ke't tua Ian nhat.
.
Gia trj nho nhat ciia V la
A.
80. B.50. C.60.
D.
40.
Ldi
gidi:
Nhan
xet: Dung dich chiia muoi NaNOs
va
H2SO4
loang
-> can
giai
theo
phuong trinh ion.
;?>
.v,
t, •5
ncu
=0,06
mol;
n^+ = 2.0,2.0,5 = 0,20
mol; n^^,
=0,2.0,2 = 0,04
mol.

Phuong trinh ion thu gpn:
3Cu
+ 8H* +
2NO3-
>
3Cu^*+
2NO
+H2O
;^Mol:
a06 ai6 a04
MjoriX^:-
Cu
tan het
-> X
CO
chiia: n^ 2+
= 0,06; n +
du
= 0,04. '
De
thu dugc luang ke't tua
Idn
nhat thi:
njs^aOH
= n
+
+
2n
2+ /
H

Cu
"NaOH
= 0,04 + 2.0,06 = 0,16
mol
y
_>
V
= 0,08
lit
=
80ml
—> Dap an A. ;
.^w
Cty
TNHH
MTV DWH
Khang Vigt
14: Hoa
tan
hoan
toan
0,1
mol FeS2
trong
200ml
dung dich HNO3 4M,
san pham thu dugc gom dung dich
X va
mot
chat

khi
thoat
ra. Dung dich
X
CO
the
hoa tan toi
da m
gam Fe. Biet
trong
cac
qua trinh tren, san pham khu
duy nhat cua
N*^
deu la NO. Gia tri ciia
m
la
A.
5,6. '
B.7,0.
C.8,4.*
'
'V',
•
Lai
gidi:
ry,
•;
D.
2,8

->
Fe(N03)3
+ 5NO +
2H2SO4
+
2H2O
0,1
0,2
phan ling hoa hoc:
FeS2
+
8HNO3
-
Mol:
0,1 0,8
Dung
dich
X
gom cac ion:
Fe3^
=
0,1 mol;
H*
= 0,4
mol;
NO
;
=
0,3 mol va
SO

]' = 0,2
mol.
^,
^, , ., , „ "
WO '-*
"'"^ * mi
Cac
phan ung hoa tan Fe:
Fe
+ 4H* +
NO;
>
Fe3*
+
NO
+
2H2O
QkHi
Mol:
0,05 4-0,4 -> 0,05
••idl,0»
j„rtS«.^^iT
'ihtvU/
Fe
+
2Fe3^
>
3Fe2-
q%
lOH

dfegfror*
,;>fefii
JfiSv)
Mol:
0,075 <- 0,15
—> mpe
= 0,125.56 = 7,0
gam
->
Dap an B.
Vi
d^ 15:
Iron
200ml
dung djch
X gom
Ba(OH)2 0,1M va
NaOH
0,3M voi
100ml
dung dich
Y
gom
Al2(S04)3
0,1M
va
H2SO4
0,1M, thu dugc
a
gam ke't

tua.
Gia
tri cua a la
A.
6,22.
B.
4,66.
C.
5,82.
D.
5,24.
Laigidi:
* ^
TrongX:
n„
2+=
0,02
mol; n,,_+
= 0,06
mol; n^^_
= 0,10
mol.«^'^->
OH"
Trong
Y:
n
3+
= 0,02
mol;
n^^j-

= 0,04
mol;
n^+ = 0,02
mol.
Cac
phuong trinh phan ung khi pha trpn:
H+
+ OH' > H2O
0,02 ^ 0,02 ,
AP*
+ 30H" >
Al(OH)3i
0,02 0,06 -> 0,02
Al(OH)3
+ OH" > AlO; +
2H2O
0,02 <- 0,02
,Ba2^
+ S04~ >
BaS04>l
0,02 -> 0,02 ^ 0,02
Mol:
'ol:
±
oh
Vgy:
a = 0,02.233 = 4,66
gam
->
Dap an B.

C&m
nang
fln
luyjn
thi dgi hgc 18
chuy6n
dg H6a hoc
-
MguySn
Van H5i
Vi
dv 16: Hoa tan
hoan
toan
9,46 gam
hon
hgp
gom Na,
K va Ba vao
nuoc,
thu
duQC dung dich
X
va 1,792 lit
khi
H2
(dktc). Dung dich
Y
gom
HCl

IM va
1^
H2SO4 0,5M. Trung
hoa
dung dich
X
boi
dung dich
Y,
tong khoi lugng
cae
muol
duoc
tao ra la
, i
A.
14,72
gam.
B. 16,14
gam.
C. 19,98
gam.
D. 17,14
gam.
Lai
giai:
1,792
=
0,08
mol.

.
'2
22,4
Cac phan ung
hoa hoc:
Na
+
H2O
1
K
+
H2O
Na+
+
OH"
+ -H2
2
K+
+
OH-
+ -Hi
H
( r{
mi'n>
tV
i
i.
Iv
P
J

r.i
I
1^x4
f
1
1,
t fin J^j ti )
Ba
+
2H2O
>
Ba^^
+
20H"
+
H2
"M
' 1
N/jflnxet:
n^„.
=
2nH,
=0,16
mol.
OH
^
Mat
khac,
nong
do

HCl
gap
doi
H2SO4
->
trong cung mot
the
tich thi
„,p.r,
ft
T<
j\riv,
*.
,
^
Trong
Y:
•
"H+
=
"HCl
+ 2nH2S04 =
2a + 2.a = 4a
mol
n
= 2a
mol;
n ,
=amol.
cr

so|"
H2O
Trung
hoa X
boi
Y: H++ OH"
•
->
4a = 0,16 -> a = 0,04
mol.
Khoi
lugng muoi thu
dugc
= 9,46 +
m^^.
+
mg^a-
V
= 9,46 +
a08.35,5
+ 0,04.96 = 16,14
gam.
;
—> Dap
an B.
Vi
d\ 17: Cho hon hop gom Ba va Al (ti le mol 1:1) vao
nuac
(du),
thu

dugc
dung djch
X va 1,12
lit khi
H2
(dktc). Dung dich
Y
gom HCl
0,5M va
H2SO4
0,1M.
Cho tu tu den het
100ml
dung dich
Y vao X,
thu
dugc
m gam ket
tiia.
Giatricuamla
^
-
a ,
i,*-^a'^.r,*•••
,
.WUt,
•-i^;
•:
:
A.

0,78
gam.
B. 1,56
gam.
C. 3,11
gam.'
'
D. 5,44.
^.•vr.,
Laigidi:
J-'
^
;
.X^CMU.
.,
"H2
= 0,05
mol.
Cac
phan ung
hoa hoc:
^OJ)
• *
S
,
Ba +
2H2O
>
Ba2*
+

20H"
+
H2
Mol:
a a 2a a
^9
Cty
TNHH
MTV
DVVH Khang Vi$t
Al
+
OH-
+
H2O
AIO2
+
-H2
2
;;i
•
Mol:
a a a 1,5a
Ta
c6:
nHj
= 2,5a
=
0,05
mol

-> a
=
0,02
mol.
Trong
X:
n^^^_
= 0,02
mol; n^^2^
= 0,02
mol; n^,Q_
= 0,02
mol
Trong
100ml
Y:
n^+=
0,07
mol; ngQ2-= O'Ol mol; n^|_ =
0,05
mol.
Khi
cho
tu tu
Y vao X:
. <
OH-
+
H+
-

0,02 -> 0,02
Mol:
AIO2
+
H+
+
H2O
>
Al(OH)3
4'
Mol:
0,02
-> 0,02
0,02
Al(OH)3
+
3H+
>
Al3+
+
3H2O
Mol:
0,01 <- 0,03
dnSrf/;
Ba^+
+
SO 4-
>
BaS04>l'
Mol:

0,01 <-
0,01
-> 0,01
->
m=
mA,(OH)3
+
mBaS04
=
0-01-78+
a01.233
= 3,11 gam.
: .m
->DapanC.
, , , , V,,,'
9.
PHl/ONG PHAP
LIEN
HE NGUYEN
T6
-
NH6M
CHLTC
a.
Npi
dung
'
So
mol
cac

nguyen
tu c6
trong nhom
chiic
luon ti 1^ thu|in voi so' mol nhom
chuc.
, torn
•
•
-
r'h-Hn
'fd
b. Cac tnrong hg(p thuang
gap
Phan
ling
hoa hpc
Moi
lien
h|
Cach
tinh
ROH
>
RONa
+
^ H2
2
no=2nH2
RCOOH

^NaHCOg
^
RCOONa
+
CO2 + H2O
^
no=2ncooH
no=
2nco2
R_NH2
)
R-NH3CI
nN
=
riHCi
Vi
Dv
MAU
Vi
dv
1
(A-09):
Khi
dot
chay
hoan
toan
m gam hon hgp
X
gom hai

ancol
no,
don
chuc,
mach
ho
thu
dugc
V
lit khi CO2
(6
dktc)
va a gam
H2O.
Bieu
thu-c
lien
h?
giiia m,
a va V la:
dm
nang
On
luygn
thi dji hpc 18
chuy6n
H6a
hpc
-
IMguySn

Van HSi
V
V
A.
m=
a- —.
. , B. m
=
2a
.
5,6
: . f;,: v'X,- 11,2 ' ; 'y }
C
m
=
2a-^.
' D. m =
a
+
-^. '
•
^^^^^
22,4
,< ,,,) 5,6
^J,,
•
^ n :;j.|iT
Nhan
xet:
Khi do't chay

cac
ancol (no, don chuc, m^ch ho),
ta
luon
c6:
a
V
"ancol
-
nH20
-
"CO2 nancol=
— -
18
22,4 - ., ,
,.^^5^
Do
X
chua
cac
ancol dan chiic
no
=
noH
=
nx
-> no
18
22,4
Bao toan

kho'i
lucmg: mx
=
mc
+
mH
+
mo
, ^
V
, a , a V , V
oM
->m=12.
+2.—
+ 16.(
)->m
= a . cfi n
.r^j,/
22,4
18 ^8 22,4^ 5,6 ^^^'^ -^"^^
-> Dap an A. Qs^* •*
"••'''A
<—• '^'Ht '
fCf-if:)^rA
Vi
du
2:
Dot
chay hoan toan
m

gam
hon
hop
Y
gom
ba
ancol
don
chuc, thuoc
cung
day
dong ding, thu dugc 37,4
gam
khi
CO2
(dktc)
va
27 gam
H2O.
Gia
triciiamla
iO,U
liJ#'
aoM
A.
27,1.
P'''rfm B. 28,6. C.23,6. ,3m
+
D.
37,1.»-

m
37
4 i
"COT^—^
=
0,85 mol->
nc =
0,85 mol.
44
27
'-'^
'
WtyfAiHi>HOXH-M
v'
nH20=
— = 1.5
mol
nH
=
3,0 mol.
^^^^^
Cac
em
can
thay rang, khi
dot
chay
Y:
nnjo
>

"cOa
^
chtia
3
ancol
no
-> nv
=
nH-o
-
nco,
= 1'5
-
0,85
=
0,65 mol.
, , ^
Do
Y
chua
cac
ancol don chuc
->
no
=
noH
=
nx
->
no

=
0,65 mol.
Bao toan
khoi
lugng:
mv
=
mc
+
mH
+
mo
=
0,85.12
+
3,0.1
+
0,65.16
= 23,6 gam.
-> Dap an
C.
Vi
dy
3:
Do't chay hoan toan mot lugng hon
hop
X
gom
hai ancol (no,
da

chuc,
mach
ho,
ciing
so
nhom
-OH) can vua
dii
V
lit khi
O2,
thu dugc
5,6
lit khi
CO2
va
6,3
gam
H2O
(cac
the
tich
khi
do 0
dktc).
Gia
trj ciia
V la a t
A. 5,60.' B.3,92.
C.

7,28.
D.
1,12.
Laigidi:
' r •
5,6 6 3
'
''^''^.ymm-Mtifbw
yi:
^C02
mol; nH20=0,35 mol.
;, . ^
Theo
bai ra,
X
chua
2
ancol
no
->
nx
=
n^jo
"
"CO2 "^ol •
Cty
TNHH
MTV DWH
Khang
Vigt

So nguyen tu
C
trung
binh
=
"'''"^
=
2,5
X
chua mgt ancol
da
chuc
c6
so nguyen tu
C
nho hon 2,5ancol
do la
C2H4(OH)2.
• '
*
po
X
chiia
2
ancol cung so'nhom -OH
-> 2
ancol deu hai chuc:
^ • '
^ no
=

noH
=
2nx
->
no
=
0,2 mol.
, .
Bao toan nguyen
to
O:
no(OH)
+
2no2
=
2ncoj
+
nH20
''''^'''^'
^ ^2.0,25
+
0,35-0,1
^Q
225 mol
^ =
0,325.22,4
=
7,28 lit.
Dap an
C.

Vi
dv 4
(CD-12):
Dot
chay hoan toan hon hg-p
X
gom
hai ancol
(no,
hai chuc,
mach
ho)
can vua
du Vi
lit khi
O2,
thu dugc V2 lit khi CO2
va a
mol
H2O.
Cac khi
deu
do 6
dieu kien
tieu
chuan. Bieu thuc lien
he
giiia
cac gia
tri

Vi,V2,
a
la
A.
Vi
=2V2 +ll,2a.
B.
Vi
=V2 -22,4a.
;
C.
Vi =
V2
+
22,4a.
D.
Vi =
2V2
-
11,2a.
, / i ,
Lbigidi:
OH ;
V2
Theo
bai ra,
X
chua
2
ancol

no ->
nx
=
nH20
'
^C02
~ ^ '
2V2
Do
X
chua
2
ancol hai chuc
->
no
=
noH
=
2nx
->
no
= 2a -
Bao toan nguyen
to O:
no
(on)
+
2
no2
=

2 nQQ^
+
VXH^Q
2V2
^ 2Vi _ 2V2
22,4
2^
22,4
->
2a - +
^=-^
= ^-^
+
a Vi =
2V2
-
11,2a
on
22,4
22,4 22,4
-> Dap an
D.
Nhan
xet:
Bai
nay cac em
can nho
voi
ancol
no

thi:
nancoi
= n^^Q
"
"CO2
'
dong
thai
bie't
ap
dung
bao
toan nguyen
to
oxi.
Vi
d^
5
(B-12):
Do't chay hoan toan
m
gam hon
hgp
X
gom
hai ancol, thu dugc
13,44 lit khi
CO2
(dktc)
va

15,3
gam
H2O.
Mat khac,
cho
m
gam
X
tac
dyng
voi
Na
(du), thu dugc 4,48 lit khi H2 (dktc).
Gia
trj cua
m la -
A. 12,9.
B.
15,3. C.12,3.
D.
16,9.
.•v^
.
Lbigidi: •
, • • -(i
=c
~J
ui
r*3Y^;^:-
^^002

=
0/6 mol;
nH20=
0,85 mol; nH2
=
0,2mol.
, ^
,a,Rt;
q? i •
Dva tren moi quan h| nguyen
to'
- nhom chuc thi
voi
ancol:
no
=
noH
no=noH=2nH2
=
0,4mol.
Meit khac:
nc=
nco2
=
0,6 mol;
nH =
2nH20= 1/7 mol.
Ca'm nang 6n luy^n thi
d<ii
hpc 18 chuy6n 6i H6a hgc - Nguyin Van H^i

Bao toan
kho'i
lugng: mx
=
mc
+
mn
+ mo
= 0,6.12
+
2.0,85.1
+
0,4.16
=
15,3 gam.
Dap an B.
Vi
du
6
(B-12): Cho hon hop
X
gom metanol, etylen glicol
va
glixerol. Do't chay
hoan toan
m
gam
X
thu dugc 6,72 lit khi
CO2

(dktc). CQng
m
gam
X
tren cho
tac dung voi
Na
du thu dugc to'i da
V
lit khi
H2
(dktc). Gia tri aia
V
la
A. 3,36.
„
B. 11,20. C.5,60 y.g; D. 6,72.
• Lmgidi:
•
. nco2 =
0'3mol
^
nc=a3mol. 0'mti^Qj
Nhan thay trong
X
cac ancol deu c6 so'cacbon
=
so'nhom
chiicl
"CllVjt

*
ijj
(
nc
=
noH
=
0,3 mol.
,r
p
Mat khac, khi cho
X
tac dung voi Na:
noH=
^^H2
~*
'^H2
~
O'^^
mol. s;
->
V =
0,15.22,4
=
3,36 lit
^
Dap an
A.
Vi
du 7: Hon hop

X
gom hai axit cacboxylic don chuc. Dot chay hoan toan 0,1
mol
X
can 0,24 mol
O2,
thu dugc
CO2
va
0,2 mol
H2O.
Cong thuc hai axit la
A. HCOOH
va
CH3COOH.
i
B.
CH2=CHCOOH
va
C2H5COOH.
io,
rm
•
-f
i
C.
CH3COOH
va
C2H5COOH.
, ;

D.
CH3COOH
va
CH2=CHCOOH.
*
3iwb fefl
loarifi
dG
Nh|n
thay
X
chua 2 axit cacboxylic don chiic
->
chua
2
nguyen tu oxi
^ no
=
2ncooH
=
2nx
—>
no
=
0,2 mol.
Bao toan nguyen to O: no
(COOH)
+ 2 no,
=
2 nco, + "HOO

,,
^
^ ^
,;,fj
fin
qP;Uv.4™-
^
->
0,2 + 2.0,24
=
2 nco2 + 0,2 ncoj
=
0/24 mol.
•
rtljiW
+ So' nguyen tu
H
trung
binh
=
—S2.=
^-^'^ =
4
_>
Loai
B va
C
(vi cac axit
"X
0,1

, chua so'nguyen tu
H > 4.
+ So' nguyen tu
C
trung
binh
= =
-5^= 2,4
->
Loai
A
(vi cac
axit chua
0,1
so nguyen tuC
< 2).
-> Dap an D.
^
'••m'r\fl•„,,••
Vi
d\
8
(B-11): Hon hgp
X
gom vinyl axetat, metyl axetat
va
etyl fomat. Dot
chay hoan toan 3,08 gam X, thu dugc 2,16 gam
H2O.
Phan tram

so
mol ciia
vinyl
axetat trong
X
la
A. 75%. B. 72,08%.
C.
27,92%. D. 25%.
Cty
TNHH
MTV DWH Khang
Vijt'
Lai
gidi:
„ = ^^=
0,12
mol
nH
=
0,24
mol.
'^'^v,
off.
r
"H2O
18
Hon
hgp
X:

vinyl axetat
(C4H6O2),
metyl axetat
(C3H6O2)
va
etyl fomat
(C3H6O2).
Cac
em
can thay rang cac chat trong
X
deu chua
6
nguyen tu
H ->
Khi dot
chay 1 mol X, thu dugc
3
mol
H2O
->
nx
=
•^riH20
=
0,04 mol.
Cac chat trong
X
deu chua
2

nguyen tu
O
no
=
2nx
=
0,08 mol.
Bao toan
kho'i
lugng cac nguyen to' trong hon hgp X, ta c6: ,.;!
>
mx
= mc
+
mH
+ mo ->
mc
=
3,08-0,24.1 -0,08.16
=
1,56gam.
-
->•
nco2 = '^c
=
0,13 mol.
' '
>
•
Nhan thay, khi dot chay

1
mol moi chat metyl axetat
va
etyl fomat deu thu
dugc
nco2
=
nH20'
"eng voi vinyl axetat
thi:
n^^^
-
n^^o
=
1-
v /
Do v^y:
nvinyi axetat
=
r\QQ^
-
nH20 0,13
-
0,12 =
0,01.
' "
->
%
nvinyi axetat
=^^.100% = 25%

-¥
Dap an D.
,
,bfniJ3»,,,A
Vi
d\
9
(A-11):
Hon hgp
X
gom axit axetic, axit fomic
va
axit oxalic. Khi cho
m
gam
X
tac dung voi
NaHCOs
(du) thi thu dugc 15,68 lit khi
CO2
(dktc). Mat
khac,
dot chay hoan toan
m
gam
X
can 8,96 lit khi
O2
(dktc), thu dugc 35,2
gam CO2

va
v
mol
H2O.
Gia
tri
cua
v
la
A. 0,2. B.0,3. C.0,6. D.0,8.
N/ion xet: Khi cho axit cacboxylic tac dung voi
NaHCOa
ta luon c6:
"C02=r»C00H
->
no(X)
=
2ncooH
=
2nco2
" «^
-> no(X)
=
2. —'-—
=
2.0,7= 1,4 mol.
r,y'
22,4
"^^
• ^

Bao toan nguyen to O:
no(X)
+
2
no2 =
2
nco2
'^H20
^14
+ 2^
=2.
^ +
nH,o
^
2,2
=
l,6
+
y y =
0,6mol.
'
'22,4 44
''•••ft ^
dm
nang
6n
luygn
thi d?! hgc 18
chuy6n
dj H6a hpc -

IMguygn
Van Hai
Lot
gidi:
Nhan
xet:
Khi
cho axit cacboxylic tac dyng voi
NaHCOa
ta luon c6: j
,3
J
nc02=
"COOH
' ' •
1 344
no(x) = 2ncooH = 2 ticoj no(x)
=
2. =
2.0,06=
0,12 mol.
i'r'f,
• 22,4
Cty
TNHH
MTV DWH
Khang
Vi^t
Bao toan nguyen to
O:

no (x)
+
2 =
2
ncoj
+ riHjO
0,12
+
2.
2,016
^
2 .4.84
+
"Hjo ^ "Hjo
=
0/08 mol.
22,4 44
mH20=
0,08.18
= 1,44 gam
—> Dap an B.
Vi
dy 11: Cho m gam hon hg-p X gom hai ancol tac dung v6i Na (du), thu duoc
) 4,48 lit
khi
H2.
Mat khac, dot chay hoan toan m gam X, thu dugc 13,44 lit
khi
CO2
va 16,2 gam

H2O.
Cac the tich do 6 dktc.
Gia
tri
ciia
m la
A.
14,5. B. 15,4. C.12,2. D. 13,8.
( •
•
' Ldfigiai:
nH2 = 0,2 mol. , " ' • • '-aw
N;'"^-"
#
I
nco2
=
0,6->nc=0,6;
DHJO^O'^
->nH=l,8 . Y .,1 4 'ri *rr
*M>
„i, v
Dya
tren moi quan
h^
nguyen to-nhom chiic thi voi ancol:
no
(X)
= noH
-> no(X)=

noH"
2nH2 = 0,4 mol.
••!«?»
m-nmi
mod «;i,ib fob ,xtM
Bao toan
kho'i
lugng: m = m(- +
mH+
mo
^
m =
0,6.12
+ 1,8.1
+
0,4.16
= 15,4 gam
-> Dap an B. • M
Vi
dy 12: Hon hgp X gom 2 amino axit no (chi c6 nhom chiic
-COOH
va -
NH2
trong phan tu), trong do ti 1^ mo :
mN
=
80:21.
De tac dung vua du voi
7,66 gam X can 100 ml dung djch
KOH

IM. Cho 7,66 gam X tac dyng vtra
du
voi dung dich
HCl,
thu dugc dung dich Y. Co can Y thu dugc m gam
muoi
khan.
Gia tri
ciia
m la
A.
11,31.
B.
10,58.
C.9,85. D. 9,12.
Lai
gidi:
^
nKOH=0,lmol.
I
Nhan xet:
'»
Tu ti
1§
khoi
lugng:
A,U,A
Q
ne
cjsGf

^-
niN
21
De
thay:
ncooH
= ^KOH =
O,lmol.
mo 80 no 80/16 10
y = = _ ^
;_y^,
y[
^
21/14 3 - .
ivl^t
khac, cac em
luu
y dya
quan
h? nguyen to - nhom chuc:
no(X)=2ncoOHno(x)=0,2mol.
<•:'•:
nN(X)=
0,06 mol.
HN
= "NH2 = "HCl
^
"HCl =
0,03 mol
.

^
^
Bao toan
khoi
lugng:
, •
r ,
m = mx+
mHci
= 7,66 +
0,06.36,5
= 9,85 gam.
OrM^'
^" " ,
_> Dap an C.
Vi
13:
Dun
nong m gam hSn hgp X gom cac
este
voi 350ml dung dich
NaOH
2M,
thu dugc dung dich Y chiia muol cua mgt axit cacboxylic don chuc va
13,9 gam hon hgp ancol Z. Cho Z tac dyng voi Na du, thu dugc 4,48 lit
khi
H2
(dktc). Co can Y, nung nong chat ran thu dugc voi
CaO
cho deh

khi
phan
ling
xay ra hoan toan, thu dugc 4,8 gam mgt chat
khi. Gia
tri cua m la
A.
40,6.
B.26,6.
C. 30,7. D. 34,5. ' "
rAv>r^
itT,'.'
".sn '
Loigidi:
.\i .life
n
= 0,7 mol; n = 0,2 mol.
H2
Nhan
xet: Cac
este
tao thanh tu cung mgt axit cacboxylic don chuc.
Khi
ancol tac dung voi Na:
-OH
+ Na
-> -ONa+
-H2
iMv^jndaEv
0,2

fa;
Mol:
0,4 <-
noH=2nH2 =0,4 mol.
M|it
khac:
n.coo-=
"OH
= 0,4 mol
"RCOONa=0,4mol.
->
nwaOH
dir
= 0,7 - 0,4 = 0,3 mol.
Phan
ling
voi toi xut
(NaOH
he't,
RCOONa
con du):
RCOONa
+
NaOH
)
RH
+ Na2C03
Mol:
0,3 <- 0,3 0,3
4,8

M
RH-
0,3
-= 16
RH
la
CH4.
Bao toan
khoi
lugng: m +
mwaOH
=
mRCooNa
+
mz
-> m =
0,4.82
+ 18,4 -
0,4.40
= 30,7 gam.
—>
Dap an C.
Ca'm
nang
6n
luygn
thi dgi hpc 18 chuy§n dg H6a hqc -
Nguygn
Van Hki
Vi

dy 14: Do't chay hoan
toan
m gam hon hgp X gom hai ancol, thu
dugc
11,2
h't khi C02 (dktc) va 12,6 gam H2O. Mat khac, cho m gam X tac dung voi
Na
(du), thu
dixgc
4,48 h't khi H2 (dktc). Gia tri cua m la m 0 - . ,„
A.
12,4. ; B. 15,2. C. 12,6. D. 13,8.
. ^
Laigidi:
rico2=^=0,5mol;nH2O=-^=0'7mol.
V j ,j rto ^rt,
^
ff,
Ta
c6: mc =
0,5.12
= 6,0 gam;
mH
= 0,7.2 = 1,4 gam. ' '' <•
H
Khi
X tac dung voi Na:
n"^-:-
'<
mcj "* '•onn ni'<»'P!' wl- -

i'-^ -OH + Na >-ONa+ t
riD,b
:,:n.b
:xmt
Mol:
0,4 0,2 .„.
no = noH = 2nH2 -> no = 0,4 mol -> mo =
0,4.16
= 6,4 gam.
Bao
toan
kho'i lugng: mx = m^ +
mpi
+
mo = 6,0 + 1,4 + 6,4 = 13,8
^
j,^
^
->Dap an D. ., .
Cty TN TV
DVVH
Khang
Vi§t
CAC
AXIT
vo CO mm
HIND
1
AXIT
CLOHIDRIC:

HCl
a, Lithuyet
+
Tfnft
dung voi kim loai, bazo,
oxit
baza, muoi. Vi du: Si n
?
f
Fe + 2HC1 >
FeCh
+ Hat ''^^^ '''
CaC03+
2HC1 >
CaCh
+ COat + H2O '^^-'^^'^-^ "'^
+
Tin?z
fc^""- Tac dung voi cac
chat
oxi hoa manh: Mn02, KMn04,
KCIO3,
K2Cr207.
Vidy:
,.,^„„ „ ;„,,.,,„/, ^
Mn02
+ 4HC1 -^-^
MnCh
+ CI2
+2H2O

"
ry^.^
^^ ^^^^
2KMn04 + 16HC1 > 2KC1 + 2MnCl2
+5CI2
+
8H2O
qfi(}
K2Cr207 + 14HC1 2KC1 + 2CrCl3 +
3Cl2
+7H2O
COH/
b. Vidumau ,
^ •cr^^'
rfi '•'i-i '
Vidul:
Cho cac phan
ling
sau:
.,;.^d Au.
-^iyti^'^^
'-Xji::;
(a) HCl
+Mn02
>
MnCh
+CI2
+2H2O.
,
,„,f^eH

(b)
2HCl
+ Fe >
FeCh
+H2. , .v,t X:
jfefb
gnii^
XiVt.
(c)
6HCl
+ 2A1 > 2AICI3 + 3H2. " ' -W! M i:i
,i/:}c::
i
(d) 16HCl + 2KMna >
2KCl
+ 2MnCl2 + SCh +
8H2O.
iM>f
§'
?.
-
Cac phan
ling
trong do
HCl
the
hi?n
tinh oxi hoa la " s
f
8 f i

A.(b),(c). B.(a),(b). C. (b), (c), (d). D.(a),(d).
• Laigidi:
Nhan
xet: Trong cac phan ung (b) va (c), so' oxi hoa cua hidro giam tu +1
(trong
HCl) xuohgO (khi H2). , ,
-> Dap an A.
Vi
dv 2: Hoa tan hoan
toan
7,6 gam hSn hgp hot
FesOA
va Cu trong
200ml
dung
djch
HCl 1,2M (loang). Sau khi cac phan ung xay ra hoan toan, thu
dugc
dung dich X (khong chua axit du). Co can X thu du^c m gam muo'i khan.
Gia
tri cua m la
A.
10,39.
B.
14,20.
C. 5,16. D.
11,10.
Lffigidi:
Gpi
so mol:

Fe304
= a; Cu = b. ' ^
•
' j
Theo bai:
232a
+ 64b = 7,6.
•
?-•<.}(
• ••: '^^j^ X :,,;. i
^han xet: chi c6
Fe304
phan ung
tryc
tiep voi axit. f| ,* g .j :
Trong X khong con axit du nen
Fe304
phan ung vua dii voi HCl:
Cac phan ung hoa hpc:
Caim
nang
On
luy$n
Ihi dgi hgc 18
chuy6n
dg H6a hgc -
Nguygn
VSn Hit
i
*

•
Fe304 + 8HC1 >
FeCh
+
2FeCh
+
2H2O
Mol: 0,03 <- 0,24 -» 0,03 0,06
-> a = 0,03 -> b = 0,01 mol.
Cu +
2FeCl3
>• CuCh +
2FeCl2
J
t arHQUii >J > TiX.
Mol: 0,01 -> 0,02 ^ 0,01 0,02
m
= 0,01.135 +
0,04.162,5+
0,05.127 = 14,2 gam -> Dap an B.
Vi
3: Day gom cac chat deu tac dung
dugc
voi dung dich HCl loang la
A.
KNO3,
CaC03,
Fe(OH)3.
^ B. FeS,
BaS04,

KOH. ,
.^.^^^
C.
AgNCte,
(NH4)2C03,
CuS. D.
NaHCOs,
FeS, CuO. ' ,,,
Lai
gidi:
Loai A vi
KNO3
khong tac dung; loai B, C vi
BaS04
va CuS khong tan
trong
dung dich HCl loang. ^ •
Dap an D. Cac phuang
trinh
hoa hoc:
^''^ ^
' iOnM>i
NaHCOa
+ HCl >
NaCl
+ H2O + CO2 ''
FeS + 2HC1 >
FeCh
+ H2S
CuO + 2HC1 > CuCh + H2O

Vi
du 4: Hoa tan hoan toan 8,55 gam hon hop gom Na, K va Ba vao nuoc, thu
duoc
dung dich X va 1,792 lit khi H2 (dktc). Dung djch Y gom HCl va
H2SO4,
ti 1$ mol tuong ung la 2:1. Trung hoa dung djch X boi dung dich Y,
tong khoi luong cac muoi dugc tao ra la
A. 13,81 gam. B. 11,39 gam. C. 15,23 gam. D. 19,07 gam.
1,792
n^,
= 0,08 mol. Cac phan ling hoa hgc: '
r,
^ 22,4 '
•
:
Na + H2O )• Na* + OH" + -Hi
2
8 K + H2O > + OH- + iH2
o
n
Ba +
2H2O
> Ba2- + 20H- + H2
Nhan xet: n^^. = 2nyi^
=0,Id
mo\.
"H*
° "HCl + 2nH2S04 = 2a + 2.a = 4a mol
n^,.
= 2a mol; n^ i_ = a mol.

Trung hoa X boi Y: H* + OH" > H2O
-> 4a = 0,16 -> a = 0,04 mol.
Khoi lugng muoi thu
dugc
= 8,55 + m + m 2- = 8,55 + 0,08.35,5 + 0,04.96
Cl
so^
• * = 15,23 gam —> Dap an C.
Trong Y:
Cty
TNHH
MTV
DVVH
Khang
Vijt
Vi
dV 5: Cho 2,13 gam hon hgp X gom Mg, Cu va Al a dang bgt tac dyng
hoan toan voi O2 thu
dugc
hon hgp Y gom cac oxit c6 khoi lugng 3,33 gam.
"The
tich
dung dich HCl 2M vua du de phan ung het voi Y la
A. 150 ml. B.SOml. C. 75 ml. D. 90 ml.
Lai
gidi:
01
bai nay, cac em rJia't thiet phai ap dung bao toan khoi lugng de h'nh khoi
lugng oxi tham gia phan ung:
mkimiiHii + moxi = moxit —^ moxi = 3,33— 2,13 = 1,2 gam.

12
noo
= — =0/0375 mol -> no =
0,075
mol -> n ^
=0,075
mol.
^-^2 32 ^
(0x11)
Khi
cho oxit
bazo
tac dung voi axit tao ra nuoc: . .,fm
i.
i
02- + 2H^ > H2O. , ^ ; ;,
Suy ra: n^+= 0,15
mol->
n^ci^ 0,15 mol
-> VHCI =
0,075
lit = 75ml Dap an C.
Vi
d\ 6: Hoa tan hoan toan 7,8 gam K vao 500ml HCl 0,2M, thu
dugc
khi H2 va
dung dich X. Co can X thu
dugc
m gam chat ran khan. Gia trj ciia m la
A. 14,90. B. 13,05. C. 7,45. D. 20,50.

•<•
, Lai gidi: iiiX = .
HK
= 0,2 mol; = 0,1 mol;
nHci
=0,1 mol. ,, ;j ,.
K + HCl > KCl +
lH2t
«o.'HBOH n.: foH ,(>
2
'h-,
Mol: 0,1 0,1 0,1
't^»^'-
Luu
y: K con du se tiep tuc phan ung voi nuoc:
>.
Z'
K
+ H2O > KOH +
iH2t
2 • OH£' -
•
-
Mol: • 0,1 0,1 r<^r ,
-> m =
mKci+
mKOH
= 74,5.0,1 + 56.0,1 = 13,05 gam Dap an B.
Vi
dy 7: Hoa tan hoan toan 8,97 gam kim loai kiem M vao SOOnil dung djch

HCl 0,2M, thu
dugc
khi hidro va dung djch X. Co c?n X thu
dugc
14,73 chat
ran
khan Y. Kim loai kiem M la
A.Rb. B. K. C.Na. D.Li.
Lai
gidi:
Jl/ion
Ob ^
Cac phan ung hoa hgc:
•<
•= , I^^/TU' »• .| ,
M
+ HCl > MCI + -H2 va M
+H2O
> MOH + -H2
2 2
N/ian xet: Bao toan khoi lugng my =
mi4
+ m^i_ + m^^^.
m^„. =14,73-8,97-0,1.35,5 = 2,21
gamn^„.
= 0,13 mol.
OH UH
73
Ca^m nang 6n luy^n thi dgi hqc 18 chuy6n dg H6a hgc - Nguygn VSn Hai
8 97

riM=nHci
+nQ^.= 0,23
->
M=-^=39(K)^
Dap an B.
Vi
dv
8: Cho
m
gam hon hop
X
gom Cu, Mg,
Fe
tac dung voi axit HCl du, thu
dugc
dung dich
Y,
448ml khi
(dktc)
va
0,64 gam chat ran.
Cho
dung dich
NaOH
du vao
Y,
loc
ket
tiia
va

nung trong khong khi toi khoi luong khong
doi, thu
dugc
1,2 gam chat ran. Gia tri ciia
m
la
,
A. 0,80. B. 1,16.
C.
1,44. D. 0,84.
Loigidi:
N^flnxef:
mcu=
0,64 gamncu
=
0,01.
r •.
Ggi
so
mol trong X:
Fe
=
x;
Mg
= y -> x + y =
nH2
=
0,02 mol.
'
So

do
phan ung
:
Fe
FeCh
>
Fe(OH)2
-^^^
lFe203
Mg
-±»£U
MgCh
>
Mg(OH)2
—^
MgO
^ mFe203
+
mMgO
= 1/2 ^ 80x + 40y = 1,2 -» x = y =
0,01
mol.
m
=
0,64
+
0,01.56
+
0,01.24
=

1,44 gam.
-> Dap an C.
'
Luu
f.
Khi
nung
ngoai
khong khi,
Fe(OH)2
chuyen thanh
Fe(OH)3
va bi
phan hiiy thanh Fe203.
Vi
du
9: Hoa tan hoan toan hon hgp
X
gom
Fe
va Mg
bang mot lugng vira dii
dung dich HCl 20%, thu
dugc
dung dich
Y.
Nong
do
cua
FeCk

trong
Y
la
15,76%.
Nong
do
phan tram cua
MgCh
trong
Y
la
A. 24,24%. B. 11,79%.
C.
28,21%.
D.
15,76%.
,
Loigidi:
Fe
+
2HC1
^
FeCh
+ H2
Mg
+
2HC1
>
MgCh
+ H2

K.i
i
Xet voi 100 gam dung dich
Y:
-> m^^^ij
=
1^,76 gam.
mMgci2
+
mH20 = 100 -15,76 ^ 84,24 gam.
Nhan xet: Lugng H2O trong
Y
cung chinh la lugng H2O c6 trong dung dich HCl
bandau.
Mat khac, do nong do HCl bang 20% mH20
=
4mHci
^
mMgCl2
+
4mHci
=
84,24.
,
Bao toan nguyen to'Cl:
31
52
2 1
2nFeci2
+

2nMgCl2
=
"HCl
"> +
~n^gClj
=
^"^HCl
mMgCl2
=11'79
^
C%(MgCl2)
=
11,79%
->
Dap an B.
Cty
TNHH
MTV DVVH Khang
Vift
'
do
^'^^ ^^"^
^
NaC\a KCl
voi
HiS04
dac, du.
jChi
thoat
ra cho hoa

tan
vao
nuoc thu
dugc
dung dich
Y.
Cho
bgt Zn du
vao
Y
thu
dugc
448ml khi (dktc). Khoi lugng
NaCl
trong
X
la
^ 0,585. B.
1,170.
C.
1,755.
D. 2,340.
'
Loigidi:
Goi
so
mol:
NaCl
=
x; KCl

=
y. Cac phuong
trinh
phan ung:
jsjaCl
+
H2SO4
—^
NaHS04
+
HClt
'''' '
jCCl
+
H2SO4
KHSO4
+
HClt
Zn
+
2HC1
>
ZnCh
+ Hat "
>''•''
^
Bao toan nguyen to:
CI" > HCl > -Hi
Ta c6: mx
=

58,5x
+
74,5y
=
2,66
va
nH2
=
0,5(x
+ y) =
0,02.
_» x
=
0,02;
y
=
0,02
->
mNaci
=
0,02.58,5
=
1,17 gam Dap an B.
Vi
du
11: Hoa tan het
m
gam hon hgp
Mg va
MgCOa

trong dung dich HCl, thu
dugc
4,48 lit hon hgp khi
X
(dktc).
Ti
khoi cua
X
so
voi
H2 la
11,5. Gia tri cvia
mla
A. 13,2. B. 10,8.
C.
19,2. D. 9,0.
Loigidi:
^-^••«:MfBrtf 6'>
Cac phuong
trinh
phan ung:
' •' i A
Mg
+
2HC1
)•
MgCh
+ H2
Mol:
a a

•t^|,\V,
MgCOs
+
2HC1
>
MgCh
+ CQ2 + H2O ''"^
Mol:
b b •
'iliiA::
Theo
bai:
M =
11,5.2
=
23 va a +
b =
0,2 mol.
, ,^ ,
Ap dung
cong
thuc cua phuong phap duong
cheo,
ta c6:
,
"H2
44
-23
2- 23
=-

—
=- ->
a = b =
0,1
mol
1
b 1
^C02
m
= m^g +
mMgcos
" " ^^'^ 8^ ^ ^ ^'
2.
AXIT
SUNFURIC:
H2SO4
Li
thuyet
Dung
d/clz
H2SO4/oflng: Tinh axit manh
'1,1 , ^
Fe
+
H2SO4
>
FeS04
+ H2t , j ,
FeS
+

H2SO4
>
FeS04
+
H2St
,
"*" Dwn^djc^
H2SO4
(fflc:
Tinh oxi hoa manh
Ngoai
tinh
axit manh, axit sunfuric dac con the hi^n
tinh
oxi hoa m^nh,
tac
diing
dugc
voi nhieu kim loai, hgp chat, :
ff;-!;! » {'"-l
C^m nang
6n
luy$n
thi dgi hpc 18
chuySn
dg H6a
hpc
-
NguySn VSn
HJi

Cu
+
2H2S04(^flc)
——>
CuS04 +SO2
+
2H2O
:/>
2Fe
+
6H2SO4
(dac)
Fe2(S04)3
+
3SO2
+ 6H2O
',' '
2FeO
+
4H2SO4 (dac)
>
Fe2(S04)3
+
SO2
+4H2O
2Fe304
+
IOH2SO4 (dac)
>
3Fe2(S04)3

+
SO2
+
IOH2O
Luu
y:
Cac kim loai Al, Fe, Cr khong tac dung
vdi
Ji2S04
dac, nguQi.
Dieu
che
So
do:
Quang pirit FeS2
hoac
S ) SO2 —^
SO3
—^
'

.• ,.H, »,I
'• ,.
H2S04.nS03
>
(n+1)
H2SO4
Cac phan ung:
S +
O2

>
SO2
4FeS2
+
IIO2
—^
2Fe203+
8SO2
••
• "ID :6i fiB'lii^H
2SO2
+
O2
2S03
,-xm:o:.fii
,
H2SO4
+
nSOs
>
H2S04.nS03
(Oleum)
^
H2S04.nS03+
nH20
>
(n+l)H2S04
b. Vi dv mau:
Vi
dy 1: Cho day cac

chat
sau: KBr, S,
Si02,
FeO, Cu va
Fe203.
So'chat
trong
day
CO
the bi oxi hoa boi dung djch
axit
H2SO4
(dac,
nong)
la
A.
4.
B.5. '
G.3.c|f:
D.2.
Lai
gidi:
Nhan
xet: Axit
H2SO4
dac, nong the hien tinh oxi hoa khi
tac
dung vai
chat
CO

tinh khu (chiia nguyen to' dang
6
muc oxi hoa tha'p).
2KBr
+
3H2SO4
2KHS04
+
Br2
+
SO2 +
2H2O
S
+
2H2SO4
—^
3SO2
+
2H2O
: '
2FeO
+
4H2SO4
—^
Fe2(S04)3
+
SO2
+
4H2O
Cu

+
2H2SO4
CuS04
+
SO2
+
2H2O
^
fH',:^::
Dap an A.
Luu
y: Khi tac dyng voi
H2SO4
d^c,
FeaOs
the hi^n tinh baza:
FeaOs
+
3H2SO4
—^
Fe2(S04)3
+
3H2O
Vi
dy
2:
Cho
oxit
ciia kim loai
M

(hoa
tri
2)
tac
dung vira dii vai dung dich
H2SO410%
(loang),
thu
dug-c
dung dich muoi
c6
nong dp bang
14,45%.
Kim
loai Mia
' ' ""
A.
Mg.
:
B. Fe.
,
C. Cu.
' ^
D. Zn.
Lcn
gidi:
Phan ling hoa hpc:
MO
+
H2SO4

>
MSO4
+ H2O *
Cty
TNHH MTV DWH
Khang ViSt
Gia thie't dung djch ban dau chiia 1 mol
H2SO4
(tiic
chiia 98 gam
H2SO4)
100
_>
Khoi
lupng
dung djch
H2SO4
=
98.
—
= 980 gam.
^^^^^•^
-=0,1445
->M
=
56
(Fe)-^
Dap an
B.
980 + M

+ 16
Vi
dy 3:
Nung
hon hop
X
gom
a
mol
Fe
va 0,015 mol Cu trong khong khi mpt
thoi
gian, thu dupe
6,32
gam chat ran
Y.
Hoa tan hoan toan
Y
bang dung
dich
H2SO4
dac nong (du), thu dupe
0,672
lit khi SO2 (san pham khu duy
nha't
6
dktc). Gia
tri
ciia
a la

A.
0,04.
B.0,05.
C. 0,07. D. 0,06.
Lm
gtat:
^
0 672
-bVT' ".'^
ncri^=—^
=0,03
mol.
, , ,, ,
i.
,
Bao toan
khoi
lupng:
mo2
=
my " "^x
=
6,32
-
56a
-
0,96
=
5,36
-

56a.
Nhan xet: ne'u dya theo phuong
trinh
phan ung se rat dai va kho
giai.
Cach
1:6
day, cac em can su dyng so do phan ung:
.tysj.j,
:
Fe,Cu(l)
> Y (2) >
Fe^3Cu^M3)
Xet su trao doi electron
6
cac giai doan:
(1) -> (3): Fe -3e > Fe*-^
nenhuong
= 3 np^ = 3a OsH »• ;
Cu
-2e > Cu*^
nenhiKmg=2ncu
=0,03
"^s:
(1) (2):
O2
+4e > 2Ct^ ^
nenhan
=
4no2

=
^^^^^^=0,67-7a
(2) (3): S**
+2e
>
SO2
nenh,^n=2nso2
=0,06.
-
Bao toan electron: 3a + 0,03 = 0,67
-
7a + 0,06
^
a = 0,07 mol
->
Dap an C.
CachZ:
Qui
doi
Y
thanh:
Fe
(a mol); Cu
(0,015
mol) va
O
(b mol).
Bao toan
khoi
lupng:

56a
+
16b
=
6,32
-
0,015.64
=
5,36.
Bao toan electron:
3a +
2.0,015
=
2b +
2.0,03
a =
0,07;
b
=
0,09
, ,, .
Dap an C.
Vi
dy
4
(B-12):
Cho
cac
chat
rieng bi^t sau:

FeS04,
AgNOa,
Na2S03,
H2S,
HI,
Fe304,
Fe203
tac
dung voi dung dich
H2SO4
dac, nong. So
truang
hpp xay
ra phan ung oxi hoa
-
khu la
- ?
Rf
'A.
6.
. B.3. C.4. ,
DJfjfi
Lai
gidi:
Nhan xef: Phan ung oxi hoa-khu xay ra khi
H2SO4
tac dung vai
chat
c6
tinh

khu
(chiia nguyen to dang 0 muc oxi hoa tha'p).
2FeS04
+
2H2SO4
Fe2(S04)3
+
SO2
+
2H2O
77
im nang 6n luy^n thi dgi hoc 18 chuygn dg H6a hqc - Nguygn Van Hi\
HlS
+
3H2SO4
' >
4SO2
+
4H2O
,
8HI
+
H2SO4
^'-^ 4I2 +
H2S
+
4H2O f-SnO'w! ;oi*:^f
'
2Fe304
+

IOH2SO4
3Fe2(S04)3
+
SO2
+
IOH2O
,
1/
Dap
an C.
Luu
y:
^eiOj,
AgNOs xay ra phan ung
trao
doi, Na2SQj khong tr.c dung:
Fe203
+
3H2SO4
Fe2(S04)3
+
3H2O
2AgN03
+
H2SO4
Ag2S04i
+
2HN03
i
d\

5:
Cho
6,08
gam hon hop gom
Li,
Na
va
Ba
vao
nuoc
(du),
thu
duoc
dung dich
X
va
1,344
h't khi H2 (dktc). Dung dich
Y
gom HCl IM va H2SO4
0,5M. Trung
hoa
dung dich
X
boi dung dich
Y,
tong khoi lugng
cac
muoi
dugctaorala

, :
A.
10,38
gam.
B. 11,09
gam.
C. 11,80
gam. D.
9,42
gam.
Lbi
gidi:
1
344
"H2
=
=
0,06
mol.
y ,,.
Cac phan ung
hoa hoc:
Li
+
H2O
> +
OH-
+ -H2
2
^,

^ ^ 1
•
*•
iiJ
Na
+
H2O
> Na^ +
OH"
+ -H2
Ba
+
2H2O
>
Ba2+
+
20H-
+ H2
Nhan
xet:
n^^. = 2nH2 =
0,12
mol.
' '''
Mat
khac,
nong
do
HCl
gap

2
Ian
H2SO4 trong cung
mot the
ti'ch
thi
"Ha
=
2nH2S04
•
Goi
nHci
=
2a -»
nH2S04
= a-
-> Trong
Y:
n
+
=
4amol;n
o_ = amol;n_. =
2amol.
. ,
,,
, H
,,.• ,,.',„
SO4 CI
ii

ion'ji
Trung hoa
X
boi Y:
H* +
OH"
>
H2O
^y^:-yyh
4a =
0,12
a =
0,03
mol.
\V
Khoi luong muoi thu
dug-c
=
6,08
+
m .
+
m
2- ' '
CI
SO^ ; ,1-
X
i
V.
',K

.
=
6,08
+
a06.35,5
+
0,03.96
=
11,09
gam.
^ Dap
an B.
,
1
dvi
6:
Hoa tan
hoan
toan
5,28
gam hon hgp bpt
Fe304
va Cu trong
80ml
dung
dich H2SO4 IM
(loang,
vua dii).
Sau
khi

cac
phan ung xay
ra
hoan
toan, thu
duoc
dung djch
X. Co
can
X
thu
du(?c
m
gam muoi khan. Gia trj
ciia
m la
A.
8,64. B.7,68. C. 15,68. D. 11,68.
Cty
TNHH IVITV DVVH Khang
Vigt
Lcngiai:
Gpi
so
mol:
Fe304
=
a;
Cu = b.
Theo

bai:
232a
+
64b
=
5,28.
;i;
jV/ian xet: chi CO
Fe304
phan
ling
true Hep voi axit.
ji b
uAi
Trong
X
khong con axit du nen
Fe304
phan
ling
vua du voi H2SO4:
Cac phan ung
hoa
hpc:
Fe304
+
4H2SO4
>
FeS04
+

Fe2(S04)3
+
4H2O • •
'
: Mol:
0,02 <- 0,08 ^ 0,02 0,02
a =
0,02
^ b
= 0,01 mol.
"' ~ ' '
CuS04
+
2FeS04
•'
->
0,01 0,02 ' ' '
m
=
0,01.160
+
0,01.400
+
0,04.152
=
11,68
gam Dap
an
D.
n.'

Vi
7:
Cho
3,68
gam hon hop gom Al, Mg
va
Zn
tac
dung voi mot luong vua
du dung dich H2SO410%,
thu
dxxgc
dung dich
X va 2,24
lit khi
H2
(dktc).
Khoi lugng dung dich
X
la
A.
101,68
gam.
B. 88,20
gam.
C. 101,48
gam. D.
97,80
gam.
Lcngiai:

Bao toan nguyen to hidro:
nH2S04
=
"H2 =
0,10
mol.
''^
" '' '
-> Khoi luong dung dich H2SO4 =
0,10.98.
^
=
98
gam.
Cu
+
Fe2(S04)3
Mol:
0,01 -> 0,01
10
Bao
toan
khoi luong:
3,68
+
98
=
mx +
0,10.2
^ mx =

101,48
-> Dap
an C.
Luu
y: Bai
nay
cac em di
quen
tru
khoi luong khi
H2
bay
ra, va
chi
tinh:
mx
=
3,68
+
98
=
101,68 va se
chgn
nham
dap an
A!
Vi
8: Hoa tan
hoan
toan

6,44 gam
hon
hgp X gom Al, Fe va Zn
bang
mot
luong
vua
du
dung dich
H2SO4
loang,
thu
dugc
2,688
lit
H2
(dktc)
va
dung
dich chua
m
gam
muoi.
Gia
tri cua
m la ' "
A.
19,04. B. 20,54. C. 17,96.
D.
14,50.

Lai
gidi:
Cach
1:
Bao
toan
nguyen
to H:
nH2S04
=
"H2
"
0,12
mol.
Bao
toan
khoi luong:
6,44
+
0,12.98
= m +
0,12.2
->
m
=
17,96 gam
-> Dap
an C.
Cach
2:

Ta c6:
nM
=
nH2S04
=
"HJ
=
0,12
mol.
Neu
1
mol
kirn
loai
M >
MSO4
thi
khoi lugng
tang
96
gam
. . ;
0,12
tang0,12.96.
,
Vay
m
=
6,44 + ai2.96
=

17,96.
I-
'
->
Dap an C. •
„.,
(^„j,,4i<);/:H:H
,
).^-
'
•'fm-E
' o
C^m
nang
6n
luygn
thi dai hgc 18
chuy6n
H6a hpc -
Nguygn
Van
H&\
Vi
du 9: Hoa tan
hoan
toan 2,81 gam hon hop X gom Fe203, FeO, CuO can
50ml
axit
H2SO4IM
(loang).

Khir
hoan
toan 2,81 gam X
bang
khi
CO
(nung
nong)
thu dugc m gam
kim
loai.
Gia
tri
ciia
m la
A.
3,24.
B.2,65.
C. 3,06. D.2,41.
aofl
' •
'Laigidi: •i-,'>r"i
Nhan
xet: Khi cho oxit kim
loai
tac
dung
voi
axit/.ion
0^~

trong
oxit se ke't
hqp voi
H*
trong
axit
tao
thanh
H2O
theo
phuong
trinh:
^
, ^
Ijjjy,
02- + 2H^ > H2O ^ ia,,j^ u S},.0-'K
Mol:
0,025
0,05
Vay:
m o = 0,025.16 = 0,4 gam
-> m kim:o?i = 2,81 - 0,40 = 2,41 gam -> Dap an D.
s^iBi
JOAI
flj 4-»f
Vi
d\ 10: Hoa tan het
2,088
gam
FexOy

bang dung dich
H2SO4
dac, nong (du),
thu dugc dung dich X va 324,8ml khi SO2 (san pham khu duy nhat, 6 dktc).
Co
can dung dich X, thu dugc m gam muo'i sunfat khan. Gia tri ciia m la
A.
5,22. B.5,40. C. 5,80. D. 4,84.
•HV
•• ' ' Laigidi:
Nhan
xet: Oxit sat
FexOy
tac dung voi
H2SO4
dac, nong -> SO2 thi
oxit
la
FeO
hoac
Fe304.
Cac
em luu y rang 1 mol FeO
hoac
Fe304
deu chua 1 mol Fe''- nen deu c6
kha
nang nhuong 1 mol electron, do do:
0 3248 '
=

npe^Oy
=
2ns02
->
"FexOy
=
2"-^^=
^'029 mol
,
ns« '
"^'^ '^^••^ ' ^
->
Mpe o = =^ = 72
(FeO).
. >0r
urn
^ n'f)) •
Bao toan nguyen to'Fe:
2FeO
> Fe2(S04)3 ; p'
mpe2(so4)3= 0/0145.400 = 5,80
gam->
Dap an C. , '
3.
AXIT
NITRIC:
HNO3
A.
Li thuyet
+

Tinh axit
manh:
Tac dung voi kim loai, bazo,
oxit
baza, muoi. Vi du:
MgO
+
2HNO3
> Mg(N03)2 + H2O
CaC03
+ 2HN03 > Ca(N03)2 +
COzt
+ H2O
+
Tinh oxi
hoa
manh
+
Tac dung voi kim loai: Axit nitric tac dung dugc voi hau het cac kim loai
(tru
Au,
Pt), ke ca kim loai c6 tinh khu ye'u nhu
Cu,
Ag,
Cu
+4HN03(f?flc) > Cu(N03)2 + 2N02T
+2H2O
3Cu
+8HNO3(/0flM^^)—->
3Cu(N03)2+

2NOT+4H2O
Litu
y:
1. Ba kim loai kha manh (Mg, Al, Zn) c6 the khu
HNO3
thanh
N2O,
N2
hoac
NH4NO3;
2.
Al,
Fe, Cr khong tan trong HNO3 d$c ngupi. '' , "' "
Tac
dung
vai
hap
chat
3FeO
+ 10HNO3
>
3Fe(N03)3
+ NO +
5H2O
'•'^^ rl-y:
3Fe2*
+NO; + 4H* >
3Fe3*
+
NO

+
2H2O
£n,;,
FeC03
+
4HN03 > Fe(N03)3
+
NO2
+
CO2
+
2H2O
*
.ciijj
FeS2
+ I8HNO3 > Fe(N03)3 +
2H2SO4
+ I5NO2 + 7H2O
Dieu che
So do: N2 > NH3 > NO > NO2 -4
HNO3
N2
+ 3H2 < > 2NH3 . ,i
4NH3 + 5O2 '""^•'^ > 4NO +
6H2O
2NO
+ O2 >
2NO2
4NO2
+ O2 +

2H2O
> 4HN03
<
'bCl
4y
Vi
du 1: Cho
3,024
gam mpt kim loai M tan het trong dung dich HNO3 loang,
thu dugc
940,8
ml khi
NxOy
(san pham khu duy nhat, a dktc) c6 ti kho'i doi
voi
H2 bang 22.
Khi
NxOy
va kim loai M la
A.NOvaMg.
B.
NChvaAl.
C.N20vaAl. D.
N20vaFe.
Lai
gidi:
Theo bai:
M^^Q
= 22.2 = 44
NxOy

la N2O Loai A va B.
f
,(>«
.,, .'
n,,o=^Jo,042™ol.
22,4
'^o^h/r
Cach
1: Phan ung hoa hgc: ^ nfo '
8M
+
10nHNO3 > 8M(N03)n
+
nN20
+
5nH20 ^
Q,;.^^
_v
8
0,336 3,024
„
n
-> n
=
3 va M
=
27
(Al)
Dap an
C.

'^'^ '
Cach
2: Nhan thay de tao dugc khi N2O thi kim loai phai kha m^nh (nhu Mg,
Al,
Zn) -> Loai D -> Dap an C.
Vi
dv 2: Hoa tan hoan toan hon hgp gom 0,04 mol
FeS2
va a mol
CU2S
vao
axit HlSf03 (vua dii), thu dugc dung dich X (chi chua hai muoi sunfat) va V
lit
khi
duy nhat NO2 (dktc).
Gia
tri ciia V la
A.
13,44.
B.
17,92.
C.
20,16.
D.
22,40.
im
nang
6n
luy^n
thi d?i hpc 18

chuySn
dS H6a hpc -
Nguyin
Van HSi
Lai
gidi:
Nhan
xet: cac em lim y la dung dich X chi chiia hai
muoi
sunfat —> sau cac
phan
ling,
S nam he't 6 dang goc sunfat.
Ta
CO
cac so do chuyen hoa:
1
1' t t-
tiVK'.
FeS2 > -Fe2(S04)3
Mol:
0,04 0,02 ' ' • *
• Cu2S >
2CuS04
,;<:J3+,
sOl-'';;
*•
(OMbU' -
>^
Wlh

>
,(fWi
Mol:
a -> 2a ^aicHS: +
(Ov^jr!
<
>
}A}W\ i r
Bao toan nguyen to S, ta c6:2
npeSz
+ i^Cu2S = "SO4 A' . •
>
->
0,04.2
+ a =
0,02.3
+ 2a ^a = 0,02. , , * /in ur
Cac phan
ling
khu: v
FeS2-15e
> Fe*3 + 2S^
Cu2S - lOe >
2Cu^2
+
Bao toan electron: Ln^Oj = 15r>FgS2
^'^'^Cu2S
nN02
=0'8mol-> V^o
=17,92

lit ->DapanB.
/i
3: Hoa tan hoan toan mpt hon hgp gom hai kim lo?ii Fe va Cu bang
dung
djch HNO3 dac nong thi thu dugc 2,24 lit khi mau nau (dktc). Neu
thay axit HNO3 bang axit H2SO4 dac, nong, du thi the
tich
khi SO2 (dktc)
thu
dugc sau phan ung la bao nhieu?
A.
3,36
lit.
B. 5,60
lit.
C. 2,24
lit.
' D. 1,12
lit.
Lot
gidi:
nN02
= 0/1 mol. Ji*
V;•
•
• M
,
Chatkhu:
Fe - 2e > Fe*3 va Cu - 2e > Cu^^
Chat oxi hoa: N^^ +ie > NO2

hoac
+2e > SO2
Bao toan electron: ng=nfj02 =
2nso2
rcitjirf
»s" ='
"502 = ^= 0,05 mol
Vso2
=
0,05.22,4
= 1,12 lit ^ Dap an D.
4: Hoa tan hoan toan 2,08 gam hon hop gom FeS va FeS2 trong dung
djch
axit HNO3 (dac, du), thu dupe
5,376
lit khi NO2 (dktc) va dung dich X.
Cho X tac dung vai dung djch
Ba(OH)2
du, Ipc ket
tiia
va nung trong khong
khi
den
khoi
lupng
khong doi thu dupe m gam
chat
ran. Gia tri ciia m la
A.
8,43. B. 8,59. C.

10,19.7
D- 6,19.
Lai
gidi:
, > ,
nN02=
1^=0,24
mol.
Gpi
so mol: FeS (a mol) va FeS2 (b mol). Ta c6: 88a + 120b = 2,08.
Cty
TNHH
MTV DWH
Khang
Vi$t
Cac phan
ling
khu:
peS_9e
> Fe*3 +
FeS2-15e
> Fe^3 + 25"*
Bao toan electron:
nN02
= ^
"FeS
+15
npeS2
-> 9a + 15b = 0,24 ^ a = 0,01; b = 0,01.
Bao toan nguyen to Fe va S:

FeS
Mol:
0,01
FeS2
Mol:
0,01
->
-Fe203
+
BaS04
2
0,005
0,01
'I it ti
y
->
-Fe2C)3
+
2BaS04
2
0,005
0,02
t III II U ' I I
-> m =
160.0,01
+
0,03.233
= 8,59 -> Dap an B. ^.4 >yf > *'l) -a /'
Vi
dv 5: Cho 1,92 gam Cu vao 60 gam dung dich

HNO3
21%, thu dupe Vml khi
NO
(san pham khvr duy nhat) va dung dich Y. The
tich
dung dich NaOH
IM
can thiet de ket tua he't ion Cu^* trong dung dich Y la
A.
60ml. B. 120 ml. C. 150ml. D. 180ml.
KI;
Loi
gidt:
.J-
.0.$-/
:
ncu = = 0,03
mol;
n^NOa = ^^'^^^^ = 0,2
mol.
Phan ung hoa hpc:
63
3Cu(N03)2
+
2NO
+
4H2O
64
3Cu +
8HNO3

Y gom:
Cu(N03)2
= 0,03 mol;
HNO3
du = 0,2 - 0,08 = 0,12 mol. ; j
Cac phan ung: H* + OH' >
H2O
va Cu^^ +
20H-
>
Cu(OH)2',
= n„+ + 2n^ 2+ =0/12 +
0,03.2
= 0,18 mol
OH
H Cu
-> VNaOH
= 0,18 lit = 180ml ^ Dap an D.
mi y: Cac em de quen phan ung
trung
hoa axit (H* + OH) va chi
tinh:
. =
2n^^2+
- 0/06 mol —>
VNaOH
= 0,6
lit
= 60ml se chpn nham A!
Vi

d\ 6: Cho 19,2 gam kim loai M (hoa tri n) tan hoan toan trong dung dich
HNO3,
thu dupe 4,48 lit khi NO (san pham khu duy nhat, 6 dktc). Kim loai
Mia:
A.
Mg. B. Al.
C. Cu.
Lot
gidi:
nNo
= 0,2 mol. Phan ung hoa hpc:
3M
+ 4nHN03 > 3M(N03)n + nNO + 2nH20
D.
Fe.
Nhan
thay:
XXM
=
3n
NO
_
0,6
mol
n
19 2
MM
= = 32n
u,o
n

n
= 2 va M = 64 (Cu) Dap an C.
83
a'm
nang
an
luy^n
thi d^i hoc 18
chuyen
H6a hgc -
Mguyen
van Hi'i
/i
du 7: Nung 2,94 gam hon hgp X gom cac kim loai Al, Zn, Mg
trong
khi oxi,
sau mot thai gian thu duoc 3,42 gam hSn hap Y. Hoa tan hoan
toan
Y vao
dung dich
HNOs
(du), thu duoc 2,016 lit khi NO2 (san pham khu duy nhat,
a dktc). So mol HNO3 da phan ung la
A.
0,20.
B.0,24.
C.0,16.
. D. 0,14.
'
Lotgtat:

Bao
toan
kh^ luc^ng: mo^ = 3,42 - 2,94 = 0,48 gam '^^^^ "^'^^^^^
-> no2 = 0,015 mol no= 0,03 mol n^.2 ^^^.j^= 0,03 mol. ''^
Cachl:Sudungsad6: X > Y ^ Muoinitrat + NO2
Lm
y: Y chua ca kim loai (con du) va oxitsgS + sOrl 4—"-
+
Khi
cho kim loai +
HNO3:
nj^^_
(muoi)
—
ngtraodoi—
fij^jQ^
—
0,09 mol. ^
+ Khi cho
oxit
+
HNOa:
1 goc
trong
oxit
se bi thay the bang 2 goc NO
3
de
tao muoi nitrat, do do: n =2n , = 0,06 mol. " «li 1
Bao

toan
nguyen to N:
nHNOs
=
"NO3
"^^o " " ^'^'^
-> Dap an B.
:ach 2: Qui doi Y thanh X va O (0,03 mol). ' ' *
Bao
toan
electron: n^^xj^^ng+nj^Q^ =0,15 mol —> n . =0,15 mol.
3
Bao
toan
N:
nnNOa"NO3
^
^'^^
~^ ^'
n du 8: De nhan biet ba
axit
dac, nguoi: HCl, H2SO4,
HNOs
dung rieng
bi$t
trong
ba lo bi ma't nhan, ta dung thuo'c thu la
A.Al.
B.Fe.
CCuO.

" D. Cu.
Laigidi:
Nhan
xet: Al, Fe khong tac dung voi H2SO4 va HNO3 dac ngupi -> Loai A
va
B. Loai C vi CuO tac dung voi ca 3
axit
-> dung dich muoi mau xanh.
Dap
an D.
nfi^-t
„,.,'i;.j
c ,, y.'
Giai
thich:
HNOs
hoa tan Cu va c6 khi NO2 mau
nau
bay ra:
CU
+ 4HN03 > Cu(N03)2 + 2NO2 + 2H2O
Dung
dich H2SO4 d^c hoa tan Cu va c6 khi khong mau,
miii
xoc
thoat
ra:
Cu
+ 2H2S04d,ic > CuS04 + SO2 + 2H2O
1 du 9: Cho 3,6 gam Mg tac dung het voi dung dich HNO3 (du), sinh ra 0,84 lit

khi
X (san pham khu duy
nha't,
a dktc).
Khi
X la
A.N2.
B.N2O.
CNC)2.
D.NO.
Laigidi:
img = 0,15 mol; nx =
0,0375
mol.
Cty
TNHH
MTV DWH
Khang
Vi?t
lil
Chat
oxi hoa:
N*'* + ne > X
Bao toan electron: 2nMg = n.nx ^
ai5.2
=
0,0375.n
^>,dn
mJ - IrV
_^

„ = 8 Khi X la N2O ^ Dap an B. " - ^'""^ - ' " '
Vi di?
10: Hoa tan hoan toan m gam Al bang
dung
dich
HNO3
loang, thu dugc
dung
dich
X (khong chua
NH4NO3)
va hon hgp khi gom 0,015 mol khi
N2O
va
0,01 mol khi NO. So mol axit
HNO3
da tham gia phan ung la
A
0,12. B. 0,15. C. 0,17. D. 0,19.
Laigidi:
each
1: Cac phan ung: * J
tl
^ >'
'
8A1
+ 3OHNO3 > 8A1(N03)3 + 3N2O + 15H2O
'
^
' '

Al
+ 4HN03 > A1(N03)3 + NO + 2H2O • '' '
uu^.x^
•
Dua
theo cac phan
ling,
nhan thay: nHN03 = 10nN2O + ^^NO '
_>
UHNOS
=
10.0,015
+
4.0,01
= 0,19 mol -> Dap an D.
Cach 2: Bao toan electron:
3nAi
=
8nN20
+ 3nNo=
8.0,015
+
3.0,01
= 0,15 mol -> nAi = 0,05 mol.
Bao toan nguyen to N:
nnNOa
3nAi(N03)3 + 2nN20 +
^^NO
1
nHN03

=
0,05.3
+
0,015.2
+ 0,01 = 0,19 mol -> Dap an D.
Vi
d^ 11: Cho
27,45
gam hon hgp gom Al, Zn va Cu tac
dung
vvra
du voi
950ml
dung
dich
HNO3 2M, thu dugc
dung
djch chua m gam
muoi
va 4,48
lit
hon hgp khi X (dktc) gom NO va N2O. Ti
khoi
ciia X so vai H2 la 18,5. Gia
tri
cua m la
.
120,45.
V ,B.
124,45.

C.
99,65.
D.
112,05.
-
Laigidi:
6 bai nay, truac het cac em can tim so mol moi khi
trong
X de thu dugc ket
qua:
ni>jo=0,l mol;
nN2o=0,l
mol. ^ - -'•> - S'KS
:«ori
f:w 7^
A\,Zn,Cu
)
Muoinitrat
+ NO + N2O ''^^^
Chat oxi hoa: N^^ + 3e > NO;
2N^-'*
+ 8e > N2O
vacothexayracaquatrinh:
2N*-^ + 8e > NH4NO3 (amol)
Khi
cho kim
loai
+
HNO3:
^'^^^ '

"NO3 "
'"""i"'
= 3 nfjo + 8 nN20 + 8
nNH4N03
=
(I'l
+ 8a)
mol.
Bao toan nguyen to
N:
UHNOS
^
"NO3
^ "^^^ ^
"N2O
+ 2nNH4N03 ,. ;
0,95.2
= 1,1 + 8a + 0,1 + 2.0,1 + 2a a = 0,05 mol.
Bao toan
khoi
lugng:
m =
m^i, zn, Cu
'"NOS ^
"^NH4N03
=
27,45
+ (1,1 +
8.0,05).62
+

0,05.80
=
124,45
gam.
85
:im
nang
On
luygn
thi dgi hgc 18
chuyfin
66 H6a hgc
-
Nguygn
Van HSi
—> Dap an B.
Nhan
xet: Can nhan ra bai toan da "giau di" san pham
NH4NO3.
Vi
dyi 12: Cho manh Cu phan ung he't voi dung dich HNOa, thu
dugc
0,896
h't
(dktc) hon hgp khi
X
(gom NO va NO2) c6 khoi lugng
bang
1,52 gam. Co
can dung djch sau phan

ling
thu
dugc
a gam muoi khan. Gia trj cua a la
A.
5,64. B. 7,52. C. 9,4.
,
D.
15,04.
Lai
gidi:
Truoc
he't cac em can tim so mol moi khi: n^o = nN02 ^ ^'^^
Bao toan
electron:
•
'''V:/'-
'
"••'••H'
"^•1\::^.,iriyP4^-^,' j>fy'{^
2ncu
= Sn^o +
"NOZ
=
0,02.3
+ 0.02 = 0,08 mol -> ncu = 0,04 mol.
, f
Bao toan nguyen to Cu: ncu(N03)2 = "cu = 0,04 mol. ;+
|
A

^
a
=
0,04.188
= 7,52 gam Dap an B.
du 13: Hoa tan
hoan
toan 8,4 gam Mg
bSng
dung dich
HNO3
vua du, thu
dugc
0,672
lit N2 bay ra
6
dktc va dung dich X. Khoi lugng muoi khan thu
dugc
khi c6 can dung dich X la
A.
51,8 gam. B. 30,1 gam. C. 55,8 gam. D. 15,04 gam.
;s
'
Lai
gidi:
'''
HMg = 0,35 mol;
txj^^
= •^^= 0,03 mol.
i*

Bai toan nay cac em c6 the giai khi vie't phuong
trinh
phan
ling.
Tuy nhien, giai
theo
phuong phap bao toan
electron
se nhanh hem.
Chatkhu: Mg
-
2e Mg^2
->
nenhucmg
=
2nMg =
0,7mol.
,
Chat
oxi hoa: 2N*''
+
lOe
>
N2 —^
nenh?n
=
10n[vj2
=
0,3mol.
Nhu

vay so mol
electron
trao
doi
chua
bang
nhau —>
"chua
on".
6
day, mot
san pham khu da
dugc
"giau di", do la su tao thanh muoi
NH4NO3:
Chat
oxi hoa:
2N*s
+ se >
NH4NO3
"••'^***W;'yi'«^
.111
'-•••"'Mol:' ''
0,4 -> a05
^'''^''l;:,
Vay: m = mMg(N03)2 + mNH4N03 =
0,35.148
+
0,05.80
= 55,8 gam Dap an C.

Lu-u
y: Nen ap dung
ngay
bao toan
electron:
2 n^g = 3 n^o + 8 nNH4N03
•
i
du 14: Hoa tan
hoan
toan 12,42 gam Al
bang
dung dich HNOa loang (du),
thu
dugc
dung dich X va 1,344 lit (a dktc) hon hgp khi
Y
gom N2O va N2.
Ti
khoi cua Y so voi hidro la 18. Co can X, thu
dugc
m gam
chat
ran khan.
Gia tri cua m la
A.
38,34.
B.
34,08.
C.

106,38.
D.
97,98.
Lai
gidi:
- -
iorM.
Ta c6: nAi = 0,46 mol; nv = 0,06 mol; My = 18.2 = 36
Cty
TNHH
MTV DVVH
Kliang
Vi§t
Theo
cong
thiic duong
cheo
ta c6:
28
-36
1
44
-36
1
nN20=
nN2 = 0,03 mol
nN20
nN2
Chatkhu: Al
- 3e

> AP
Chat
oxi hoa:
2N*5
+ Se >
N2O
va
CO
the
xayra
qua
trinh:
2N*^
+ 8e
1
} (
2N^5
+
lOe
f
I
->
N2
-> NH4Na (a mol)
Baotoanelectron:3nAi
=
8nN2O+10nN2+8nNH4N03
->
3.0,46
=

8.0,03
+
10.a03
+ 8a a = 0,105 mol.
^^''^'
- -
V$y:m = mAi(N03)3 + mNH4N03 =
a46.213
+
0,105.80
=
106,38gam.
''
•'••'''^
-> Dap an C.
Lim
y: Ba kim loai kha manh (Mg, Al, Zn) tac dung voi axit
HNO3,
c6 the tao
thanh muoi
NH4NO3!
Vi
dv 15: Hon hgp X gom Mg, Al va Zn. Cho m gam X tac dung
VvJ-i
dung dich
HCl
du, thu
dugc
36,45
gam muoi clorua. Mat

khac,
hoa tan he't m gam
X
trong dung dich
HNO3
du, thu
dugc
55 gam muoi nitrat kim loai. Gia tri
cua m la
.
A.
5,8.
B.8,7.
C.11,6.
D. 14,5.
; :
Lai
gidi:
• ,
V-\T->
'•••T
Nhan
xet: Do goc
Or
va NO
3
deu c6 dien tich 1- nen
klii
ket hgp voi cung
mot lugng ion kim loai giong nhau thi: n^|_ =

"NO3
"
^
("^^O
•*•
^ '
ra
\
ao toan khoi lugng, ta c6: m
+
35,5a
=
36,45;
m + 62a = 55."
^ , ,
-> a = 0,7 mol
^
m =
36,45
-
35,5.0,7
= 11,6 gam
-> Dap an C.
4. AXIT PHOTPHORIC:
H3PO4
a. Li
thuyet
H3PO4
+
NaOH

H3PO4
+
2NaOH
H3PO4
+
3NaOH
mol
1:1
mol 1:2
->
NaH2P04
+
H2O
->
Na2HP04
+
2H2O
mol 1:3
Neu:
nH3P04
>
Na3P04
+
3H2O
a = 1 :
NaH2P04
: :\ vnU
W'l
0 ,-
'en

/Tir
',-,->•")
:'t[
ill, :
':>'Y
tV
:•'::„
1< a < 2 :
NaH2P04
+
Na2HP04
, , ,
2 <a < 3 :
Na2HP04
+
Na3P04
i
b. VI Dy MAU
^
Vi du
1
(B-08):
Cho 0,1 mol P2O5 vao dung djch
chua
0,35 mol KOH. Sau khi
phan ung xay ra
hoan
toan, dung dich thu
dugc
c6 cac

chat
la
j
A.
K3PO4
va K2HPO4. B. K2HPO4, KH2PO4. IA nuu
C.K3P04vaKOH.
D.
H3PO4
va
KH2PO4.
'•Q-;
87
Cim
nang On luy^n thi dgi hoc 18
chuy6n
dg H(5a hpc
-
Nguygn
Van HSi
Lai
gidi:
Nhanxet:
P2O5
+
3H2O
>
2H3PO4
^ , '
nH3P04

=
2ni^05
=
0'2
mol.
'' '
->
J}mH_ ^ 1 75 ^
tao2
muoi:
KH^PavaK2HPO4.
nH3P04
0,2
digji:.^
—> Dap an B. •*
r^^hnvl
•
~ -M -*
'•n4i;
;finri:i
©up jvi v?,?: dvlt 65
Lieu
y:
Cac em can
chuyen
P2OS
thanh
H3PO4
va xet ti If mol nhu tren.
Vi

dv 2
(B-09):
Cho
100ml
dung djch KOH 1,5M vao
200ml
dung dich
H3PO4
0,5M, thu
dugc
dung dich
X. Co
can dung dich
X
thu
dugc
m
gam muoi
khan. Gia tri
ciia
m la
A.
15,5 gam. B. 18,2 gam. C. 12,8 gam. D. 16,4 gam.
V,
. Lai
gidi:
^. "KOH
=
mol;
nH3P04

= 0,10 . ^
Cach
1: NMn
xet
= 1,5
->
tao
2
muoi:
KH2P04va
K2HPO4.
=
• -
"H3PO4
H3PO4
+
KOH
>
KH2PO4
+ H2O
Mol:
a -> a a f ,0,
H3PO4
+
2KOH
>K2HP04+2H20
Mol:
b -> 2b -> b
''^'r
•!'>

• U:
Ta c6:
a
+ b = 0,1;
a
+ 2b = 0,15
->
a = b = 0,05 mol.
->
m
=
0,05.136
+
0,05.174
= 15,5 gam Dap an A.
Cach
2: Ap dung bao
toan
khoi lugng cho
so
do phan
ling:
H3PO4
+ KOH >
Muoi
+
H2O
~^
"^H3P04
+ r"KOH=n^ + rflH20

Cdc em luu
y:
"OH-(KOH)
=
'^^l'
"H-^(H3P04)
=
^'^^ "^"l
^ '^H20 = 0,15 mol.
^
m
=
0,1.98
+
0,15.56
-
0,15.18
=
15,5 gam
->
Dap an A.
Vi
dv 3:
Cho
100ml
dung dich
H3PO4
a
mol/1 vao
100ml

dung dich KOH 2M
thu
dugc
dung dich
Y c6
chua
14,95 gam hon hgp muoi. Gia tri cua
a la
A.
0,75.
B.
1,00.
C.0,50.
D. 0,80.
Lai
gidi:
NMn
xet:
Vi
Y
chua
hon hop muoi
c6
muoi axit
->
KOH het.
''OH-(KOH)
=°'2;
"H+(H3P04)
=°'3^ ^

OH^O
=
"KOH
= 0,2.
So do phan ung:
H3PO4
+
KOH
>
Muoi
+
H2O.
Bao
toan
khoi lugng:
0,la.98
+
0,2.56
= 14,95 +
0,2.18
^
a
=
0,75->
Dap an A.
D
1J
Cty
TNHH MTV DWH
Khang

Vi?t
Vi
dv
4= Cho
21,3
gam
P2O5
vao dung djch
c6
chua
a
gam NaOH, thu
dugc
dung dich
c6
chua
28,4 gam
Na2HP04
va
b
gam
Na3P04.
Gia tri
ciia
a, b
Ian
lugt
la
A.
28; 8,2. B. 20; 16,4. C. 28; 16,4. D. 20

;
8,2.
Lot
gidi:
2;i
3 28 4 '"b
=-^ = 0,15 mol; nN32HP04 =
7^
=
0'2
mol.
„
^ ,,.
^^^.^^
P2OS
+
3H2O
>
2H3PO4
nH3P04 =
2np205
=
0,3mol.
Bao
toan
nguyen
to P:
nH3P04
=
nNa2HP04

+
nNa3P04
0,3 = 0,2 + —
->
b
= 16,4 gam. •
. . ,
So do phan ung:
H3PO4
+
KOH
>
Muoi
+
H2O.
Bao
toan
khoi lugng:
0,3.98
+
a
= 28,4 +
b a
= 28 gam Dap an
C.
Vi
du
5:
Cho
0,05

mol
P2O5
vao dung dich
chua
0,4
mol NaOH. Sau khi phan
ung xay ra
hoan
toan,
dung dich thu
dugc
c6 cac
chat
la
A.Na3P04vaNa2HP04.
B.
Na2HP04, NaH2P04.
C.
Na3P04
va NaOH. D.
H3PO4
va
NaH2P04.
i,t:.,dMte
Lai
gidi:
,,f,„:
'
Nhanxet:
P2O5

+
3H2O
>
2H3PO4
,i
Sij,
v«x
;^fi!j
ut;.a(,j 'MO ul ' >.
-> nH3PO4 =
2np2O5
=
0,l mol.
r;
4,
i]±JaOH.
^
M=
4 >
3 ^
NaOH
du va tao 1 muoi
Na3HP04.
,.
nH3P04
0,1 • '
—> Dap an C.
Luu
y:
Cac em can

chuyen
P2O5
thanh
H3PO4
va xet ti If mol nhu tren.
5. BAI TAP 6N
LUYEN
Bai
1:
Hoa tan het
1,04
gam hon hop
X
gom Mg,
Fe
bang
40
gam dung dich
HCl
7,3% thu
dugc
dung dich
Y va
0,672
lit khi H2 (dktc). Nong dp
%
ciia
HCl
trong
Y la

A.
2,43%.
B.2,18%.
C.
1,83%.
D.
1,78%.
Bai
2:
Hoa tan
hoan
toan
h6n hgp
X
gom
Fe va
Zn
bang
mgt lugng vua dii
dung dich HCl C%, thu
dugc
dung dich Y. Nong do phan tram
ciia
FeCh
va
ZnCh trong
Y
Ian lugt la 8,05% va
8,63%.
Gia tri cua

C la
A.
5.
B.15. C.20.
D.
10.
Bai
3:
Hoa tan
hoan
toan
7,8
gam h6n hgp gom Mg va Al
bang
dung dich HCl
du.
Sau phan ung thu
dugc
dung djch
c6
khoi lugng tang them
7,0
gam
so
voi
ban dau. Khoi lugng
ciia
Al trong hon hgp ban dau
la
A.

5,40 gam. B. 2,70 gam. C. 1,35 gam. D. 4,05 gam.
Ca'm
nang 6n luygn thi d^i hgc 18 chuyfin d6 H6a hgc - NguySn van Hai
Bai
4: Dot
nong
2,80 gam
hon
hgp X
gom Cu,
Zn va Mg
trong khi
oxi
du,
thu
dugc
4,08
gam hon hgp
oxit
Y. De
hoa tan het
Y
can toi
thieu Vml dung
djch H2SO4IM.
Gia
tri ciia
V la
A.
80.

r?S
CIB. 60. C. 100.
g
D.
120.
^e;/-
Bai
5:
Hoa tan
hoan
toan
m
gam hon hgp gom
Mg, Al,
Zn
trong dung dich
H2SO4 loang,
du
thu
dugc
0,672
lit khi H2 (dktc)
va 3,92 gam
hon hgp muoi
sunfat.
Gia
tri
cua m la
A.
2,48. B. 1,84. C. 1,04.

D.
0,98.
Bai
6: Cho 6,45
gam hon hgp hai kim loai
X
va
Y
(deu
c6
hoa
tri
2) tac
dung voi
dung dich H2SO4 loang du.
Sau
khi phan ung
ket
thuc, thu
dugc
1,12
lit khi
(dktc)
va
3,2
gam
kim loai. Lugng kim loai
nay
phan
ung vua du vai

3,55
gam khi CI2 khi dot nong.
Cac
kim loai
X
va Y
Ian lugt
la
A.
Zn va Mg.
B.
Zn va Cu. C. Mg
va
Cu. D. Mg
va Pb.
Bai
7:
Nung hon hgp
X
gom
0,28
mol
Fe
va
a
mol
Cu
trong khong khi mgt thoi
gian, thu
dugc

25,28 gam
chat
ran
Y. Hoa tan
hoan
toan
Y
bang
dung dich
HNO3
loang (du),
thu
dugc
1,792
lit khi NO (san
pham
khu duy
nhat
6
dktc).
Gia
tri
cua
a
la
A.
0,03. B.0,10. C.0,06.
'
D.
0,08.

^
Bai
8: Cho 10,8 gam
hon hgp gom
Fe304 va
Cu
vao
dung dich H2SO4 loang du.
Sau khi
cac
phan
ling
xay
ra
hoan
toan,
thu
dugc
dung dich
X
chiia
m
gam muo'i
va
chat
ran
Y.
Hoa tan het
Y
trong dung djch HNO3 loang,

sinh
ra 0,448
lit khi NO
(san
pham khu duy nhat,
a
dktc). Gia
tri
ciia
m la
A.
16,80.
B.
18,32. C. 13,92. D. 18,48.
Bai
9:
Hoa tan
hoan
toan
m
gam
oxit
FexOy
bang
H2SO4
dac
nong,
thu
dugc
dung dich

chua
4
gam mgt
loai muoi
sat
duy
nhat
va 0,224
lit
SO2
(dktc).
Cong
thuc ciia oxit
sat va gia
tri
m
Ian lugt
la
A.
FeOval,44.
B. Fe304 va 4,64. C. Fe203 va 3,20.
D.
Fe304 va 2,32.
Bai
10:
Hoa
tan
hoan
toan
4,4

gam
hon hgp
X
gom
Fe,
Cu, Ag trong dung dich
H2SO4
dac,
nong thu
dugc
2,24
lit khi SO2
(san
pham khu duy nhat,
a
dktc)
va dung dich
Y. Co can
Y
thu
dugc
m
gam muoi khan.
Gia
tri ciia
m la
A.
14,2. B. 9,2. C.9,3.
^
D. 14,0.

Bai
11:
Nhiet phan
hoan
toan
6,32
gam
KMn04, toan
bg
lugng khi O2 sinh
ra
cho phan ung
het
vai
3,6
gam hon
hgp
X
gom
Fe
va
Cu, thu
dugc
hon
hgp
Y.
Hoa tan het
Y
trong dung dich H2SO4
dac

(du),
thu
dugc
0,784 lit SO2
(dktc). Thanh phan
%
khoi lugng
Fe
trong
X la
A.
46,67%. B. 53,33%. C.
31,11%.
D.
69,89%.
Bai
12: Cho 0,015
mol mgt loai
chat
oleum
vao
nude, thu
dugc
200ml
dung dich
X.
De
trung hoa
X can
dung

200ml
dung dich NaOH 0,3M. Oleum
do la
A.
H2SO4.2SO3.
B.
H2SO4.3SO3.
C.
H2SO4.4SO3.
D.
H2SO4.SO3.
4
Cty
TNHH
MTV DVVH Khang
Vi§t
Bai
13: Cho
m
gam hon hgp
X
gom
Al,
Cu vao
dung dich HCl (du),
sau
khi
ket thiic phan
ung
sinh

ra 3,36
lit khi
(a
dktc).
Neu cho
m
gam hon hgp
X
tren
vao mot
lugng
du
axit nitric
(dac,
ngugi),
sau
khi
ket
thiic phan
ling
sinh
ra 6,72
lit khi NO2
(san
pham khu duy nhat,
a
dktc).
Gia
tri
cua m la

A.
11.5-
'yimn
12,3.
C. 10,5.
D.
15,6.
Bai
14:
Hoa tan
hoan
toan
1,23
gam hon hgp
X
gom Cu
va
Al
vao
dung dich
HNO3
dac,
nong
thu
dugc
1,344
lit khi
NO2
(san
pham

khu duy
nhat,
6
dktc)
va
dung dich
Y. Sue
khi NH3 (du)
vao
Y,
thu
dugc
m
gam
ket
tiia.
phan tram khoi lugng
cua
Cu trong hon hgp
X
va gia
tri
cua m
Ian lugt
la
A.
21,95% va 2,25. B. 78,05% va 2,25.
C.
21,95% va 0,78. D. 78,05% va 0,78.
Bai

15: Hoa tan het 20,6 gam hon
hgp gom Cu,
Fe,
Zn
trong dung dich HNO3
du,
thu
dugc
dung dich
Z
(khong
chua
NH4NO3)
va
hon hgp khi gom
0,15
mol
NO
va 0,25
mol NO2.
Co can
dung dich
Z
thu
dugc
khoi lugng muoi
khan
la
"'*
;^*^

bv.X'-^
;iosTJ'4io
.
A.
45,4
gam
B. 64,0
gam.
C.
51,6
gam. D.
57,8
gam.
Bai
16:
Hoa
tan
hoan
toan hon hgp gom
9,75
gam
Zn va
9,6
gam Cu vao
200ml
dung dich
X
chua
dong thoi HNO3 2M
va

H2SO4IM.
Sau
khi phan ung
xay
ra
hoan
toan thu
dugc
khi
NO (san
pham khu duy
nhat)
va
dung djch
Y.
Kho'i
lugng muo'i CO trong
Y la:
A.
38,55
gam.
B. 50,95
gam.
C. 63,35
gam. D.
43,50
gam.
Bai
17: Cho
3,2

gam bgt Cu
tac
dung
voi
100ml
dung dich
X
hon hgp gom
HNO3
IM
va
H2SO4 0,1M.
Sau khi
cac
phan
ling
xay
ra
hoan
toan, sinh
ra
V lit khi
NO (san
pham khu duy nhat,
6
dktc). Gia
tri
ciia
V
la

A.
0,784. B. 0,896. C. 0,672. D. 1,344.
Bai
18: Hoa tan
hoan
toan
0,06
mol FezOs
va
0,1
mol
FeS2
trong
260ml
dung
dich HNO3
4M,
san
pham
thu
dugc
gom
dung dich
X va
mot
chat
khi
thoat
ra.
Dung dich

X c6
the hoa tan toi
da m
gam
Cu.
Biet
trong
cac
qua
trinh
tren,
san
pham khu duy
nhat
ciia
N*"*
deu
la
NO.
Gia
tri ciia
m
la
A.
9,60.
B.
16,64. C. 11,52. D.
14,03
Bai
19: Cho 12,45

gam hon hgp
X
gom
Fe,
Mg, Zn vao dung dich HNO3 du, thu
dugc
dung dich khong
chua
NH4NO3
va hon hgp
khi
gom
0,2
mol
NO
va
0,1 mol NO2. Dot
chay
12,45 gam
hon hgp
X
trong khi
clo du
thu
dugc
bao
nhieu gam muo'i
clorua?
)U!Ji
A.

37,3
B.23,1.
C. 30,2.
Doj D. 16,0.
Bai
20:
Cho
3,84
gam Cu
phan
ung voi
80ml
dung dich
chua
HNO3
IM
va
H2SO4 0,5M,
sau
phan ung thu
dugc
V
ml khi NO (dktc).
Gia
tri ciia
V
la
A.
896. B. 560. C. 448.
D.

336.
Q1
dm
nang
On
luygn
thi dai hpc 18 chuySn dg H6a hpc -
Nguygn
van Hai
Bai 21:
Cho m gam hot Cu kim
loai
vao
200ml
dung
djch
HNOa
2M,
c6
khi
NO
thoat ra. De hoa tan het hot Cu, can them tiep 100ml dung
dich
HCl
0,8M,
dong
thoi
Cling
c6
khi

NO thoat ra. Gia
tri
cua m la
A.
9,60 gam.
B.
3,84 gam. C. 10,24 gam. D. 11,52 gam.
Bai
22:Hoa tan hoan toan 0,01 mol
FeS2
trong 42 gam dung
dich
HNOs 37,5%
theo phan ung:
FeS2
+
HNO3
>
Fe(N03)3 +
Isr02
+
H2SO4
+ H2O
Cho V ml dung
dich
NaOH 2M vao dung
dich
sau phan ung de thu dugc
lugng
ket

tiia
Ion nhat. Gia trj nho nhat cua
V
la
A.
40ml.
B.
70ml.
C.
80ml.
' D.
100ml.
'''^
,
H '
.'.•••A
6. HU'6NG DAN
-
LCJI GIAI
7,3 : i :
40.
Bail:
nHn =
1^ =
0,08
mol; nH,
=
=
0,03
mol.

Ik
i
Hri
=
u,uo mui;
11
HI
36,5 "2 22,4
Ggi
so'mol:
Mg =
x va Fe
=
y
mol.
Ta CO
so
do:
Mg ) H2; Fe "^^^ > H2
. nH2
=
X
+
y
=
0,03
mol
->
nnci
= '^^^2 ^ ^'^^

XfVay:
nHci
di, =
a08 - 0,06
=
a02 mol.
t
mv =
40
+
1,04 - 0,03.2
=
40,98 gam
n
02 36 5 ^'^^
^
C%Hci=
' ^ ^ .100% =1,78% -> Dap an
D.
40,98 ^
Bai 2:
Phan
ung hoa hgc:
Fe + 2HC1 > FeCh + H2
Zn
+ 2HC1 > ZnCh + H2
Xet vai
100 gam dung
dich
Y:

mFeCl2
=
gam va
niz^ciz
=
8,63 gam ,
"1
->
mH20
(Y) =
100 - 8,05 - 8,63
=
83,32
gam.
Nhan
xet: Lugng
H2O
trong
Y
ciing
chinh la lugng
H2O
c6 trong dung
dich
HCl
ban dau.
Mat
khac, do nong do
HCl
bang C%

m
HCl
_
mH20
100-C
C
83,32.C
Bao toan nguyen
to'CI:
FeCl2
+2nz,
Dap an D.
2.8,06
2.8,63
83,32.C ^ ^„
2npea2
+2nznc.2
=nHCl
^
-1^^1^=36,5(100-0
^
=
Cty
TNHH
MTV
DVVH
Khang
Vigt
Bai
3:

2A1
+ 6HC1 > 2A1C13 +
3H2
^^' hS^ "
Mg
+ 2HC1 >
MgCh
+ H2 :
Khoi
lugng dung
dich
tang them
=
Dugc - Mat ->
7 =
7,8 -
mH2
' '
-> mH2 =
0'8 gam ->
nH2 =
0,4 mol. .
^
i^^^ A
Ggi
so
mol: Al = x; Mg = y:
• •
?
«^

Ta c6:
27x
+
24y
=
7,8; l,5x
+
y
=
0,4
x =
0,2
mol;
y
= 0,1 mol.
''
^
' '
-> mAi =
0,2.27
=
5,4 gam -> Dap an
A.
]
•=
'
•
Bai 4:
Bao toan
kho'i

lugng ta
c6: mo2 =
4,08 - 2,80
=
1,28 gam •
1
78
Vay:
no = -—
=
0,08
mol
-> n 2-
=
0,08
mol.
*
>
^(S"' • •
16
Khi
cho
oxit
kim
loai
tac dung voi
axit,
ion
0^~
trong

oxit
se ket hgp voi
H*
trong
axit
tao thanh
H2O
theo phuong
trinh:
:-m.ms
nbiji
.1
02- + 2H* > H2O ,
,,;,;.3««B(-^
Mol:
0,08 0,16 "
"
' i; V
1
n no
^
,4r
• <•
«
Vay:
nH2S04 =
2"H^
=0,08
mol
^

V^d
H2SO4
=
^
=
0,08
lit =
80ml
Dap an
A. >')
n^oJ c: '
Bai
5:
N/ifln
xet:
Bao toan nguyen
to'H:
H2SO4
> Hi ;« ,
Mol:
0,03 <- 0,03
'^muiBi':"
Bao toan
khoi
lugng cho so
do:
<,><i'>|
Kim
loai
+

H2SO4
)•
Muoi
+ H2 s <
Ta c6:
mkimiogi
+
maxit
=
mmuoi
+ m^^^
->
m = 0,03.2
+
3,92 -
0,03.98
=
1,04 gam Dap an
C.
' '*. ;''
Nhan
xet:
Khi
cho
X
va
Y
vao dung
dich
H2SO4

loang du tha'y con
lai
3,2 gam
kim
loai
khong tan -> c6 mgt
kim
loai
dung sau H trong day dien hoa (gia
su do la
Y)
->
mv =
3,2 gam;
mx =
6,45 - 3,2
=
3,25 gam.
X +
H2SO4
>
XSO4
+ H2 '
3 25 ' i
-> nv = nH-, =
0,05
mol-»
Mx = —— = 65
^
XlaZn.

Y + CI2 —!—>
YCI2
, , .
3 2
->
nY=
na2 =
0'05mol -> Mx=
-rhrz
=
64:
-^YlaCu.
->
Dap an
B.
, ,,
•
93
dm
nang
6n
luygn
thi d^i hpc 18
chuyen
6i H6a hgc -
NguySrr
Van Hai
1 792
Bai 7: 11^0= O'O^ mol.
Cachl:

Bao toan
khoi
lucmg: = my - mx =
25,28
-
0.28.56
- 64a = 9,6 - 64a. >
So do phan ung:
-j^yr:^],,)
'^'^M'''!^^
:P".
¥e,Cn{l) > Y(2) > Fe-^
Cu-^S)
Xet su trao doi eleclion o cac giai doan: ; £3
— >
le'-* -> nenhirong=3npp
=0,84
>
Cu*^ ^ nenhirong=2ncy = 2a i
U)^(3): Fo V -
Cu
-2e
(1)^.(2):
O. +4e ^
20-^->
nenhjn
=
4no2
= •^^^•^^= 1,2 - 8a
^

;Y|V
8
^2) ^(3): N** +3e >
NOnenhjn
= 3nNo = 0,24.
ir-;:!
i
•' Bao toan electron: 0,84 + 2a = 1,2 - 8a + 0,24 -> a = 0,06 mol.
-> Dap an C. • t
viy/i
Cach
2:
Qui
doi Y thanh: Fe = 0,28 mol; Cu = a mol va O = b mol.
Bao toan
khoi
luong: 64a + 16b = 8,58 -
0,28.56
= 9,6.
Bao toan electron:
3.0,28
+ 2.a = 2b + 0,24 -> a = 0,06; b = 0,36.
' -»Dap anC. ,, . • ': '
.• •
f
Bai 8:
Phan
ling
hoa hoc:
Fe304

+ 4H2SO4
Mol:
a
Fe2(S04)3
+ Cu
Mol:
a a
Fe2(S04)3
+
FeS04
+
4H2O
a a
->
2FeS04
+
CuS04
u^^m *
2a a
Chat ran Y la Cu du. Phan ung hoa tan Cu bang HNO3 loang:
3Cu + 8HNO3 > 3Cu(N03)2 + 2NO +
4H2O
Mol:
0,03 <<- 0,03 <- 0,02
232a
+ 64(a + 0,03) = 10,8 gam ^ a = 0,03 mol.
Vay: m =
152.3.0,03
+160.0,03
= 18,48 gam Dap an D.

4 0 224
Bai 9:
nFe2(S04)3
= =
"SO2
= = 0'°! mol. ,.:
Nhan
xei: Oxit sk
FexOy
tac dung voi
H2SO4
dac, nong
SO2
thi oxit la FeO
hoac
Fe304.
Trong
1 mol FeO
hoac
Fe304
deu chua 1 mol Fe*^ nen deu c6 kha nang
nhuong
1 mol electron, do do:
=
npe^Oy
= 2nso2npe^Oy = 2.0,01= 0,02 mol
Cty
TIMHH
MTV
DVVH

Khang
Vi$t
Bao toan nguyen to Fe: 2FexOy >
xFe2(S04)3
0,02
x.0,01
->
X.0,01
= 0,01 x = 1 -> oxit sat la FeO. ,i , ,
^ m =
72.0,02
= 1,44 gam Dap an A.
2
24 \' * t : ' f , . f
Bai 10:
nsoj
= ^ =
^jj-h_L:
GQI
cong thiic rhung cua cac kim loai la R. ' '
2R
+
2n}Lb04{dac)
'" > R2(S04)„ +
nS02
+ 2nH20
mol:
0,2 <- ai ^ 0,2 r '
i5ao
toan

khoi
hmng. 4,4 +
98.0,2
= m +
0,1.64
+
18.0,2
i .?
>^
in = 14 gam -> Dap an D.
rarh
2: Khi cho kim loai tac dung voi axit
F12S04
(dac, nong), so mol go'c sunfat
2
24
tao muoi duQ-c
tinh
nhu sau: n 2- =
"SOT
= —— = 0,1 mol.
SO4
22,4 J, ,5,^
Do vay: m = 4,4 +
0,1.96
= 14 gam -> Dap an D. •
Bai 11: 2KMn04 K2Mn04 + Mn02 + O2 .1'
"°2^i-i§^'^'°^"'°^'"S°2=^=0'035mol. ^
• '•^•.^•^.•'^.fr.:
Nhqn

xet: Bai nay neu dua theo phuong
trinh
phan ung se rat dai dong va kho
giai.
6 day, cac em can su dung so do phan ung: *'"'' £
•
Fe(amol),Cu(bmol)(l) > Y (2) > Fe^^ (3) •''''^^^^
Tu-(l)->(3):
Chatkhu:
Fe - 3e > Vv*^ Cu - 2e > Cu^^
Chat oxi hoa: O2 + 4e >20
2
+ 2e
>• SO2
^*^'''
Bao toan electron: .
v^iiiiBi
3nFe+2ncu=4no2
+
2nso2-^
3x + 2y = ai5. • ? ^
Mat
khac: 56x + 64y = 3,6 ^ x = 0,03; y = 0,03 ->
%mFe
=
46,67%.
Dap an A > '
.''ft'
Bai 12:
GQI

cong thiic oleum la
H2S04.nS03.
» t fu it
H2S04.nS03
+ nH20 >
(n+l)H2S04
Mol:
0,015 ,
0,015(n+l)
, , ,
H2SO4
+ 2NaOH >
Na2S04
+
2H2O
Mol:
0,015(n+l)
0,03(n+l)
->
0,03.(n+l)
=
0,2.0,3->
n = l ^ Oleum la
H2S04.S03^
Dap an D.
Bai 13:
n^^=^^=
0,15 mol; n^o2 = g = 0,3 mol. ^^^^^ ^J^^
95
Cam

naiKj
on luyeii
ilii
liai
iioc
18
chuy6n
d6 H6a hpc -
Nguyjn
VSn Hii
Nhan
xet: Khi cho X + HCl (du), chi c6 Al phan ung (Cu dung sau H).
Khi
cho X + HNO3 (dac, ngupi), chi c6 Cu phan ung (Al bi thu dong hoa,
khongtan). ,.
Cach
1: Giai
theo
phuong
trinh
2A1 + 6HC1 >
2A1C13
+ 3H2 n^i = ^n^^ = 0,1 mol. ^
Cu + 4HN03 >
Cu(N03)2
+
2NO2
+
2H2O
^ ncu = -"NOZ = 0,15 mol.

^ m =
0,1.27
+
0,15.64
= 12,3 gam Dap an B.
m-Uoi
+
Cach
2: Ap dung bao toan electron: ' "
3nAi
= 2 nH2 ^ n^i = 0,1 mol;
2ncu
= nN02 ^ ncu = 0,15 mol. '
m
=
0,1.27
+
0,15.64
= 12,3 gam Dap an B. ^ ' ""''^ = «!
Goi
ncu
= X mol;
nAi
= y mol. Ta c6: 64x + 27y = 1,23 gam. -(/ufc
«6wm
OJ;v
+ So do phan ung:
Cu >
Cu(N03)2
> Cu(OH)2i > [Cu(NH3)4](OH)2 '^^

Al
>
A1(N03)3
>
Al(OH)34'
'•^'^'••^^
'-^'^
+ Bao toan electron:
2ncu
+ 3nAi = nNo, -> 2x + 3y = 0,06 mol. ,
S<,?/'
->
X = 0,015 mol; y = 0,01 mol -> %mcu =
^^^^^.100%
=
78,05%.
1,23
Vay: m = mAi(OH)3 =
0,01.78
= 0,78 gam -> Dap an D. f;
Cdc em can
lieu
y: Cu(OH)2 tan trong dung dich NHs du.
BailS:
, ("f), tf)
Nhan
xet: Day la bai toan c6 nhieu
chat
khu (3 kim loai) va tgo ra 2 san pham
khi

nen can ap dung dinh luat bao toan electron.
Chatkhu:
Cu-2e
> Cu"2.
pe-3e
> Fe«;
Zn-2e
^Zn*2
Chat
oxi hoa: N^^ + 3e ^O; N*^ + le > NO2.
Bao toan electron:
netraodoi
— 3njyjQ +
1.
n[sjQ2 = 0,7 mol.
Den day, cac em can luu y la khi cho kim loai tac dung voi HNOs, so mol goc
NO
3 nam trong muoi dugc
tinh
nhu sau:
n
(muoi)
= n„
trao
doi = 0,7 mol. ' '' '
N03
Bao toan khol lupng: m = m^^ pg
"^MO-
^ "^^'^
0,7.62

= 64 gam. ^
-> Dap an B. \/^
Bai 16: n2n= 0,15 mol; n^u^ 0,15 mol.
Nhan
xet: Khi axit HNOs c6 mat dong thoi voi axit H2SO4 loang thi lugng H*
trong
dung djch la do 2 axit phan li ra —> giai
theo
phuang
trinh
ion.
96
Cty TNIIH MTV DVVH
Khang
Vi§t
=
nHNOa
+ 2nH2S04 = 0,4 + 2.a2 = 0,S mol
,p
^
-3
Trong X: j _ = 0,4 mol; n^ o_ = 0,2 mol. i:;n<
nin^''
NO3
SO4 ' ,
••' 'Hi "yi.H~'. /rai
JICL.
' i/iih
Phuong
trinh

ion
riit
gpn:
^
, .
3Zn + 8H^ +
2NO3
>2>Zn^*
+ 2NO + 4H2O,
Mol:
0,15 ^ 0,4 -> ai
oii^d.'
3Cu + 8H* + 2NO; >
3Cu2*
+ 2NO +
4H2O
o) O' ?;
Mol:
0,15 -> 0,4 -> 0,1 , , 1,:^
=.««„,m
* : Inm ££.0 - ,,->?i
H"^
tham gia phan ung het. ' " , -
_> Khoi
lupng
muoi trong Y = 9,75 + 9,6 +
0,2.96
+
0,2.62
=

50,95
gam. . , ,.
^ Dap anB. ^ ^.^^^ ^^^^ ^
^ ^
3 2
nr^,
= = 0,05 mol. n' r
>
N/jflM
xet: Khi axit
HNO3
c6 mat dong thai voi axit
H2SO4
loang thi lug-ng
H* trong dung djch la do 2 axit phan li ra giai
theo
phuong
trinh
ion.
[=
nHN03
+ 2nH2S04 = 04 +
2.0,01
= 0,12 mol , ^ ,^
Trong X: \ ^^J. ^
Q^Q^
^^^^
.i.,££.Kl,0
= V «
NO3

SO|
Phuang
trinh
ion
riit
ggn: t <> , . ;A - ; •-t-r ^j^^;
3Cu + 8H* +
2NO3
>
3Cu2-
+ 2NO +
4H2O
.5.', , ;
Mol:
0,045
0,12 -> 0,03 -> 0,03
V =
a03.22,4
=
0,672
litDap an C. .OI^S * 'tia + y >
BailS:
,
•/••••e'
'.fOU'"
\ . .
Phan
ling
hoa hpc:
FeS2

+
8HNO3
>
Fe(N03)3
+ 5NO +
2H2SO4
+
2H2O
^
Mol:
0,1 0,8 0,1 0,2
Fe203
+
6HNO3
>
2Fe(Na)3
+
3H2O
, .a nil
qftCl
'
Mol:
0,06 0,24 Q,\2 '
Dung
dich X gom cac ion: , <> ,
Fe3*
= 0,22 mol; = 0,4 mol; NO
^
= 0,66 mol va SO
]'

= 0,2 mol.
Cac phan ung hoa tan Fe: ^, IQ
<,
;
3Cu + 8H* + 2NO; )•
3Cu2^
+ 2NO +
4H2O
Mol: 0,15 <- 0,4 iVj/'V '
Cu +
2Fe3*
> Cu^* +
2Fe2*
^,.rid
ao:u.,.j,
i:;^!
::M;;t:
^
Mol:
0,11
<-0,22
.•''i^*''.^'^MM'^^'
-
'.''^'^
mcu
=
0,26.64
= 16,64 gam->
DapanB.^^^
, o-iH'W]JM-!<>.:/

97
Cim
nang
On luyQn thi dai hgc 18 chuySn dg H6a hgc
-
Mydygn
van Hi\
Bai 19:
N/ifln A:e^:
Khi X
tac
dung voi moi chat oxi hoa
HNO3
va
Ch,
cac
kim
loai
deu
duQic
dua
len muc oxi
hoa cao
nhat —>
So
mol electron trao doi trong
hai
truong
hgp la
bang nhau.

Chat
oxi hoa:
N^s + 3e ^ NO; N*^ + ie •—-> NOz.
Chat
oxi hoa:
Ch + 2e > 2C\-
Bao toan electron:
3nNo+nN02
=2nci2
^2nci2
=
0,2.3
+
0,1
= 0,7
mol
-)•
nci2
=
0,35 mol
->
mmu6i
=
mx
+
mQj
=
12,45 +
0,35.71
=

37,3 gam.
•
-> Dap an
A. '^^^^^
• ^ '
Bai20:ncu
=
a06mol.
, '^^
uimrDi
N^flM
xef:
Svr
c6 mat
axit
H2SO4
loang
se
"dong gop" them
lu^ng
trong
i.
dung
dich.
Khi
do,
lugng
la do 2
axit phan li
ra nen

can
giai
theo phuong
trinh
ion.
^
Ta c6:
n^+ =
nHNOg
+
2nH2S04 =
0,U
mol;
n^^_
= n^NO^
= 0,08
mol.
3Cu
+ 8H^
+2NO;
>•
3Cu2*
+ 2NO +
4H2O
•> ^i-ou
=
Mol:
0,06 <- 0,16^ 0,04 -> . - ,v 0,04 1
->
V =

0,04.22,4.1000
=
896ml
^ Dap an
A.
' •
;X
gncn
Bai 21:
Nh|n xet:
Theo
bai,
ta
coi
Cu
tan vua
du
trong hon
hgp
HNO3
va
HCl.
*
Taco:
n„+
=0,2.2+
0,1.0,8
= 0,48
mol; n,,^_
=

0,2.2
= 0,4
mol.
H
NO3
3Cu
+ 8H* +
2N03-
>
3Cu2^
+ 2N0 +
4H2O
, A U ^
Nhan
thay
^ < nen
tinh
so
mol
Cu
theo H^
•••
T
(:3n
. 3
0 48
'!-!
Ta c6: ncu
= —^ =
0,18 mol

-> m =
0,18.64
=
11,52 gam
8 8
Dap
an
D. ,()r/f?Py
4.
^r\>=,'4
Bai 22:
„
,
42.0,375
„ ,
Ta co: nHNOg
=
———
= 0,25
mol.
•'«
DO
FeS2
+
I8HNO3
>
Fe(N03)3
+
15N02
+

2H2SO4
+
7H2O
^
Mol:
0,01^ 0,18 0,01 0,02
Dung
dich
sau
phan ung gom:
7 ' ' ' ' ' ^'-^^
Fe(N03)3
=
0,01 mol;
HNa =
0,07 mol;
H2SO4
= 0,02 mol.
'1.0 dm.
De thu dugc ket
tiia
Ion nha't:
"NaOH
^
3"Fe(N03)3+r>HN03 + 2nH2S04= 0,14 mol
VNaOH
=
0,07
lit
=

70ml
Dap an
B.
08
Cty
TNHH
MTV
DVVH Khang Vijt
TiXH
CHAT
CUA CAC HIttROXIT
^
HIDROXIT
KIM LOAI
KIEM,
KIEM
THO
^ ^
^^^^^^^
'
Neu:
= a ^
nco2
rt-'.
, *
j,.Uthuyet
' ' ;
j;,,^,
.
Hidroxit

kim
loai
kiem
•'••^'•^g.^jjj
''^
''•'.yt
> ; ,
+
Tac
dung
vol oxit
axit
udl >K ;v
C02
+
2NaOH
>Na2C03
+H2O
o»U.;
Mt^O
^iO.^'.uog
;< rr.
CO2
+
NaOH
>•
NaHCOs
j
rififi^J
rmt

fcrb
rr!t;;<
:
a
< 1 :
NaHC03
1<
a < 2 :
NaHCOs
+
NazCOg
a
>2
:Na2C03
+
Tac
dung
voi
dung
dich
muoi:
Fe(N03)3
+
3NaOH
>
Fe(OH)3
i +
3NaN03
ji^f^
.^^^j,

,^
+
Tac
dung
voi
hgp
chat
luang
tinh
^OD"
AI2O3
+
2NaOH
>
2NaA102
+
H2O
.JBnocfofe)
;o:
Al(OH)3
+
NaOH
>
NaA102
+
2H2O
/.X)
:(«:>* rimii
+
Dieu

che ' M
2NaCl
+
2H2O
^^'^ >
2NaOH
+ Cht + HzT
j^Y/bib^-,:
Hidroxit
kim
loai
kiem
tho - >' tn
r
.x^ry
•\r,m<'f 0
-
,I'-VT;
'•(' !}
y.;
CO2
+
Ca(OH)2
>
CaCOsi
+H2O
2CO2
+
Ca(OH)2
>

Ca(HC03)2
'•
Tinh
so
mol:
n =
2nca(OH)2
*
Neu:
= a
nco2
a
< 1
:
Ca(HC03)2
^>-n^
jiSt-S)
t
1<
a < 2
:
CaCOs
+
Ca(HC03)2
M£t,0 ^^K'
,
a
>2
:CaC03
b.Vidv

mau
Vi dvi 1:
Cho cac
chat:
NaHC03,
CO, Al(OH)3,
Fe(OH)3,
HF,
CI2, NH4CI.
So chat
tac dung dugc voi dung dich
NaOH
loang
6 dieu
kif n
thuong Ik , :
A.
4.
, , B.5. C.3. D.6.
Lbi
giai:
Nhan
xet: Day la cau hoi
tong
hqip,
lien
quan den
kien
thuc lop 10
(HF,

CI2),
lop 11 (CO,
NH4CI)
va
lop
12
(NaHC03,
Al(OH)3,
Fe(OH)3).
Cac
chalt xay ra phan ung: •,
.^^j^j^
^i,.
v
(
^
NaHC03
+
NaOH
> Na2C03 +
H2O
3a'm
nang
On
luy^n
thi dgi hpc 18
chuy6n
dg H6a hpc -
Nguygn
Van Hi\

Al(OH)3 +
NaOH
——> NaA102 + 2H2O
CI2
+ 2NaOH >
NaCI
+
NaClO
+ H2O ;.|^| „
NH4CI
+
NaOH
>
NaCl
+ NH3 + H2O "^'^^
HF .
NaOH
> NaF . H2O ' ^'•W
m.^>rfmimm^^^^
-^DapanB. - • ' ,
^n^'^^'
Luu
y: CO khong tac dung vai dung dich
NaOH.
*
Vi dv 2
(CD-12):
Hap thu hoan toan 0,336 h't khi
CO2
(dktc)

vao 200ml dung
dich
X gom
NaOH
0,1M va KOH 0,1M thu
dugc
dung djch Y. Co c^n Y thu
dugc
a gam chat ran khan. Gia trj ciia a la * '' XkM T .siX)
A. 2,58. B.2,22.
'Ht,.:
C.2,31. D. 2,44.
•
A;;,,.:.,
Lmgidi:
In = n^aOH + "KOH = 0,02 + 0,02 = 0,04 mol
TrongX:
\"
K^^s.^
.iv ,-',
v::;.!
n
+
= 0,02 mol; n+ = 0,02 mol.
Nhan thay: = = ^ > 2 OH" du va phan ling chi tao ra
muoi
cacbonat. 1 . ' '
>
I'
Phuong

trinh
ion: CO2 + 20H- >• CO3" + H2O 1
i,.!' ?',
Mol: 0,015 0,03 0,015 ''^ '
Dung dich Y gom cac ion: '
Na* = 0,02 mol; = 0,02 mol; OH" =
0,01
mol va CO
3'
= 0,015 mol. \i
Bao toan khoi lugfng: a = m
+
+ m+ + m +m 7
^'"^1':)
° Na* OH' CO;'
=
0,02.23
+ 0,02.39 + 0,01.17 + 0,015.60 , ,
^ m = =
2,31
gam -> Dap an C.
/i
d^ 3
(B-12):
Syc 4,48 lit khi
CO2
(dktc)
vao 1 lit dung dich Z chiia hon hgp
Ba(OH)
2

0,12M va
NaOH
0,06M. Sau khi cac phan ung xay ra hoan toan thu
duQfc m gam ket tua. Gia trj cua m la
A. 19,70. B. 23,64. C. 7,88. ^ D. 13,79.
, Ldtigidi:
roH'"
"NaOH + 2nBa(OH)2 = +
2.0,12
= 0,3 mol
Tronc Z: < •
"2+= 0,12 mol; n^,+ = 0,06 mol.
Nh$n thay: = M = 1,5 > 1 _^ phan ung t^o ra 2 muoi:
cacbonat
va
, "CO2 0,2 ,y
hidrocacbonat. '» I s;
Ap
dung
cong
thuc
tinh
nhanh:
n^^j-
=
n^j^-
- ncoj = 0,3 - 0,2 = 0,1 mol.
Cty
TNHH
MTV DWH

Khang Vljt
Nhan thay: r\^^2-< "332+ ^ ngacOa = "co2-"
= 0,1.197 = 19,7gam. ' '
Dap an A.
Vi
dy 4: Hap thy hoan toan
2,688
lit khi CO2 (a dktc) vao 900ml dung djch
Ba(OH)2 a
mol/1,
thu
dugc
11,82 gam ket tua. Gia trj ciia a la '''®r ''^
A. 0,03. B.0,04.
C.0,05.
D. 0,lO.
Laigidi: j .,-1 ,, . . ,,,,
nco2
=042
mol; nBaC03 = 0.06 mol. -
C^chl:Giaitheo phuong
trinh
Cf'J 'fiMS • "j '
Ba(OH)2 + CO2 >
BaCOs
+ H2O' *
Mol: 0,06 0,06 ^ 0,06 ' ^ -
^^•'''^
Ba(OH)2
+

2CO2
>
Ba(HC03)2
,
Mol: 0,03 0,06
-> riBa(OH)2 = 0,09 mol ^ a = 0,1 M Dap an D.
each
2: Bao toan nguyen to C:
'^'^
^ " " ' '
.
j,
*
^'»"
'
nc02
= "BaCOa + 2nBa(HC03)2 nBa{HC03)2 = 0.03. ^ ^ ^
Bao toan nguyen to Ba: > fj < 10 - ; '
nBa(OH)2 = "BaC03 + nBa(HC03)2 = 0,09 -> a = 0,1M. ,
->DapanD. '
Vi
dv 5: Trpn 100ml dung dich X gom
NaOH
0,2M va Ba(OH)2 0,1M voi 100ml
dung dich Y gom H2SO4 0,1M va
MgS04
0,2M, thu
dugc
m gam ket
tua.

Gia
trj
cua m la
A. 2,91. B.7,57. C. 3,49. D. 1,16.
Laigidi:
TrongX:
n^^j* = 0,01 mol;
n^^^^
= 0,02mol;
n^^,
= 0,04mol.• -'J*,
TrongY:
n„ 2+= 0,02 mol; n
o_
= 0,03 mol; n+= 0,02mol.
Cac phuong
trinh
phan img khi pha
tion:
i 5(HO)
H^
+ OH" > H2O +
'^liO:':':
Mol: 0,02 -> 0,02 , ^p ,
^
Ba2^
+ SO^" > BaS04i ^,
,, da*:0
Mol: 0,01 -> 0,01 -» 0,01 • #i(;>\ii^,:'^ViV', •
Mg2*

+ 20H' >
Mg(0H)2i
rv- +
iKKS
•
Mol: 0,01 ^ 0,02 ^ 0,01
V$y: a = 0,01.58 + 0,01.233 = 2,91 gam
-> Dap an A. . i
101
dm
nang
6n
luygn
thi dgi hpc 18
chuy6n
66 H6a hpc -
Nguygn
Van HSi
Cty
TNHH
MTV
DVVH Khang Vijt
Vi
6: Cho 5,35 gam hon hop gom Na, K va Ba vao
nuoc
(dn), thu
duoc
dung dich X va 1,12 h't khi H2 (dktc). Dung djch Y gom HCl va
H2SO4
c6 ti

1$ mol tuang ung la 1:2. Trung hoa dung dich X bai dung dich Y, tong khoi
• lugng cac muoi
dugc
tao ra la
A.
9,40 gam. f ,3'B. 8,69 gam. C. 7,98 gam. D. 9,90 gam.
Laigidi:
•
ivCK,.!
' •XOiO ,/
=
TT—
= 0,05 mol. Cac phan ung hoa hoc: , 1:. i, ,
, •
,
4
^,,<j^,^fi'-iom<:r:d-
:jo:>'''
2Na +
2H2O
>
2Na*
+ 20H- + H2 -a:
j,,nmd^^,:.^fiiD
:M.
2K
+ 2H2O > 2K^ + 20H- + H2 . .
Ba ^+
2H2O
> Ba^^ + 20H- + H2

^f^,^,
^0,0 :!oK
Nhan
xet: • ^ ,
"oH 2"H,=0,,mol.
^„ ^^^^ ^^^^^
dO,0
Gpi
nHci=a
2504
^2a.
Trong Y: n^+ = 5a mol; = 2a mol; n^^. = a mol.
aaoi
okH
i£.ffas;.
Trung hoa X bai Y: + OH- > H2O
5a = 0,1 -> a = 0,02 mol.
Khoi lupng muol thu
dugc
= 5,35 +
m|l°+
' C^'" "
oHO)«3"
so^
rrrtH.'
1
jov M.i,0s(]H.U)Bii = 5,35 +
0,02.35,5
+
0,04.96

=
9,90gam,,i
jgyjj s
:,K)->
DapanD.Ti:jY;itib
ijfB vi/t:k,,U
R,,>GgM, CVI-MLAJ AJdiK
rw^'j;
y',ffo|f> grw:!)
2.
HIDROXIT LI/ONG TINH
,3
a. Li thuyet
Cac hidroxit luong
tinh:
Zn(OH)2, Al(OH)3, Sn(OH)2, Pb(OH)2, Cr(OH)3 ,
*
Tinh
luang
tinh
Al(OH)3
+ NaOH >
NaAI02
+
2H2O
Al(OH)3
+ 3HC1 > AICI3 +
3H2O
u
fK-Hc

Zn(OH)2
+ 2NaOH >
Na2Zn02
+ 2H2O
' •001 £0.0
^
'fio-iT
Zn(OH)2 + 2HC1
-> ZnCh +
2H2O
;foM
luu y: Cac
hidroxit:
Zn(OH)2,
Cu(OH)2
tan
trong
dung
dich
NH3.
*
Dieu
cheAKOWs
'teM v
AICI3
+
3NH3
+
3H2O
> Al(OH)3i +

3NH4CI
AICI3
+
3NaOH
> Al(OH)3 + 3NaCl •
NaA102
+ CO
2
+2H2O
> Al(OH)3l
+NaHC03
J, Vi drf mau
Vi
1: Cho mot mau hgp kim
Na-Ba
(ti If mol 1:1) tac dyng v6i
nuac
(du),
thu
dugc
dung dich X va 3,36 lit H2 (a dktc). Cho
400ml
dung dich
AK(S04)3
0,lM vao X, sau khi phan ung
hoan
toan,
thu
dugc
a gam ket tua. Gia tri

cua a la
A.
24,86.
B.
29,54.
C.
27,98.
Cac phan ung hoa hoc:
2Na
+ 2H2O > 2NaOH + H2
Mol:
X X 0,5x
Ba +
2H2O
>
Ba(OH)2
+ H2
Mol:
X -> X X
D.
34,20.
- ¥
Theobai:
nH2 = 0,15 -> l,5x = 0,15 -> x = 0,l.
Dung dich X gom: Na* = 0,1 mol; Ba^- = 0,1 mol; OH- = 0,3 mol.
Khi
trgn X vai dung dich gom: AP* = 0,08 mol; SO l~ = 0,12 mol: >
f f
>
Ba2*

+ SO
2-
->
Mol:
0,1
Al^ + 30H- •
Mol:
0,08 -> 0,24
Al(OH)3
+ OH-
Mol:
0,06 <- 0,06
BaS04i
0,1
->
Al(OH)3i
0,08
—> AIO2 +
2H2O
•61 X mfb
pmb,f/.tt-r
a =
0,1.233
+
0,02.78
=
24,86
gam ^ Dap an A.
Vi
d^ 2

(A-12):
Hoa tan
hoan
toan
m gam hon hgp gom
Na20
va
AI2O3
vao
nuoc
thu
dugc
dung dich X trong
suo't.
Them
tfr tir dung dich HCl IM vao
X, khi het
100ml
thi bat dau
xuat
hien ket tiia; khi het
300ml
ho?c
700ml
thi
deu thu
dugc
a gam ket tiia. Gia tri
cvia
m va a Ian lugt la

''^^^"
A.
27,7 va 15,6. B. 35,9 va 23,4. C. 56,3 va 23,4. D. 55,4 va 15,6.
Laigidi:
Cac phan
ling
hoa hgc:
^•1
TOY
-gnv
nkti^
•>••
Na20+
H2O
»
2NaOH
AI2O3
+
2NaOH
>
2NaA102
+ H2O
Nhan
xet: dung djch X trong
suot
AhOs
tan het.
NaOH + HCl
Mol:
0,1 <- 0,1

—)•
NaCl + H2O
NaAlCfe + HCl + H2O
Mol:
0,2
AI(OH)34'
0,2
+ NaCl
,14
103
elm
nang
fln
luy^n
thi dgi hpc 18
chuy6n
ai Hda hpc -
Nguyjn
Van
Hai
—>
a =
0,2.78
= 15,6 gam ->
Loai
phuang an B va C.
Nhan thay:
Khi
cho 700ml dung
dich

HCl vao
X,
toan bp
NaA102
se chuyen
het thanh ket tua
Al(OH)3,
va sau do bi hoa tan mpt phan trong
HCl:
NaAlOz
+ HCl +
H2O
>
Al(OH)3i
+
NaCl
iiek;
Al(OH)3
+ 3HC1 >
AICI3
+
3H2O
*
Mol:
^ 0,6-x
^iM.;n^^:'K
'Vi^eorf.'^^n'if
nidq^ifi
De thay:
nAi(OH)3

=
0'2 mol -> x - ^'^^ ^ = 0,2 -> x = 0,3
mol.
x iolv.
Tu
day suy ra: NazO
=
0,2
mol;
AI2O3
=
0,15 mol.
^
m = 0,2.62 + 0,15.102 = 27,7 gam ' ^
->
Dap an
A.
"
Vi d\
3: Nho tu tir dung
djch
NaOH den du vao dung
dich
X. Sau khi cac
phan
ling
xay ra hoan toan chi thu dugc dung
dich
trong suo't. Chat tan
trong

dung
djch
X la
UM^ll
i-'
r T
;
i
C
A.CUSO4.
B.Fe(N03)3.
C.
AlCla.
D.
Ca(HC03)2.
Lbi
giai:
Logi
A
vi:
CuS04 + 2NaOH >
Cu(OH)2>l
{xanh)
+ Na2S04
|„|,ji
Loai
B
vi:
FeCb +
3NaOH

>
Fe(OH)34
(ndu
do) +
3NaCl
Loai
D vi
Ca(HC03)2
+
2NaOH
> CaCOsi' (trdng) + NazCOs +
2H2O
Dap an C.
Dau
tien tao ket
tiia,
sau do ket tua tan, thu
dupe
dung
dich
trong suo't:,
AICI3
+
3NaOH
>
Al(OH)3i
+
3NaCl
Al(OH)3
+ NaOH >

NaA102
+
2H2O
f,
i,f, frnWr t«i^ (fbf
,X
Vi d\
4
(A-11):
Cho cac chat: NaOH,
Sn(0H)2,
Pb(OH)2,
Al(OH)3,
Cr(OH)3.
So
chat
CO
tinh
chat ludng
ti'nh
la
^ ^
A.
3. B.l. C.2. ' D.4.
LMgidi:
)
Cac
hidroxit
luong
tinh

la:
Sn(OH)2,
Pb(OH)2,
Al(OH)3,
Cr(OH)3.
Cac phan ung voi NaOH: „ ,;,
Zn(OH)2
+
2NaOH
>
Na2Zn02
+
2H2O
Pb(OH)2
+
2NaOH
>
Na2Pb02
+
2H2O
Al(OH)3
+ NaOH >
NaA102
+
2H2O
Cr(OH)3
+ NaOH >
NaCr02
+
2H2O

->
Dap an
D.
.j^l^
•OdA
Cty
TNHH
MTV DWH
Khang
Vi$t
Vi
5: Cho 500ml dung
djch
Ba(OH)2 0,1M vao V lit dung
dich
Al2(S04)3
0,1M;
sau khi cac phan ung ket
thiic
thu
duQC
12,045 gam ket
tiia.
Gia tri
ciiaVla
'^^-'-^
'
A.
0,30. B.0,10. C.0,20. '
''^^^^

D. 0,15.
Laigiai:
.^'^.t.O
Cac phan ung hoa hpc: » .
1
>
3Ba(OH)2
+ Al2(S04)3 > 3BaS04i +
2Al(OH)3>l
> (1)
Ba(OH)2
+
2Al(OH)3
>
Ba(A102)2
+
4H2O
' ' (2)
Tnmng
hop 1: Chi xay ra phan ung (1)
—>
Ba(OH)2
het. ^
Ket
tiia
gom: BaS04 = 0,05 mol va
Al(OH)3
=— mol. ^, ^
• ^.
,

->
Khoi
luong ket
tiia
=
0,05.233
+—.78
=
14,25 gam
Loai.
. /
3
,
Truong
hop 2: Xay ra phan
ling
(1) va (2). ^
3Ba(OH)2
+ Al2(S04)3 > 3BaS044' +
2Al(OH)3i
(1)
Ba(OH)2
+
2Al(OH)3
>
Ba(A102)2
+
4H2O
'^'^'^^Qf^'f^)
Mol:

a05-3a
ai-6a
v.^r;;
:
Ket
tiia
gom: BaS04 = 3a mol;
Al(OH)3
= 2a - (0,l-6a) = (8a-0,l) mol.
->
Khoi
lu(?ng ket
tiia
=
3a.233
+ 78(8a - 0,1) = 12,045 gam
->
a = 0,015 mol -> V = 0,15
lit
-> Dap an D.
Vi d\
6: Hoa tan hoan toan m gam Al vao Vml dung
djch
HCl
IM,
thu
dupe
dung
dich
Y. Them tii tu dung

djch
KOH IM vao Y, khi het 100ml thi bat
dau xua't hi^n ket
tiia;
khi het 400ml
hoac
800ml thi deu thu
dupe
lugng ket
tiia
bang nhau. Gia tri
ciia
a va V Ian
lupt
la
A.
2,7 va 500. B.2,7va400. C.5,4va700. D.5,4va600.
Laigiai:
Cac phan
ling
hoa hpc:
^^ ^^j'
Al
+ 3HC1 >
AICI3
+ |H2 , .OOc,
6yf>!ic

KOH
.HCl —> KCl

+H2O
S«.MS,0
:
Mol:
0,1 0,1 , y;. ,
Vay
trong dung
djch
X,
so'mol
HCl
=
0,1
mol. g
AICI3
+
3KOH
>Al(OH)34
+
3KCl,|,,,r
Mol:
ai <- as ai / i/;
->
So mol ket
tiia
bang nhau va bSng
0,1
mol.
Nhan thay: Khi cho 800ml dung
dich

KOH vao X, toan bp
AICI3
se chuyen
het thanh ket
tiia
Al(OH)3,
va sau do bi hoa tan mpt phan trong
KOH:
dm
nang
On
luyjn
thi dgi
hQC
18
chuyen
dS H6a hpc
-
NguySn
Van
HSi
AlCb
+ 3KOH > Al(OH)34. + 3KC1
Mol:
X 3x X , .
Al(OH)3
+ KOH >
KAIO2
+
2H2O

'\ "^^^
Mol:
0,7-3x
<-
0,7-3x
or
<>i
-> r, H t. n /
De
thay:
nAi(OH)3
=
04
mol ^ x - (0,7 - 3x) = 0,1 ^ x = 0,2 mol.
Tir
day suy ra ban dau: Al = 0,2 mol; HCl = 0,7 mc*
ij.iu
;m»iq
'ifiO
->
m =
0,2.27
=
5,4 gam va V = 700ml , ,
^
,,.:.IS-«A
,
+
£(HO)Rf*i
-^DapanC.

.,^ ^.u.)$B4
^(1-!0)!A1
i(HGfey
3.
BAI TAP ON
LUYEN
^ ' ^ ^ *^ ^
Bai
1 (A-12): Cho day cac
chat:
Al,
Al(OH)3,
Zn(OH)2,
NaHC03,
Na2S04. So
chat
trong day vua
phan
ung
dugc
voi
dung
dich HCl, vira
phan
xing
du<?c
vai
dung
dich NaOH la
'

A. 5. B.4. C.3. D.2.
Bai
2
(B-09):
Iron 100ml
dung
dich X gom
H2SO4
0,05M va HCl 0,1M voi 100ml
dung
dich Y gom NaOH 0,2M va Ba(OH)2 0,1M, thu
dugc
dung
djch Z.
Dung dich Z c6 pH la
A.
13,0. B. 1,2. C. 1,0. . D. 12,8.
Bai
3: Trpn 100ml
dung
dich X gom
Al2(S04)3
0,1M va
H2SO4
0,1M voi 100ml
dung
dich Y gom
Ba(OH)2
0,1M va NaOH 0,7M, thu
dugc

a gam ket hia.
Gia tri cua a la , ,
A.
1,56. , ,
B.3,ll.""'f'
C.0,78.
D. 3,89. *'^t
Bai
4: Hoa tan
hoan
toan
5,35 gam hon hgp gom Na, K va Ba vao
rtuac,
thu
dugc
dung
dich X va 1,12 lit khi
H2
(dktc). Dung dich Y gom HCl va
H2SO4
*'
vai ti
1$
mol tuong ung la 1:2. Trung hoa
dung
dich X boi
dung
dich Y, tong
kho'i
lugng cac muo'i

dugc
tao ra la
A.
9,19 gam. B. 10,61 gam. C. 9,90 gam. D. 7,98 gam.
Bai
5: Cho Vml
dung
dich KOH IM vao coc
dung
150ml
dung
dich AlCb IM
ta thu
dugc
7,8 gam ket
tiia.
Gia tri nho
nhat
va Ion
nhat
ciia V Ian lugt la
A.
300 va 500. B. 300 va 600. C. 200 va 400. D. 200 va 500.
Bai
6: Cho Vml
dung
dich NaOH IM vao 100ml
dung
dich
chua

dong
thai
HCl
0,2M va ZnCh 0,2M. Gia tri Ion
nhat
cua V de thu
dugc
0,99 gam ket
tuala
A.
60. ' ^ B.90. C.70. * D. 80. '
Bai
7: Cho V lit
dung
dich NaOH 2M vao
dung
djch
chua
0,1 mol
Al2(S04)3
va
0,1 mol
H2SO4
den khi
phan
ung
hoan
toan,
thu
dugc

7,8 gam ket
tiia.
Gia
trj
Ion
nhat
ciia V de thu
dugc
lugng ket tua tren la
A.
0,45. B.0,35. C.0,25. D. 0,05.
"^'^^^
Cty
TNHH
MTV DWH
Khang
Vigt
8: Y la
dung
dich Zn(N03)2 x
mol/1.
Khi cho 30ml
dung
dich KOH 2M va
khi
cho 70ml
dung
dich KOH 2M vao 100ml
dung
dich Y, deu thu

dugc
lugng ket tua
bang
nhau.
Gia tri cua x la
A.
0,3. B.0,2.
C.0,5.
D. 0,4. *
Bai
9: Y la
dung
dich
chua
x mol AlCb. Khi cho 60ml
dung
dich NaOH IM va
khi
cho 140ml
dung
dich NaOH IM vao Y, deu thu
dugc
lugng ke't tua
bang
nhau.
Gia tri cua x la
A.
0,03. B. 0,04.
C.0,02.
D. 0,07.

Bai
10: Hoa tan
hoan
toan
2,7 gam Al trong 400ml
dung
dich HCl IM, thu
dugc
dung
dich X. Cho tung gigt den het V lit
dung
dich NaOH IM vao X,
thu
dugc
3,9 gam ket
tiia.
Gia tri Ian
nhat
ciia V la ,
j^.
>:,
A.
0,25. B.0,35.
C.0,45.
D. 0,05. '
4.
Hl/^NG
DAN
- L6I
GIAI

Bai
1:
i!5
:•(
qfid
;
Cac
chat
thoa
man la:
Al,
Al(OH)3,
Zn(OH)2,
NaHCOs.
Cac
phan
ung vai NaOH:
Al
+ NaOH +
H2O
)-NaA102+-H2
2
Na2C03 +
H2O
•
NaA102
+2H2O
> Na2Zn02
+
2H2O

2A1
+ 6HC1 -> 2AlCb +
3H2t
NaHC03
+ 2HC1 NaCl +
H2O + CO2
Al(OH)3
+ 3HC1 ^
AlCb
+
3H2O
Zn(OH)2
+ 2HC1
^ ZnCb
+
2H2O
NaHCOs
+ NaOH
Al(OH)3
+ NaOH
Zn(OH)2
+ 2NaOH
Dap an B.
Bai2:
Nhan
xet: Bai nay chiia nhieu
chat
dien li nen giai dua
theo
phuong

trinh
ion.
Do bai
toan
chi hoi ve pH -» chi
quan
tam den
H*
va OH".
" "HCl +
2nH2S04
= O'Ol +
2.0,005
= 0,02 mol
^OH-
=
'^NaOH+2nBa(OH)2
= 0'02 +
2.0,01
= 0,04 mol
Trong X:
n„,.
= 0,01 mol; n o_ =
0,005
mol.
(_1 so^
Trong Y:
OH-
n„
2+

= 0/01
mol; n
+
= 0,02
mol.
Ba
Na
Phuong
trinh ion: H* + OH" >
H2O
0,2
=0,1=10
-1
[H1
——
= 10'" pH = 13 ^ Dap an A.
[OH-]
Bai3:
^' 0' n ~v •
Trong
X:
n
^
,3+
= 0,02 mol; n_
^_
= 0,04 mol; n^+ = 0,02mol. ,,
dm
nang
On

luy^n
thi dgi hpc 18
chuySn
dg H6a
hqc
-
Nguyjn
Van H5i
Trong
Y: n^^2+ =
O'Ol
mol; 11^^^+ =
0,08
mol;
n^^_ = 0,09
mol.
Cac
phuong trinh phan ung khi pha trpn:
_
'.MvJ
J.ii:-dn
.
^''lA'^tt&v'
+
OH" > HiO .,, ^, f,f
Mol:
0,02 -> 0,02 \ . ,, , , '
AF
+ 30H- >
Al(OH)34

.d^uksnuivimuM
fc h
Mol:
0,02 -> a06 ^ 0,02
nurmiO
u«rin ^m,
Al(OH)3
+ OH" >
AIO2
+
2H2O
J .S(W) A
irfi'Mol:
0,01 0,01
tjfc'.sjnoTf Ji<*i iTSiig
v^i'j'ieol
nidri,
r^^^^^^
Ba2*
+ SO4" >
BaS044
J'mrt
ofi:)
.X
ritijb
^rnib:^>ir'.
ni
Mol: 0,01 0,01 -> 0,01 '
Vay:
a = 0,01.78 + 0,01.233 = 3,11 gam

-> Dap
an B.
Bai
4:
nH2
=0,05
mol.
**"'
Cac
phan ung
hoa hoc:
Na
+ H2O > Na* +
OH"
+ ^Hi ^ ,
'"X)
K +
H2O
> +
OH-
+
iH2 ''ffrmsM
Ba
+
2H2O
> Ba^* + 20H- + H2
N/ifl«
xet:
n^„.
=

2nHo
= 0,16
mol.
Ggiso'mol:
nHci^^n^ol
~^
^1^2504^
2a. r i
-» Trong
Y: n^+= 5a
mol; ng^2-=
2a
mol;
n^j.
=
a
mol.
v:!
TrunghoaXboi Y:
H* + OH- > H2O
.^j^,,
^
5a
=
0,1 -> a
=
0,02
mol.
I , T
,,

Khoi
lugng muoi
thu
dupe
= 5,35 + + ^^2-
'.
! s'- ^ = 5,35 + 0,02.35,5 +
a04.96
= 9,90
gam.
f
Dap
an C.
Bai
5:
Nhan
xet:
Day
la
bai
tap
tong quat ve hidroxit luong tinh.
Cachl:
nAi(OH)3
=
0,10
mol;
n^a^ = 0,15
mol.
Trwang/lop

1:
Tim
gia tri
nho nha't cua
V:
•
3KOH
+
AlCb
>
Al(OH)3l
+ 3KC1
Mol:
0,3 <- aiO 0,10 ,
j^,
VKOH=^=0'31it
=
300ml.
a
.:,ior.^
1
Cty
TNHH MTV DWH
Khang
Vijt
0,3 vsHOM'iv;/./
Tritcmg
hap 2:
Tim gia tri Ion
nha't

ciia
V:
3KOH
+
AICI3
>
Al(OH)3i
+ 3KC1
Mol:
0,45 <- 0,15 0,15 i
Al(OH)3
+ KOH >
KAIO2
+
2H2O
/, r
,i,
.j,;';
Mol:
0,05 -> 0,05 " :
VKOH=
^^^^
=
0,5
lit
=
500ml >
Dap an A.^
3'"'^^
^

X^'"^^
gach2l
^p
dung
cong
thiic
tinh
nhanh
HKOH min
=
3"Al(OH)3
=
0,30 ->
VKQH min =
^
=
0,3
lit =
300ml.
nKOHmax=
4n^,3+
"
nAl(OH)3
=
4.0,15-0,10
=
0,50 ^
VKOH max =
0,5
lit

=
500
ml.
Luu
y:
Doi voi
hidroxit
luong
tinh
Zn(OH)2,
cac
cong
thuc
tinh
nhanh la:
nKOHmin
=
2nzn(OH)2
^3
FIKOH
max=
4n^^2+
'
2nzn(OH)2
Bai
6:
nHci
=0,02
mol;
nznci2

= 0'02
mol;
nzr,(OH)2
= 0,01
mol.
Cachl:
Hci
.
NaoH
—. H20
''^;^'";*'f
^
Mol:
0,02 ^ 0,02
•:,#.0^^,,r,<)0,0«„,,,,,:o,,,.
^
ZnCh
+
2NaOH
>
Zn(OH)2@
+
2NaCI
:'
Mol: 0,02 <r- 0,04 -> a02 *
<r"'-^"if':H;.)iiA«£
=«WMWOB1<"
Zn(OH)2
+
2NaOH

>
Na2Zn02
+
2H2O
Mol:
0,01 0,02
^n^^.
= 0,02 + 0,04 + 0,02 = 0,08
mol. ">»ra
W-U'"'
n^aOH
=
"QP^-=
0,08
mol
^ V = 0,08
lit-> Dap
an
D
w-:,^,/'^oiA"
*
£ach2:
Tinh
nhanh: n^^o^^3^= 4n^^2+
-
2nz„(OH)2
= 4.0,02 - 2.0,01 = 0,06
mol.
nNaOH=
"HCl

+
HNaOHmax=
0,02 + 0,06 = 0,08
molV
= 80
ml.
•
-
I,
i
Bai7:
nH2S04
=
0,10mol;
nAi2(S04)3
=
0,10mol;
nAi(OH)3
=
0,10mol
,
Cachl:
^.v
2NaOH
+
H2SO4
>
Na2S04
+
2H2O

fjCiBl'l
Mol:
0,20 <- 0,10 fo In?
Al2(S04)3
+
6NaOH
>
2Al(OH)3^
+
3Na2S04
)IA
Mol:
0,10 -> 0,6 -> 0,20 {
<i
(,
Al(OH)3
+
NaOH
>
NaA102
+ H2O f
,(irfi),A
lol:
0,10 -> aiO • *'
n^^.
=
0,20
+
0,60
+ 0,10 =

0,90
mol
V
=
—
=
0,45
lit
^
Dap an
A.