Cẩm nang ôn luyện thi đại học môn Hóa học 18 chuyên đề

TS.
NGUYEN
VAN
HAI (Chu bien)
NGUYEN
NAM
TRUNG
-
TRAN
THE NGA
-
NGUYEN
THj THU HA
Cam
naligFluyen
thi
dai
hoc
18
CHUYEN
OE
^
HOA HOC
^
He thons cac phUdng phap
siai
nhanh
bai
tap hoa hoc.
^
Duns

cho on
tap
va
thi
tot nshiep
THPT.,.
^
Luyen
thi
vao
Dai
hoc va Cao dans.
rW VJEWTIWHSINHTHUAN
Dm
MHA
yi
lAT
RAN
RAI
HOC QUOC GIA HA NOl
MUC
LUC
huyen
di 1. Cac
phucmg phap giai nhanh
3
huyen
de 2. Cac
axit
v6 ca

diln
hinh
71
huydn
de 3.
Tinh
chat
cua cac
hidroxit
99
huySn
di 4.
Tinh
chat
cua cac
mu6'i
v6 ca 111
huyen d^
5.
Tinh
chat
cua cac
oxit
131
huyen
di 6. Cac
nguydn
to
phi kim di^n hinh
148

huydn
d6 7.
Dai cufotng
v6
kim loai
173
huyen d^
8. An
mon kim loai
-
Di^u chS' kim loai
190
huyen
de 9.
Kim loai kiem
-
Kiem
tho -
Nhom
204
huyen
de 10. Sat -
Cr6m
-
Dong
231
huyen
dl 11. Cac li
thuy^t
co

ban ciia
hoa hoc 253
huyen
de 12.
Su dien
li -
Axit, bazof, muoi
268
huyen
de 13.
Ancol
-
Phenol
285
huyen
14.
Andehit
-
Axit cacboxylic
301
huyen
de 15.
Este
-
Lipit
325
huyen
d6 16.
Cacbohidrat
341

huyen
d6 17.
Amin
-
Amino
axit
-
Peptit
352
huyen
di 18.
Polime
va vat
lieu
polime
370
e
thi
thir dai
hoc 378
•
*
Cty
TNHH MTV DWH
Khang
Vi^t
rhuyen
de 1 ? ,
CAC
PHlrtl^NC

PHAP
GIAI
NHANH
1.
PHl/ONG
PHAP BAO
TOAN
KHOI
Ll/QNG
. ^ . ,
a. Ngidung
« ^.
Tong khoi
lu<?ng cac
chat
tham gia phan ung
=
Tong khoi luong
cac
chat
tao
thanh sau phan ung.
,
m^i^^mm,
-"^-''i''!'^
-
b. H# qua
Khoi
lugng muoi = Tong khoi luong
cac

ion
tao
muoi.
Khoi
luong dung dich
sau
phan
ung =
Khoi
\ugng
dung dich
truoc
phan
ling
-
(khoi luong
chat
ket tua
+
chat
bay hoi)
vi
DU
MAU
Vi
du 1: Hoa tan
hoan
toan
1,794 gam
kim loai kiem

M vao
400ml
dung dich
HCl
0,1M.
Co can
dung dich
sau
phan ving
thu
du(?c
3,86
chat
ran
khan
Y.
Kim
logi kiem
M la
y,,?j,,
,
A.
Na.
B.
K. C. Rb. D.
Li.
yv.
Laigidi:
, ,
Cac phan ung hoa hoc:

M
+
HCl
>
MCI
+ -Hi -
2
M
+H2O
>
MOH
+ ^Hi
2 .
Nhan
xet:
Bao
toan
khoi
lu^ng
-> my =
mM+
m
J.
+ m^^^.
•
^
m^^. =3,86-1,794-0,04.35,5 = 0,646 gam ^
n^^^.
= 0,038
mol.

1
794
-nM.nHa-n„„.
=
0,078
^M.^.23
(Na).^ ^a:, ,
—> Dap an A.
Vi
2:
Cho
100ml
dung dich H3PO4
a
molA
vao
100ml
dung dich KOH
2M
thu
dugc
dung dich
Y c6
chua
15,44 gam
hon hop muoi. Gia tri
cua a la
A.
0,75.
B.1,00.

C.0,50.
D.
0,80.
Lai
gidi:
Nhan
xet:
Vi
Y
chua
hSn
hop muoi
->
chac
chan
c6
chiia muoi
axit
-> KOH phan ung
het.
Sadophanung: H3PO4
+
KOH
>
Muoi
+
H2O.
, i
Bao
toan

khoi luong:
0,la.98
+ 0,2.56 = 15,44
+
0,2.18 a =
0,80.^•'
' ^^
-> Dap anD.
| . .
dm
nang
fln
luy$n thi
d<ii
hgc 18
chuy§n
H6a
hpc
- Nguygn Van H5i
Vi
3: Dot chay hoan
toan
17,4 gam h6n hop
Mg,
Zn va
Al
trong
khi
O2
(du)

thu
du(?c 30,2 gam hon
hgp
oxit.
The
tich
khi
O2
(dktc)
da
tham
gia
phan
ling
la
A.
4,48
lit.
B. 8,96
lit.
C. 17,92
lit.
D. 11,20
lit.
Lmgidi:
'''''''
6
bai nay, cac em
khong
the

giai
bang vi?c
viet
phuong
trinh
phan
utig
va
datsomol.
:, ,' , , "
O
day,
cac em can ap
dung
bao
toan
khoi
lugng:
mkimM
+
moxi =moxit
->
moxi
= 30,2 -17,4 = 12,8 gam.
-> =^^= 0'4
mol
V02
=
8,96
lit

^
Dap an B.
Vi
dv
4
(A-08): Cho 2,13 gam hon
hgp X
gom Mg, Cu va Al
6
dang
bot tac
dyng
hoan
toan
voi
O2
thu
dugc hon hop Y gom cac
oxit
c6
khoi
luong
3,33
gam.
The
tich
dung
dich
HCl
2M

vira
dii
de
phan
ling
het
voi
Y la
A.
60ml.
B.
45ml.
C.
75ml.
D.
80ml.
Lai
gtat:
Nhan xet: bai nay cac em nhat
thiet
phai
ap
dung
bao
toan
khoi
lugng
de
tinh
kho'i

lugng
oxi
tham
gia
phan
ling:
mkimioai
+
moxi
=
moxit
—>
moxi
= 3,33
—
2,13 = 1,2 gam.
1
2
^
=
TIT
=
0,0375
mol no =
0,075
mol
'
^
V-(oxit)
=

0-075
mol.
; - ^^^^
Khi
cho
oxit
bazo tac
dung
vai
axit
tao ra nuoc:
02-
+
2H^
>
H2O.
Suy ra: n^+ = 0,15
mol
->
nnci
= 0,15
mol
->
VHCI
=
0,075
lit
=
75ml.
-> Dap an C.

Vi
dv
5
(A-12): Hoa tan hoan
toan
2,43 gam hon hgp gom Mg va Zn vao mpt
,
lugng
vira
dii
dung
djch
H2SO4
loang,
sau
phan
ling
thu dugc 1,12 lit H2
(dktc)
va
dung
dich
X
chiia
m gam
muoi.
Gia
tri
ciia
m la

^
A.
4,83
gam.
B. 5,83
gam.
C. 7,33
gam.
D. 7,23 gam.
Lai
giai:
O
bai nay, cac em c6 the
giai
chi
tiet
dua theo
phan
ling
hoa hgc.
Nhan xet: nH2S04 =
r'H2
=
0,05
mol.
So do
phan
ling:
Kimloai
+

H2SO4
>
Muoisunfat
+
H2
Bao
toan
kho'i
lugng
cho so do
tren:
^
6I>''•
->• m = 2,43 +
0,05.98
-
0,05.2
= 7,23 gam
'
ii£0tOB>;
->DapanD.
, ,
^'^ ^^-^^^
' ^
Cty
TNHH
MTV DWH
Khanq
Vigt
Vi

dv 6 (CD-08): Hoa tan het 7,74 gam hon hgp bgt
Mg,
Al
bang 500 ml dung
dich
hon hgp
HCl
IM
va
H2SO4
0,28M thu dugc dung dich X va
8,736
lit khi
H2
(dktc).
Co
can dung djch
X
thu dugc lugng muoi khan
la
A.
38,93
gam.
B.
103,85
gam.
C.
25,95
gam.
D.

77,86
gam.
Lai
giai:
Nhan xet: Bai nay neu cac em giai
theo
phuong trinh phan ling se gap kho khan
vi
can viet den
4
phuong trinh.
Truoc
het,
can xem
axit
c6 phan ling
het
hay
khong
bang each
so
sanh
so
mol H" ban dau
va so
mol khi H2 bay
ra
theo
ti 1$: 2H* >
H2.

nj^+
=
HHCI
+
2nH2S04 = 0,5.1 +
2.0,5.0,28
= 0,78 mol.
, , ,
Mat khac: nu,
= 0,39
mol
-> n^+
het
-> X
chi chiia
cac
muoi.
, '
Khoi
lugng muoi = m^|3+ +
^^^^2+
+ ^^i- +
^^^2-
'
=
7,74 +
0,5.35,5
+
0,14.96
=

38,93
gam. ,;jv^^^y.,_
-¥
Dap an
A.
' > -
Vi
dv
7:
Nung
hon
hgp
bgt
gom
15,2
gam
Cr203 va m
gam
Al
a
nhi?t
dg
cao.
Sau
khi phan ung hoan toan, thu dugc
23,3
gam hon hgp ran
X. Cho X
phan
ling

voi
axit
HCl
(du)
thoat
ra V
lit khi H2
(6
dktc).
Gia
tri ciia
V la
A.
7,84.
B.4,48.
C. 3,36. D.
10,08.
Lai
giai:
Nhan xet: Bai nay
truoc
he't
cac em can tinh dugc so mol
Al
ban dau.
Lim
y
rang,
trong
phan

ung
nhift nhom, thuong
ap
dung dinh luat
bao
toan
khoi lugng, cia
the:
mc^03
+
i"Al=mx
mAi=
23,3 -15,2 = 8,1 gam -> n^r 0,3
mol.
.0
Phuong
trinh hoa hgc: Cr203
+
2A1
> 2Cr +
Al2C)3. (j^p.^v, ^
Mol:
ai 0,2 0,2 0,1 ' '.
X gom:
Al
du = 0,1 mol; Cr = 0,2 mol.
^^-^
3
Khi
X tac

d\ing voi
axit:
Al ^
-H2
va Cr->
H2
-> nH2 = 0,35 mol
VH2=
7,84 lit -» Dap an A.
Vi
d\
8
(B-09): Cho 100ml dung djch
KOH
1,5M vao 200ml dung djch
H3PO4
0,5M, thu dugc dung djch
X. Co c^n X
thu dugc khoi lugng
chat
ran khan
la
A.
15,5 gam. B. 18,2 gam. C. 12,8 gam. D. 16,4 gam.
Lai
giai:
"KOH
= 0,15
mol;
nH3P04

= 0,10 .
Nhan xet: Cac em c6
the xet ti le
mol
KOH
va
H3PO4,
sau do viet 2 phuong
trinh
phan ung
va dat so
mol
de
giai.
Ca'm
nang
6n luyfn thi d^i hpc 18
ctiuySn
dg H6a hpc - Nguygn Van HSi
Tuy nhien,
ne'u
ap
dung
bao
toan kho'i lugng
cho
so
do:
H3PO4
+ KOH >

Muoi
+ H2O
Taco:
mH3P04 +
rnKOH=
niHjO
:
J^y^^yr^^
"H2O=0,15.
,
mx =
0,1.98+ 0,15.56-0,15.18
=
15,5
gam
—>
Dap an A.
Vi
du
9:
Dun
nong
hon hgp
khi
X
gom
0,06
mol C2H2
va 0,04
mol

H2 vol xiic
tac
Ni,
sau
mot
thoi gian
thu
dugc
hon hgp
khi
Y. Dan
toan
bg
Y
Igi
tu tu
qua binh dung dung dich brom
(du) thi
con lai 0,448
lit
h6n hgp
khi
Z (6
dktc) CO
ti
kho'i
so
voi O2
la 0,5.
Khoi lugng binh dung dich brom tang

la
V
-
A.
1,04
gam.
B.
1,32
gam.
C. 1,64
gam.
D.
1,20
gam.
I'
•
Laigiai:
;;(;';
Nhan
xet: Bao
toan khoi lugng:
mY
= mx =
0,06.26
+
0,04.2 = 1,64
gam.
0
448
or,,

Mat
khac:
nz=
= 0,02
mol
va
Mz=
0,5.32
=
16 ,
22,4
mz =
0,02.16
=
0,32
gam.
Luu
y:
Khoi lugng binh brom tang = khoi lugng
cac
hidrocacbon
bi
hap
thy.
Bao
toan khoi lugng,
ta
c6:
Kho'i
lugng binh brom tang

=
mv
-
mz =
1,64
-
0,32 = 1,32
gam.
,.j
.j-^,^,
•^i^^.j^.^
,
Dap an
B.
Vi
du
10
(A-10):
Dun
nong
hon hgp
khi
X
gom
0,02
mol
C2H2
va 0,03
mol H2
trong

mot
binh
kin
(xuc
tac
Ni),
thu
dugc
hon
hgp
khi
Y.
Cho
Y
Igi
tu
tu
vao binh dung djch brom (du),
sau
khi
ke't
thiic
cac
phan
ung,
khoi lugng
binh
tang
m
gam

va
c6
280ml
hon hgp
khi
Z
(dktc)
thoat
ra.
Ti
kho'i
aia
Z so
voi
H2
la
ia08.
Gia
tri ciia
m
la
•
,iaw;.
:
,
%
A.a585.
B.
0,620. C. 0,205.
D.

0,328.
Lbi
gidi:
Nhan
xet: Bao
toan khoi lugng: mv = mx =
0,02.26
+
0,03.2
=
0,58
gam.
,
Mat
khac:
nz=
^
=
0,0125
mol
va
Mz=
10,08.2
=
20,16
22,4
?
mz =
0,0125.20,16
=

0,252
gam.
Nhanthay:
'r^^.^
i
i,*!.'.a
"
;
.m»s;c,ci
;,/•>
•
Khoi
lugng binh brom tang
=
khoi lugng
cac
hidrocacbon
bi hap
thu.
Bao
toan khoi lugng:
m
= mv
-
mz =
0,58
-
0,252
=
0,328

gam.
—>^
Dap anD.
Cty TKi,
1TV
DVVH
Khang
Vi$t
Vi
du
11
(B-12): Hon hgp
X
gom
0,15
mol
vinylaxetilen
va
0,6 mol
H2.
Nung
nong
hon hgp
X
(xiic
tac
Ni)
mot
thoi gian, thu
dugc

hon hgp
Y c6
ti khoi
so
voi
H2
bang
10.
Din hon hgp
Y
qua
dung dich brom du,
sau
khi phan
ung
xay
ra
hoan
toan, khoi lugng brom tham
gia
phan ung
la
• ) ; ,
A.
0
gam.
B. 24
gam.
C. 8
gam.

D.
16
gam.
Lai
gidi:
Bao
toan khoi lugng:
mY
= mx =
0,15.52+ 0,6.2
=
9
gam.
5 ,
Mat
khac:
MY=
10.2
=
20
"2=
=
0/45
mol
" '
^
•
Nhan
xet:
vinylaxetilen

c6
chua
3
lien
ke't
71
—> So'
mol lien
ke't
K
ban
dau
=
3.0,15
=
0,45
mol.
•
,,^.J^:n.
Gia
su so
mol H2
tham
gia
phan umg
= a
mol
So
mol
lien

ke't
n
phan
ling
=
a
mol.
Mat
khac,
so
mol khi giam =
so
mol H2 phan ung = nx-nY
nkfusi'^'^
->
a
=
0,75
-
0,45
=
0,3
mol.
->
So
mol lien
ke't
7i du =
0,45
-

0,3
=
0,15
mol =
So'mol
Br2 phan ung.
Khoi
lugng brom phan ung =
0,15.160
=
24
gam.
,
—>
Dap an
B.
Vi
du
12:
Hon
hgp
X
gom 0,1
mol etilen,
0,2
mol
axetilen
va
0,5
mol

H2.
Nung
nong
hon hgp
X
(xiic
tac
Ni)
mgt
thoi gian,
thu
dugc
hon hgp
Y c6
ti
khoi
so voi
H2
bang
10.
Dan hon hgp
Y
qua
dung dich brom du,
sau
khi phan
ling
xay
ra
hoan

toan, tha'y
c6 m
gam
brom tham
gia
phan ung.
Gia trj
cua
A.
16.
B.32. C.8. D.24.
Laigiai:
"',';
,,s*^->r::
':^:s,rf'^~
Bao
toan khoi lugng:
mY
= mx =
0,1.28
+
0,2.26
+
0,5.2
=
9
gam.
Mat
khac:
MY

=10.2
=
20
nz==
0,45
mol
" '
Nhan
xet:
etilen
c6
chiia
1
lien
ke't
71,
axetilen
c6
chiia
2
lien
ke't
71
-> So
mol
lien
ke't
7t
ban
dau

= 1.0,1
+
2.0,2
=
0,5
mol.
Gia
su so
mol H2
tham
gia
phan
ung
= a
mol
->
So'
mol lien
ke't
n
phan
ling
=
a
mol.
Mat
khac,
so
mol khi giam =
so

mol H2 phan ung
=
nx
-
nY
^ ^
->
a
= =
0,8
-
0,45
=
0,35
mol.
->
So
mol lien
ke't
7t
du =
0,5
-
0,35
=
0,15
mol =
So
mol Br2 phan ung.
Khoi

iugng brom phan ung =
0,15.160
=
24
gam.
—>
Dap
anD.
'
7
dm
nang
On
luygn
thi dgi hpc 18 chuySn dg H6a hpc -
Nguygn
van HJi
Vi
dv 13:
Cho
2,1
gam h6n hgp
X gom
hai amin
(no, don
chiic, dong dang
ke
tiep)
phan ling
het voi

dung dich HCl (du),
thu
du(?c
3,925 gam hon hgp
muoi. Cong thuc cua hai amin trong
X
la
A.
CH3NH2 va C2H5NH2. B.
C2H5NH2
va C3H7NH2.
Y^ic
C.
C3H7NH2
va
C4H9NH2.
i
'>
•
D.
CH3NH2
va
(CH3)3N.
S 0 >.
Lot
gidi:
Nhan
xet:
Loai
D vi

hai amin khong dong dang ke tiep. Cac phuong
an con
lai
deu la amin don chuc, bac I.
Khi
cho
X tac
dung voi
axit
HCl:
R-NH2
+ HCl >
R-NH3CI
'
i<-
Bao
toan
kho'i
lugng: mamin
+
mnci
=
mmuoi
•
>.
' ;
mnci
=
3,925 - 2,1 = 1,825
gam

-»
nnci =
0,05
mol =
namm
^
Mamin
= 7^
=
42 ^
R-NH2 =
42.
0,05
->
R
=
26 -»
Hai
goc
hidrocabon la
CH3-
va
C2H5
*
Dap
an A.
Vi
dij
14:
Dot chay hoan

toan
m
gam hon hgp
X
gom
3
ancol (don chuc,
thugc
Cling
day
dong dang),
thu
dugc
26,4 gam
khi CO2
(dktc)
va 19,8
gam H2O.
Neu
thuc hien phan ling
ete hoa m gam X
(hieu suat
100%) thi
tong
khoi
lugng
ete
thu
dugc
la

A.
8,40
gam. B.
14,80
gam. C.
10,92
gam. D.
12,90
gam.
Lai
gidi:
Nhan
xet: n^jo =
1/1 >
^C02 =
0,6 ^ X
gom
cac
ancol no, mach ho va:
"X =nH20
-rtC02
=1,1-0,6
=
0,5mol
->
n(0)x
=nx =0,5
Bao
toan
kho'i lugng:

mx =
mc
+
niH
+ mo =
0,6.12
+
2,2.1
+
0,5.16
=
17,4
gam.
Trong phan ung
tao ete
thi:
nx
=
Zn^jO
i^H20
=
0,25
mol.
Bao
toan
kho'i lugng:
"^x =
r^ete
+
nHjO

mete
=
17,4 - 0,25.18
=
12,90
gam -> Dap an D.
Vi
du
15
(B-08):
Cho
3,6 gam
axit
cacboxylic
X (no, don
chiic)
tac
dung hoan
toan
voi
500
ml dung dich
gom
KOH
0,12M
va
NaOH
0,12M.
Co
can

dung
dich
thu
dugc
8,28
gam
chat
ran khan. Cong thiic cua
X
la
A.C2H5COOH.
B.CH3COOH.
C.HCOOH.
D.C3H7COOH.
Lai
gidi:
Nhan
xet:
Theo bai,
axit
phan ung hoan
toar\i baza
c6 the con
du,
do
vay neu giai dua vao phuong trinh phan ung se rat kho khan.
Cty
TNHH
MTV DWH Khang
Vijt

Tuy
nhien,
bai nay
dugc
giai nhanh chong
khi ap
dung
bao
toan
khoi
lugng: maxlt
+
mbazo=
mmuol
+
mnuoc
_^
mH20=
3,6+ 0,06.56+ 0,06.40-8,28
=
l,08gam^>
nH20=
0,06
mol.
3
6
Mat khac:
nH,o=
Max.t
=

=
60
Axit la
CH3COOH
^
0,06
•
.
.
V
-> Dap an B.
Vi
du 16: Dot
chay hoan
toan
hon hgp X gom hai
este
dong phan
can
diing
7,84
lit khi O2, thu
dugc
6,72
lit khi
CO2
va
5,4
gam H2O. Neu cho
m

gam
X
tac
dung
vua
dii
voi
dung dich
NaOH,
c6 can
dung dich
sau
phan ung
thi
thu
dugc
7,85
gam
chat
ran khan
Y va
hon hgp ancol Z. Thyc hi|n phan
ling
ete hoa
hoan
toan
Z, thu
dugc
m gam ete.
Cac

the
tich khi
deu do 6
dieu
ki?n
tieu chuan. Gia tri cua
m
la
^
A.
2,65.
B.3,70.
C.
5,15. ' " " D. 3,25.
Lai
gidi:
n
02
=
0,35
mol;
n
CO2
=
0,3
mol; n =
0,3
mol.
Nhan
xet:

UQQ^
=
nH20
^^^^
este
deu no, dan
chuc
->
Cong thiic
cua 2
este
la CnH2n02.
* ' • "' ••.:
Bao
toan
nguyen
to'oxi:
2n
^^te
2n
=
2n
CO2
^
H2O
~^ "
este
0,1
mol.
_>

n
=
^^^22.
= = 3 2
este
trong
X c6
cong thuc phan
tu
la
C3H6O2.
Heste
0,1
-> Cong thuc cau
tao 2
este
la:
HCOOC2H5
va
CH3COOCH3.
Phan ling hoa hgc:
HCOOC2H5
+
NaOH
—^
HCOONa
+
C2H5OH
CH3COOCH3
+

NaOH
—!—>
CHsCOONa
+
CH3OH
. '
De
tha'y:
nNaOH
=
neste
=
nz
=
0,2
mol.
^
Bao
toan
kho'i lugng: meste+mNaOH = mv
+
mz
->
0,1.74
+
0,1.40 =7,85 + mz ^ mz =3,55
gam.
Ht:k{OCC;i
Nhan
xet:

Trong phan ung
ete
hoa
ta c6:
1
„ , n
,mw>'^''iOfU
nAoi
•
i
H2O
=
T
nzn
H20
=
0,05
mol.
2
; y
7
.
Bao
toan
khoi lugng: mz = mete
+
m
H20
-> m = mete
=

3,55 - 0,05.18
=
2,65
gam
->
Dap an A.
,s :sv
Vi
d^ 17: Xa
phong
hoa
hoan
toan
0,1
mol
este
X (dan
chiic, mach
ho)
bang
100
g^m dung dich MOH
11,2%
(M la
kirn
loai kiem). Co can dung dich sau
phan ling,
thu
dugc
15,4 gam

chat
ran khan, dong thai ngung
tu
phan hoi
bay
ra
tha'y
tao
thanh
92
gam
chat
long. Cong thiic cua
X
la
9
Ca'm
nang
On
luygn
thi dgi hpc 18
chuyen
dg H6a hpc -
Nguygn
VSn Hii
A.
CH3COOCH3.
B.
CH3COOC3H7.4
C. CH3COOC2H5.

D.
C2H5COOC2H5.
Lot
gidi:
Nhan
xet:
Khoi
lugng MOH bang
11,2 gam " " ' '' •
->
kho'i
luang dung moi
(nuac)
= 100 - 11,2 = 88,8
gam.
a
rif/vi
•«
gfi;
Phan ung
hoa hoc:
iv;y£;f{;-^,';f0<;j;
^
rfsB<.m
RCOOR'
+ MOH
)• RCOOM
+ R'OH fti fO !d>i Oi
S^'- Mol: 0,1 0,1 0,1 0,1 ^
N/ian

f/ifli/:
Cha't long
sau
khi ngung
tu gom
nuoc
va
ancol
nen:
mancoi
= 93,4 -
mnuoc
= 92 - 88,8 = 3,2 gam.
^MROH
=
32^>R'
=
15(R'lanh6mCH3-)".''*^'
^ .„
Bao
toan
khoi
lugng:
meste
= 15,4
+
3,2 - 11,2 = 7,4 gam
^
Meste
= 74RCOOCHs = 74 ^ R la CH3-

^
Este la
CH3COOCH3
^ Dap an A.
Vi
dv 18: Cho 700 gam
cha't
beo c6 chi so
axit
la 8 tac
dung voi dung dich KOH
du,
sau
phan
ung,
khoi
lugng muo'i
thu
dugc
la 764,6 gam.
Kho'i lugng
KOH
da
tham
gia
phan ung
la
A.
140
gam.

B. 175
gam.
C. 280
gam.
D. 350
gam.
Laigiai:
Chi
so
axit
cua
chat
beo = ^"^^^ ^ 8== ^"^g^
rn^hatbeoCg)
700
m^oH ^
5600 mg = 5,6 gam nmn = 0,1
mol.
Phan
ling
hoa hoc: '
' RCOOH
+ KOH >
RCOOK
+ H2O
Mol:
0,1 0,1
*
(RCOO)3C3H.s
+

3KOH
>
3RCOOK
+
C3H5(OH)3
Mol:
3x -> 3x
Bao
toan
kho'i
lugng: mchat beo
+
mKOH
=
mmuoi
+
mgUxeroi +
m
'
-> 700 + (5,6 + 56.3x) = 764,6 + 92.x +
0,1.18
->
x = 0,8 mol
mKOH
= (0,1 +
0,8.3).56
= 140 gam.
Dap
an A.
Vi

19
(B-08):
Xa
phong
hoa
hoan toan
17,24 gam
chat
beo can
vvra
du
0,06 mol
NaOH.
Co can
dung dich
sau
phan
ung thu
dugc
m gam xa
^ phong.
Gia
tri ciia
m la
-'"^^^
*
A.
16,68. B. 18,24. C. 17,80. D. 18,38.
Cty
TIMHH

MTV
DVVH
Khang
Vigt
Lmgidi:
Ggi
cong thuc
cua
chat
beo la
(RCOO)3C3H5.
,
<
'i-b
tfW 1,. i,
Khi
cho
chat
beo tac
dung voi NaOH:
(RCOO)3C3Hs
+
3NaOH
>
3RCOONa
+
C3Hs(OH)3
Mol:
0,06 0,02
Ap

dung
bao
toan
kho'i
lugng:
mch.nt
beo +
mN.nOH =
mxa
phong
(muoi)
+
mgiixeroi
->
mx.i
phong
= 17,24 + 0,06.40 - 0,02.92 = 17,8 gam Dap an C. ' * "• '
Vi
dv 20: Xa
phong
hoa
hoan toan
m gam mot
triglixerit
bang KOH
thu
dugc
0,92 gam
glixerol
va 9,58 gam hon hgp

muo'i
cua
axit linoleic
va
axit oleic.
Gia tri
cua m la
A.
9,94. B. 9,38. C. 10,50. ' " D. 8,82. :k'^'
Laigiai:
VinV
ngiixe((ji
= 0,01 mol •>• m. :,3f' !
aijfi
(RCOO)3C3H5
+
3KOH
>
3RCOOK
+
C3H5(OH)3
Mol:
0,03 <- 0,01
Bao
toan
khoi
lugng:
m + 0,03.56 = 9,58 + 0,92 -> m = 8,82 gam.
->
Dap an D.

Vi
du 21:
Aminoaxit
X
chua
mgt
nhom -NH2.
Cho 10,3 gam X tac
dung voi axit
HCl
(du),
thu
dugc
13,95 gam
muo'i khan. Cong thuc
ca'u tao thu ggn cua X
A.
CH3CH2CH(NH2)COOH.
B.
H2NCH2CH2COOH.
C. CH3CH(NH2)COOH.
D.
H2NCH2COOH.
Laigiai:
Nhan
xet: Cac dap an deu cho
aminoaxit chua
mgt
nhom -COOH
ggi

cong thuc
cua X la
H2N-R-CC)OH.
-
j||||vi.^^^
^g,;
>
Khi
cho X tac
dung voi axit HCl:
f'^^,
H2N-R-COOH
+ HCl
CIH3N-R-COOH
.CJ^jinV
Bao
toan
kho'i
lugng:
mammoaxit
+
mHci
=
mmuoi
-> mHci=
13,95 - 10,3 = 3,65 gam ->
UHCI
=
0/1
=

^
Mx = — = 103 ^-
H2N-R-COOH
= 103. - '^-'^
0,1 •
••••vv,;::fv;„
^
R = 42
—>•
Goc
hidrocacbon
la
-C3H6-
Dap an A.
Vi
du 22: Dun
nong
m gam hon hgp gom a mol
tetrapeptit mach
ho X va 2a
mol
tripeptit
mach
ho Y voi
600ml dung djch NaOH
IM (vua
du).
Sau khi
cac
phan

ung ke't
thiic,
c6 can
dung dich
thu
dugc
72,48 gam
muoi khan
cua
c^c
amino axit
deu c6 mgt
nhom -COOH
va mgt
nhom
-NH2
trong
phan tu.
Gia
tri
aia m la
A.
51,72. B. 54,30. C. 66,00. , D. 44,48.
•
• •
• • 11
dm
nang On luy?n thi dgi hpc 18 chuySn
6i
H6a hpc

-
Nguygn Van
HJi
Lai
gidi:
Nhan
xet:
Khi
cho cac
peptit
tac
dung
voi
NaOH
thi
c6 cac
lien
ket
peptit
va
nhom
-COOH
(cua
amino axit dau C) tham gia phan ung.
Theo bai,
a
mol
tetrapeptit
mach
ha X c6 a.3 = 3a

mol
lien
ket peptit.
2a mol tripeptit mach
ho Y c6 2a.2 = 4a
mol
lien
ket peptit
Tong
so'mol nhom cacboxyl-COOH bang
a + 2a =3a.
Cac
phan
ling
nit ggn
cua
lien
ke't peptit
va
nhom
-COOH
la:
'
'yinit
-CO-NH-
+
NaOH
>
-COONa
+

H2N-
,
•
nnMQh
u 7a 7a
-COOH
+
NaOH
>
-COONa
+
H2O
, ^
Mol:
3a 3a 3a M' A
Theo
bai:
nwaOH
= 7a + 3a = 10a = 0,6 —> a = 0,06 mol.
Bao
toan
khoi
lugng:
m +
mNaOH
=
mmuai
+
m '
i( >.:-i(jin

->
m = 72,48 + 0,06.3.18 - 0,6.40 = 51,72 gam
Dap
an A.
. ,,
Vi
d\
23:
Dun nong
m gam
hon hop
gom a
mol dipeptit mach
ho X va 2a
mol
tripeptit mach
ho Y voi
400ml
dung djch HCl
IM
(vira
du). Sau khi cac
phan ung
ket
thuc,
c6 can
dung dich thu dugc
48,1 gam
muoi khan
cua cac

amino axit
deu c6 mpt
nhom
-COOH
va mpt
nhom
-NH2
trong phan
ttr.
Gia
tri
ciia
m la
A.
33,5.
B.
29,0.
C.
30,8.
D.
28,1.
'
' , Lai
gidi:
Nhan
xet:
Khi
cho cac
peptit
tac

dung
voi
HCl
thi c6 cac
lien
ket
peptit
va
nhom
H2N-
(ciia amino axit dau
N)
tham
gia
phan ung.
Theo bai,
a
mol dipeptit mach
ho X c6 a.l = a
mol
lien
ket peptit.
2a mol tripeptit mach
ho Y c6 2a.2 = 4a
mol
lien
ket peptit
>
Tong
so

mol nhom amino
H2N-
bang
a
+
2a = 3a. '
f
f
Cac
phan ung riit gpn ciia
lien
ket peptit
va
nhom
H2N-
la:
,
-CO-NH-
+
HCl
+
H2O
) COOH
+
CIH3N-
Mol:
5a 5a 5a
H2N-
+ HCl >
CIH3N-

Mol:
3a 3a
Theo bai:
nHci
= 5a + 3a = 8a = 0,4 —> a = 0,05
mol.
V'
Bao
toan
khoi
lupng:
m +
mHci
+
m =
mmuoi
m
= 48,1 - 0,4.36,5 - 0,25.18
=
29
gam.
—> Dap an B.
10
'-vi'
./\
Cty
TNHH MTV DWH Khang Vi^t
./.5„
A I,I
2.

PHLfONG
PHAP BAG
TOAN
NGUYEN
TO
a. NQidung
Trong
cac
phan ung hoa hpc,
cac
nguyen
to
luon dupe
bao
toan.
b, H? qua
Tong
so
mol nguyen
tu
ciia
mpt
nguyen
to c6
trong
cac
chat
truoc
phan ung
va

sau phan ung luon bang nhau. o,^, r'
LKU
y:
Can
xac
dinh diing
va day
dii
cac
chat
c6
chiia nguyen
to
dang
xet 6
truoc
va
sau phan
ling.
, ,
VIDVMAU
Vi
du 1
(A-08): Hoa
tan
hoan
toan
0,3
mol hon hpp
gom

Al
va
AUCs
vao
dung
dich
KOH (du),
thu
dupe
a mol hon hpp
khi
va
dung dich
X.
Sue khi
CO2
(du)
vao X,
lupng
ket tua
thu dupe la
46,8
gam. Gia
tri cua a la
A.
0,55.
B.0,60.
C.0,40.
D.
0,45.

Lai
gidi:
Cach
1: 6
bai nay,
cac em c6 the
giai dua
vao cac
phuong trinh phan ung:
Al
+ KOH +
H2O
>
KAIO2
+ •|H2
AI4C3+
4KOH
+
4H2O
>
4KA102
+
3CH4
rH-ff-+
cllV^-:
KAIO2
+ CO2
+
2H2O
>

Al(OH)3
+
KHCO3
, . ,
Cach
2: So do
phan ung:^
.^^j^,.
Al,
AI4C3
)
KAIO2
^^02^^20
^
Al(OH)3
jw, ,0-!:
Bao
toan
nguyen
to
Al:
^,
"Al
+4nAi4C3
=
I^Al(OH)3
+
^^A\4C3=
Theo bai: n^i
+nAi4C3

=
0'3
nAi
=0,2;
nAi4C3
= 0'l-
'
Vay: a =
|nAi
+
3nAi4C3
= ^,6 Dap an B.
Vi
dv 2:
Hoa
tan
hoan
toan
hon hpp
gom 0,12
mol
FeS2
va a
mol
CU2S
vao
axit
HNO3
(vua
dii),

thu
dupe dung djch
X (chi
chiia
hai
muoi sunfat)
va
khi
duy nhat NO. Gia
tri cua a la
A.
0,075.
B.0,12.
C.0,06.
"
D.0,04.
'
Lai
gidi:
Nhan
xet:
Cac
em
luu
y la
dung djch
X
chi chua hai muoi sunfat
-> sau cac
phan ung,

S
nam
het a
dang
goc
sunfat.
Ta
CO
cac so do
chuyen hoa:
; i
^
FeS2
>
iFe2(S04)3
i'ouU-,,r!>!
•
.
•
Mol: 0,12 0,06
'ih.^1bm^l^^.,
:
^;
].,>f:v.,
rv-;v:-'-
C^m nang
6n
luygn
Ihi dgi hpc 18
chuyfin

ii Hoa hpc -
Nguyin
van Hi!
.
rt:.
Cu2S
>
2CuS04
Mol- a 2a , ^^^^ , .
j,|,|,j3i);,
ii^,
Bao
toan
nguyen
to
S, ta
c6:
2
npeSj
+
"CuzS
= "504
-
H
1.
a
->
0,12.2 + a
=
0,06.3

+ 2a -> a
=
0,06 ->
Dap
an
C.
Vi
du
3
(A-12):
Cho
18,4
gam hon hop
X
gom Cu2S, CuS, FeS2
va
FeS
tac
dung
het
voi
HNO3
(dac
nong,
du) thu
duoc
V h't
khi chi
c6 NO2 (6
dktc,

san
pham khu
duy
nha't)
va
dung dich
Y. Cho
toan
bo Y vao mot
lugng
du
dung dich BaCl2,
thu
dugc
46,6 gam ket
tiia;
con
khi
cho
toan
bp Y tac
dung voi dung dich NH3
du
thu
dugc
10,7
gam ket tua. Gia
tri
cua
V

la
,^
A.
38,08.
B.
24,64.
C.
16,8.
D.
11,2.
NMn xet: Dung dich
Y
chiia
cac
ion: Fe^
Cu^^
SO4 , va NO3.
Khi
cho
dung dich
Y
+ dung dich BaCh:
Ba2*
+
SO^"
—^
BaS04i
/'1 v >
Bao
toan

nguyen to'S: ng
(X)
= nBaS04 ^
mol. ,
Khi
cho
Y
+ dung djch NHa du:
Fe3*
+
3NH3
+
3H2O
>
Fe(OH)3i
Cu2*
+
2NH3
+
2H2O
>
Cu(OH)2i
Luu y: Cu(OH)2
tan
trong
NH3 du
tao
thanh phuc
chat:
Cu(OH)2

+
4NH3
—^
[Cu(NH3)4](OH)2
"
10 7 '
Bao
toan
nguyen
to
Fe: npe
(X)= nFe(OH)3
=
Bao
toan
khoi
lugng:
mx
= m^u + mpg + mg
-» mcu
= 18,4 - 0,1.56 - 0,2.32
=
6,4
gam
->
ncu
(X)=
—
=
0'^

i"ol-
Qui
doi
X ve
hon hgp gom
cac
don
chat:
Fe =
0,1
mol; Cu =
0,1
mol
va
S =
0,2
mol.
A
iKt/tij:
-» ne
(X)
= 2ncu + Snpe + 6ns =
1,7
mol
n^o^ = (x)
=
1,7
mol
->
V

=
1,7.22,4
=
38,08
lit
^
Dap an A.
Vi
dv 4
(B-10):
Mot loai phan
supephotphat
kep
c6
chua
69,62%
Ca(H2P04)2,
con
lai
gom cac
chat
khong chiia
photpho.
Dp dinh duong ciia loai phan Ian nay la
A.
48,52%.
B.
39,76%.
C.
42,25%.

D.
45,75%.
Loigidi:
6
day,
cac em can
nho
la
dp dinh duong
cua
phan
supephotphat
dupe tinh
I
theo
% khoi lupng cua P2O5.
•;•
NMn
xet:
1 mol
Ca(H2P04)2
hay
1
mol
P2O5 deu
chua
2
mol
P.,»,u
MM.

mm
Cty
TNHH
MTV DWH
Khang Vi$t
Bao
toan
nguyen
to P
theo
so do:
Ca(H2P04)2
< >
P2O5
^
i, n
<ti
uis 1' , '
Khoi
lupng mol:
234
gam
142
gam
o.
'"'!(,
,
%
khoi lupng:
69,62% x%

69,62.142 vr^' '
—>x=—
=
42,25 ->
Dap
an
C.
,.
Vi
dvi
5
(A-12):
Mot
loai phan
kali
c6
thanh phan chinh
la
KCl
(con lai la cac
tap
chat
khong chua kali) dupe
san
xua't
tu
quang xinvinit
c6 dp
dinh
duong 55%. Phan

tram
khoi lupng ciia
KCl
trong
loai phan
kali
do
la
A.
95,51%.
B.
65,75%.
C.
87,18%.
D.
88,52%.
Loigidi:
<
*=
-
6 day, cac em can nho la dp
dinh duong
cua
phan
kali
dupe tinh
theo
%
khoi
lupng cua

K2O.
„,,
Ap
dung
bao
toan
nguyen to'K,
ta
CO
so do:
KCl
<—>
-!-K20
v;
V
Kho'i
lupng mol:
74,5
gam
47
gam
>
%
khoi lupng:
x% <- 55%
,,!'
r .
^
X =
=

87,18 ->
Dap an C.
* ' ' '
47
^
Vi
dy 6:
Dot
5,6 gam Fe
trong
khong khi,
thu
dupe hon hpp
chat
ran
X.
Cho
X
tac dung voi dung dich
FiNOs
loang (du),
thu
dupe khi NO (san pham khu
duy
nha't)
va
dung djch chua
m
gam muo'i. Gia
trj

cua
m la
A.
18,0.
B.22,4.
C.
15,6.
D.
24,2. ;
npg=0,lmol.
Nhan
xet:
Bai nay
ne'u
dua
theo
phuong trinh phan
ung se rat dai
dong
va
ton nhieu thoi gian.
O
day,
cac
em can su dung
so do
phan ung:
Fe
> X
(Fe, FeO,

Fe203,
Fe304)
>
Fe(N03)3
va
ap
dung dinh luat bao
toan
nguyen
to'Fe:
;
j
,0
Fe > Fe(N03)3
„ ,
.0 f:5
Mol:
0,1 0,1
m
=0,1.242
=
24,2
gam
^
Dap an D.
Vi
du
7:
Cho
31,2 gam

hon hpp
gom
Al, Cu
va Ag tac
dung vua dii voi
900ml
dung dich
HNO3
1,5M, thu
dupe dung djch chua
m gam
muoi
va 4,48 lit
hon hpp khi
X
(dktc)
gom
NO
va
N2O.
Ti khoi ciia
X so
voi H2
la 16,75.
Gia
tri ciia
m la ' ' '
A.
98,3.
B.97,2.

C.
96,3.
D.
91,0.
Ca'm
nang
fln
luygn
thi dji hgc 18
chuy6n
dg H6a hpc -
Nguyin
Van Hai
bai nay, truoc het cac em can tim so' mol moi khi trong X de thu dugc ket
qua:
nNo=
mol; n^jo"^ O'OS i^ol-
Al,Cu,Ag
) Muoinitrat + NO + N2O „ /, / '
Chatoxihoa: N^^ + 3e > NO;
2N*5
+ ge > N2O
va
CO
the xay ra ca qua
trinh:
ZN*' + 8e >
NH4NO3
(a mol)
Khi

cho kim loai + HNOs:
n^^Q-(muoi)
= ng
trao
J6i = 3 nfyio + 8
nivj20
+ 8 nfvjH^NOs ~ 0'^^
Bao toan nguyen to N: nHNOa = + "NO + 2
nN20
+
2nNH4N03
0,9.1,5
= 0,85 + 8a + 0,15 +
2.0,05
+ 2a ^ a =
0,025
mol. ' iTs-> K!)
Bao toan khoi lugng: m =
m^i,
cu, Ag + m^^^ +
mNH4N03
'
= 31,2 + (0,85 +
8.0,025).62
+
0,025.80
= 98,3 gam.
-> Dap an A.
NMn
xet: Bai nay da "giau di" san pham

NH4NO3.
Vi
du 8: Hoa tan het 7,8 gam hon hgp gom Al va
AI2O3
bang dung dich HCl
(du),
thu dugc V lit khi H2 (dktc) va dung djch X. Nho dung dich NHs du
vao X, IQC ket tua va dem nung den kho'i lugng khong doi thu dugc
10,2 gam
chat
ran. Gia tri cua V la
A.
2,24. B. 3,36. C. 5,60. D. 4,48. '
i';
' .
Ldigidi:
G<?i
so mol: Al = x;
AI2O3
= y. Ta c6: 27x + 102y = 7,8. ' 'Atii ./v
So do phan ung: ^
,
AlAl^Og
)AlCl3
"NH3.H2O
> Ai(OH)3_'%
AI2O3
Bao toan nguyen to
Al:
x + 2y =

2nAi203
-> x + 2y = 2.^^ = 0,2.
X
= 0,1; y = 0,05 nH2 = -HAI = 0,15 mol ^ V = 3,36
lit.
3
2'
Dap an B. ' '
Vi
dv 9
(B-12):
Nung nong 46,6 gam hon hgp gom Al va
Cr203
(trong dieu
J
kifn
khong c6 khong khi) den khi phan ung xay ra hoan toan. Chia hon
I
hgp thu dugc sau phan ung thanh hai phan bang nhau.
Phan
mot phan
i
ung vua du voi 300 ml dung djch NaOH IM (loang). De hoa tan het phan
hai
can vira du dung dich chiia a mol HCl. Gia trj cua a la
A.
0,9. B. 1,3.
C.0,5.
D. 1,5.
Cty

TNHH
MTV DWH
Khang
Vi§t
Laigidi:
hlhan
xet: Khi cho phan 1 tac dyng voi NaOH, tat ca Al va
AI2O3
deu tac
dyng
va chuyen thanh
NaA102.
Do vay, bao toan nguyen to
Al,
ta c6:
nAi=
njvjaOH
= 0/3
niol.
_^ Ban dau:
nAi
= 2.0,3 = 0,6 mol
"€1203=
^^^-r^^^=
0,2 mol. .
Cr203
+ 2A1 —^
AI2O3
+ 2Cr , ,
Mol:

0,2 ^
a4:r,^
,R>0,2;;
OA : ^ ,
Cac
chat
trong phan 2: n^i = 0,1 mol;
ncr=
0,2 mol;
0^1203
= 0,1 mol.
->
HHCI
=3nAi
+2ncr+6nAi203
= l'3mol. ^DapanB.
W dv 10: Hoa tan hoan toan 8,16 gam hon hgp gom
Fe304
va
FeS2
trong dung
dich
axit HNO3 (dac, du), thu dugc
4,032
lit khi NO2 (dktc) va dung djch X.
Cho X tac dung voi dung dich
Ba(OH)2
du, Igc ket tua va nung trong khong
khi
den khoi lugng khong doi thu dugc m gam

cha't
ran. Gia trj cua m la
A.
12,66.
B.
11,06.
C.
11,86.
D.
20,66.
Laigiduii
^ ' ' |>,s£' V - '
= ^= 0,18 mol. vM^b n^B .
22,4 ,
Ggi
so mol:
Fe304
(a mol) va
FeS2
(b mol). Ta c6:
232a
+ 120b = 8,16.
Cac phan
ling
khvr:
Fe304
- le > 3Fe^3 *~ - -
,0''
^
FeS2-15e

> Fe^^ + 2S^ - li^a.W/
Bao toan electron: . . '',;t»r,'
"NOz =
nFe304
+15"FeSz
a + 15b = ai8 -» a = a03; b = aOl. '[ _
Bao toan nguyen to Fe va S: U m
fiiJ5
ifi BfO > i
Fe304
>
^Fe203
-8
2
Mol:
0,03
0,045
808,£
1
( \.1.0 »
FeS2
>
-Fe203
+
2BaS04
2
Mol:
0,01
0,005
0,02

-> m =
160.0,05
+
0,02.233
= 12,66 -> Dap an A.
Vi
11
(A-08):
Hon hgp X gom propan, propen va propin. Ti khoi cua X so
voi
H2 Ja 21,2. Khi dot
chay
hoan toan 0,1 mol hon hgp X, tong khoi lugng
cuaCCh
va H2O thu dugc la .
A.
20,40
gam. B. 18,60 SB^JHUZ^Ji^^^j^^^ff^fpf^^
Ca'm
nang
On
luyjn
thi dgi hgc 18
chuyfin
dg H6a hgc -
Nguygn
Van H5i
Laigiai:
NMn
xet:

Cac em can
nh|in
ra cac
chat
trong
hon hop X deu c6
chiia
3
nguyen
tu
cacbon.
Nhu
vay, khi do't
chay
0,1
mol
X
thu
dugc
0,3
mol
CO2
-> nc = 0,3
mol.
Mat
khac:
Mx = 21,2.2 = 42,2 ->
mx
=
42,4.0,1

= 4,24
gam.
' " ' '"*V
'
Bao
toan
kho'i
lugng,
ta c6:
mx
= mc +
mn.
'>rn
6,(J
" f'
>
. - ^
1
/•-(.'"
M^'
«f^
'
->
mH
= 4,24 - 0,3.12 = 0,64
gam
^
nn
= 0,64
mol -> nn^o

= 0/32
mol.
Tong
khoi
lugng
CO2
va
H2O
bang
0,3.44 + 0,32.18 = 18,96
gam.
-> Dap
an C.
Vi
d\ 12:
Do't
chay
hoan toan
mpt the
tich khi thien nhien
gom
metan, etan,
propan
bang
oxi
khong khi (trong khong khi,
oxi
chiem
20% the
tich),

thu
dugc
7,84
lit khi
CO2
(a
dktc)
va 9,9 gam
H2O.
The
tich khong khi
(a
dktc)
.X
nho nhat
can
diing
de
dot
chay
hoan toan luong khi thien nhien tren
la
:gfi A.
70,0
lit.
ir
sv B B. 78,4
lit.
C. 84,0
lit.

D.
56,0
lit.
Ij
Lmgidi:
>
100
•JIA
«;:.«,)
jfW
nc02
=-^=035
mol; nH2O=^=0'55 mol.
Nhan
xet:
Ban
dau, nguyen to' oxi
0
dang
O2
tu
do,
con sau
phan
ling
chay
thi
chuyen he't
vao
CO2

va
H2O.
Bao toan nguyen
to
O,
ta c6:
2 noj =
2
TYQQ^
+
n^jjo
•
= ^'^^-^^^'^^
= 0,625
mol
VQJ
= 0,625.22,4 = 14,0
lit.
Vay:
Vkhongkhi
=5Vo2
= 70,0
litDap
an
A.
s at w
;.::;;;»•*
Vi
du 13
(A-10): Dot

chay
hoan toan
m gam
hon hgp
X
gom
ba
ancol don chiic,
thupc cung
day
dong dang,
thu
dugc
3,808 lit khi
CO2
(dktc)
va 5,4 gam
H2O.
Gia
tri ciia
m la
A.
5,42. B.5,72. C. 4,72., D. 7,42.
Laigiai:
' ^
ri
one K
A
.
.it •

•
;loM
•
nc02
=^= 0^17
mol; nH20=^=
0,30
mol.
//,4
10 .„ 4 ••
Cac
em can
thay rang, khi
dot
chay
X:
n^^^Q
>
UQQ.^
-> X
chiia
3
ancol
no
nx
=
nj^^Q
- n^Q^ = 0,30 - 0,17
=
0,13

mol.
Do
X
chua
cac
ancol
dan
chiic
^ no =
noH
= nx no = 0,13
mol.
5'
j Bao
toan
khoi
lugng trong
X, ta c6:
mx=mc
+ mH + mo = 0,17.12 + 0.60.1 + 0,13.16 = 4,72 gam
->'Dap
an
C.M/»tj;;,
t'-vs'jji'ii
lii^^
i, >., t-
Cty TNHH
MTV DWH
Khang ViQt
Vi

14
(B-10):
Do't
chay
hoan toan mpt lupng hon hgip
X
gom
hai
ancol
( deu
no,
da
chiic, mach
ho, c6
cung
so
nhom -OH)
can
vira
du V
lit khi
O2,
thu
dupe
11,2 lit
khi
CO2
va 12,6 gam
H2O
(cac the

tich khi
do a
dktc).
Gia trj
ciia
V la i
A.
14,56. B. 15,68. C. 11,20. D. 4,48.
Laigiai:
11'2
, 12,6 ,
"CO2=^=0'5"^ol;nH2O=^=0,7mol.
^
Theo
bai ra,
X
chiia
2
ancol
no ->
nx
=
nH20
"
^COi
^ ^,2
mol.
-> So'nguyen tvr
C
trung

binh
= = 2,5 -> X
chua mpt ancol
da
chuc
c6
so nguyen tu
C
nho hon
2,5
-> ancol
do la
C2H4(OH)2.
Do
X
chua hai ancol cung so' nhom
chuc
(hai chiic)
-> no =
noH
= 2nx
^ no =
0,4
mol.
Bao toan nguyen
to' O,
ta c6:
no
(OH)
+

2
no^ =
2
TXQQ.^
+
nyi^^Q
2.0,5
+ 0,7-0,4 ,
^^i^rW
'
no2=—
^ ^ = 0,65
mol
^ V02
= 0,65.22,4 = 14,56
lit
^
Dap
an A.
Vi
dy 15:
Hon hpp
X gom hai
axit cacboxylic
don
chuc.
Do't
chay
hoan toan
0,1 mol

X can 0,24
mol
O2,
thu dupe
CO2
va 0,2
mol
H2O.
Cong thiic
hai
axit
la •
timA
•
A.
HCOOH
va
C2H5COOH.
B. CH2=CHCOOH
va
CH2=C(CH3)COOH.
' '
C.
CHsCOOH
va
C2H5COOH.
D.
CH3COOH
va
CH2=CHCOOH.

Laigiai:
H!"\,0-,.
Bao toan nguyen
to O, ta c6: 2nx + 2
no,
= 2
nco2
"H2O
2.0,1
+ 2.0,24-0,2 ,
^C02=—
Y ^ =
0'24mol.
Nhan
xet:
rxyi^Q
<
rxQOj
Hon hpp
X
chua
it
nhat mpt axit khong
no ->
Logii
A va C.
-»
So
nguyen
tu C

trung
binh
=
"^^^
= 2,4 -> X
chua
mpt
axit
c6 so
nguyen tu
C
nho hon
2,4 ->
Lo^i
B
(Hai axit deu chua
so'cacbon
> 2,4).
—> Dap anD.
19
Ca'm
nang
6n
luy$n
thi dgi hpc 18
chuyen
dg H6a
hgc
-
IMguySn

VSn
H5i
Vi
dv
16
(A-12): Hon hg-p M gom mpt anken va hai amin no, don
chuc,
mach
"
h6 X
va
Y la
dong dang ke tiep (Mx <
MY).
Do't
chay
hoan toan mot lugng
r
M can dung
4,536
lit
O2
(dktc) thu
dugc
H2O,
N2
va 2,24 lit
CO2
(dktc). Chat
Yla

A.
Etylamin.
'' [
'2
:
•'XlS:',U,'0
B.
Propylamin.
-
,(•)
.A
C. Butylamin.
. ,
D. Etylmetylamin.
'!
•
.'' . . ,
Laigidi:
n
.i
. ., .
ii t»
4536
224
n ;fom
t;,i
-
no, =^^=
0,2025
mol; nro, =—= 0,1 mol.

.
•
•
^
22,4 ^2 22,4
£Dtjd3X,P'!J6do'?riT
, Bao toan nguyen to'O, ta
c6:
6: 2
no2
= 2
ncoj
+
"HJO
^HJO
=
0/205
mol.
Nh^n
thay, khi dot
chay
1
mol anken thu
du(?c
XXQQ^
=
nH20' vol hai
amin
no, don chijfc thi nHjO"'^C02
1'^

ph"^°''^8
^""^^'^'^"'^S-
CxH2x.3N
+ O2 —^
XCO2
+
(x+l,5)H20
+
0,5N2 ^
'
0,205-0,1
sot
os8
Dovay: nH20"'^C02
=l/5namin
->
namin=—^
'— =0,07mol.
-> So'nguyen tu C trung binh trong
M
=—^^<—^22_=
~
1^43.
Y
"M
I^aniin
0,07
f,
-> M
chua

mQt
chat
c6
so' nguyen tu
C
nho hon 1,43 -> do phai
la
amin
ii
CH3NH2
(X). O day cac em can luu y: anken
chua
tir 2 nguyen tu C tra len!
->
Amin
ke tiep la
C2H5NH2
(Y) Dap an A.
Vi
d\
17:
Hon hgp
X
gom mpt anken
va
hai ancol (no, don chiic, mach ho).
Dot
chay
hoan toan mpt lugng
X

can vua du V lit khi
O2,
thu duQC 15,68 lit
'' khi
CO2
va 18 gam
H2O.
Cac the tich do
6
dktc. Gia tri cua V la
•
*^'
•
A.
22,40.
B.
17,92.
C.
16,24.
D.
23,52.
nco2
=
0,7mol;
nH20=l
mol.
mt?iV6J
1
NMn
xet: Khi

dot chay anken thu
duQC
UQQ^
=
XIH^Q
,
con khi
dot
chay ancol
"ancol
=
"H2O
~
^C02
•
Do v^y:
nancol
=
"HZO
- ^coz = 1 - 0,7 = 0,3
mol.
, „,.
Do ancol don chuc nen
UQ (ancon
^
"ancol
=0,3
mol.
Bao
toan

nguyen
to O:
n
o
(ancol)
+
2no2
=
2nco2
+
nH20
2.0,7 + 1-0,3
,
-> HQ^
= =
1,05
mol
2
V02
=
1,05.22,4
=
23,52
lit
^
Dap an D.
Cty
TNHH
MTV DWH
Khang

Vi$t
Vi
dv 18: Hon hgp X gom hai amino axit no (chi
c6
nhom
chuc
-COOH va
-NH2
trong
phan tu), trong do ti 1§ mo
:
mN
=
80:21.
De
tac
d\ing vvra du voi
3,83
gam hon hgp
X
can 30ml dung dich HCl IM. Mat
khac,
do't
chay
hoan toan
3,83 gam hon hgp
X
can 3,192 lit
O2
(dktc). Dan toan bp san pham

chay
(CO2,
H2O
va
N2)
vao nude voi trong du thi kho'i lugng ket
tiia
thu dugc la
A.
13 gam.
"
B. 20 gam. C. 15 gam. D. 10 gam.
Laigidi:
HHCI
= 0,03 mol; no2
=
ai425
mol.
•
C>D
< -
-
1
^.
rno^SO
np _ 80/16 10 '
21
^ n^
21/14 ~
3

•
M|it
khac:
nN=
nNH2
=
"HCI
(x)=
0,03
mol
->
no(X)= 0,1 mol.
^ Ggiso'mol: n^
(x) =
a
i^ioJ
va nH(x)=bmol.
Bao toan khoi lugng: mx
=
m^ +
mj^ +
mo
+T^-H
-»
mc
+
niH
= 3,83
-
0,1.16

-
0,03.14
= 1,81
man xet: Ixx ti 1? khoi lugng:
Bao toan nguyen to O: n
Q (x)
+
2no2
=
2nco2
+
^HjO
12a+ b = 1,81.
'fl
iiAA
US'
^ 0,1 +
2.0,1425
= 2a + 0,5b
->
4a + b = 0,77
^
a = 0,13.
,. ,. .
Bao toan nguyen to C:
nc(X)=nco2
= '^CaC03
=0,13^
mcaCOa =13 gam.
-> Dap an A.

3. PHl/ONG PHAP
TANG-GIAM
KHOI
Ll/pNG
,
ffl>|
{
a. Npi dung
Khi
tham gia phan
ling
hoa hgc, nguyen tu (nhom nguyen tu) cua
chat
ban
dau dugc thay the (ho^c cgng hgp) bang nguyen tu (nhom nguyen tit) mai
de tao thanh san pham.
^
Do do, khoi lugng cua
chat
tao thanh
c6
the tang len hay giam di do chenh
l|ch
khoi lugng mol cua cac nguyen tu (nhom nguyen tu).
Dua
vao sy
tang
hay
giam
nay c6 the xac

djnh
so
mol
cac
chat
trong
phuang
trinh
hoa hgc, tu do
c6
the giai nhanh nhieu bai toan.
S\f thay the (cpng hgfp)
Bien doi khoi lugmg
(tinh
cho
1
mol)
C03
> CI2
Tang 71-60 = 11 gam
0-2
(oxit) ""''^"^
> SO^
Tang 96-16 = 80 gam
CO
—^
CO2; H2
H2O
Tang 16 gam
-OH

> -ONa
Tang 23-1 = 22 gam
_COOH
>
-COONa
Tang
23-1=22
gam
-NH2
> -NHsCl
Tang 36,5 gam
21
dm
nang fln luygn thi dgi hgc 18 chuySn de H6a hpc -
Nguyjn
Van Hi\
VI
DU MAU
Vi
d\ 1
(A-08):
Cho V lit hon hop khi (dktc) gom CO va H2 phan ung vol mot
lugng du hon hop ran gom CuO va
Fe304
nung nong. Sau khi cac phan ung
xay ra
hoan
toan, khoi lugng hon hgp ran giam 0,32 gam. Gia tri ciia V la
A.
0,448.

B.
0,112.
C.
0,224.
D.
0,560.
Lot
giai:
1
mol CO —*° > CO2 Kho'i lugng
chat
ran giam 16 gam. ^^^m
1
mol
H2
—*° >
H2O
—> Kho'i lugng
chat
ran giam 16 gam.
-> 1 mol (CO va
H2)
^° >
(CO2
va
H2O)
giam 16 gam.
Theobai:
0,02 mol <- giam 0,32 gam.
-> V =

0,02.22,4
=
0,448
lit.
—> Dap an A.
Vi
d\ 2: Hon hgp Y gom FeO,
Fe203
va CuO. Hoa tan
hoan
toan 6,8 gam Y
bang
dung djch HCI (du), thu
dugc
dung dich
chua
12,3 gam muoi. Mat
khac,
neu khu
hoan
toan 6,8 gam Y
bang
CO (du), thu
dugc
m gam kim
loai.
Gia tri cua m la :
A.
5,2.
B.5,6.

C.6,0.
D. 4,8. ^
Led
giai:
Nhan
xet: Khi cho Y + HCl thi oxit chuyen thanh muoi clorua va mgt ion
0^~ trong oxit
dugc
thay the
bang
hai ion CI".
, 1 mol O^- > 2 mol CI" -> Khoi lugng tang 71 -16 = 55 gam.
a mol <— tang 12,3 - 6,8 = 5,5 gam.
^ a = — = 0,1 mol -> no (Y) = 0^1 mol. ^
Khi
cho Y tac dung voi CO thi cac oxit deu bi khu thanh kim loai. ^
Bao toan khoi lugng: my = m + mo -> m = 6,8 -
0,1.16
= 5,2 gam.
-> Dap an A. ' '
"
"*
Vi
dvi 3: Hon hgp X gom CuO va
Fe203.
Hoa tan
hoan
toan 22 gam X
bang
dung dich

H2SO4,
thu
dugc
52 gam muoi. Mat
khac,
neu khu het 22 gam X
bang
CO (du), dan hon hgp khi thu
dugc
vao dung dich Ca(OH)2 (du), tao
thanh m gam ket tua. Gia trj cua m la
A.
45,5.
B.37,5.
C. 40,5. D. 50,0.
Lai
giai:
Nhan
xet: Khi cho X +
H2SO4
thi oxit chuyen thanh muoi
sunfat
va mgt ion
0-2
trong oxit
dugc
thay the
bang
mgt ion SO 4".
Cty

TNHH
MTV
DVVH
Khang Vi§t
1
mol 0-2 > 1 mol SO l~ Khoi lugng tang 96 - 16 = 80 gam. ff
a mol tang 52 - 22 = 30 gam.
f.
•
_^ a = — =
0,375mol
-> no
(X)
=
0,375
mol. '" ' '' •' '
80
Khi
cho 22 gam X tac dung voi CO
thi:
no (X) = n^oj = riCaCOa =
0375
mol.
-> m
=0,375.100
= 37,5 gam.
—> Dap an B '
'''t*''-t>
i?>';
f

vcr-f
vi KT
*
•'•H'>I
Vi
dv 4
(A-10):
Dot
chay
hoan
toan mgt lugng
hidrocacbon
X. Hap thu het san
pham
chay
vao dung dich Ba(OH)2 du, tao ra
29,55
gam ket
tiia,
dung dich
sau phan ung c6 khoi lugng giam 19,35 gam so voi ban dau. Cong
thuc
phan tu cua X la
A.C3H8.
B.C2H6.
C.C3H4.
D.C3H6.
Lot
giai:
Khi

hap thu
hoan
toan san pham
chay
gom
CO2
va hoi
H2O
vao dung dich
Ba(OH)2
du:
Ba(OH)2 +
CO2
>
BaCOs
+
H2O.
Mol:
0,15 <- 0,15 -f^'
Khoi
lugng dung dich giam = Mat -
Dugc
(r\
= mBaCOs -
(mco2
+
n^Hoo)
=
19,35gam.
->

mH20=
29,55
-
0,15.44
- 19,35 = 3,6 gam ->
nH20
= 0,2 mol. ' *"
Vay
hidrocacbon
X c6: So C : so'H = nc : nn = 0,15 : 0,4 = 3 :>8.
^ Dap an A.
„.,,.n
,,a ,,
Vi
dy 5
(B-12):
Cho 21 gam hon hgp X gom glyxin va axit
axetic
tac dung vua
dii
vol dung djch KOH, thu
dugc
dung djch Y chiia 32,4 gam muoi. Cho Y
tac dung voi dung dich HCl du, thu
dugc
dung djch
chua
m gam muoi.
Gia trj cua m la
A.

44,65.
B.
50,65.
C.
22,30.
^ D.
33,50.
Ldigiai:
Glyxin
(H2N-CH2-COOH)
= a mol va axit
axetic
(CH3COOH) = b mol.
Khi
cho X tac dung voi dung dich KOH: • •' •
:> 1 mol -COOH ^ 1 mo!
-COOK
-> khoi lugng tang 38 gam.
Theobai:
0,3 mol <-
tang32,4-21
= 11,4 gam
-> a + b = 0,3 mol.
Mat
khac:
75a + 60b = 21 a = 0,2 mol; b = 0,1 mol.
Khi
cho Y + HCl du, thu
dugc
cac muoi:

CIH3N-CH2-COOH
(0,2 mol) va
KCl
(0,3 mol)
theo
cac phan ung:
23
elm
nang fln luygn thi dgi hgc 18
chuy6n
66
H6a
hgc
- Nguygn Van
Hi\
H2N-CH2-COOK
+
2HC1
>
ClH3N-CH2-CC)OH
+
KCl
' CH3COOK
+
HCl
>
CHCOOH
+
KCl
•,l.>m^V

Vay:
m
=
111,5.0,2
+
74,5.0,3
=
44,65
gam
-»
Dap an A.
Luu
y:
Trong bai nay
cac em de
quen KCl khi
tinh
kho'i lugng muo'i
va
chpn
nham phirang
an C!
Vi
6
(A-12): Dot
chay
hoan
toan
4,64
gam mpt

hidrocacbon
X
(chat
khi
a
dieu
ki^n
thuong)
roi dem
toan
bp san
pham
chay
hap thu het vao
binh dvmg
dung dich
Ba(OH)2.
Sau
cac
phan
ling
thu
dugc
39,4 gam ket tua va
khoi
^j.
lugng phan dung dich giam bot
19,912
gam.
Cong

thiic phan tu ciia
X la
^.
A.
CH4.
, B.
C3H4.
C.
C4H10.
D.
C2H4.
Loigidi:
Kho'i
lugng dung dich giam
=
Mat
-
Dugc
=
mBaCOa
-
(mc02
+
^H20)
= 19,912
gam.
^
d-
^
mco2

+ mH20=
39,4-19,912
=
19,488
gam.
^
12
2 -
M9tkhac:mx
= mc +
mH=
—niro,+—^H-JO
=4,64.
'*
mco2
=15,312
gam
(0,348
mol); mnjo
=4,176
gam
(0,232
mol).
Vay
hidrocacbon
X c6: So C
:
so H
=
nc

:
nH
=
0,348
:
0,464
= 3:4
->
X
la
C3H4
Dap an B.
Lim
y:
Ci day bai toan khong cho dung dich
Ba(OH)2
du nen neu
cac
em viet
phuang
trinh
hap thu:
Ba(OH)2
+
CO2
>
BaCOa
+
H2O.
va suy

ra
ncQ2
=
TI^^QQ^
=
0,2
mol thi
se
bi mac sai lam
ngay!
Vi
di;i
7:
Hon hgp
Y
gom axit
axetic,
axit acrylic
va
axit adipic.
Cho 16,9
gam
Y
tac dyng vua dti vai dung djch NaOH, thu
dugc
22,4
gam muo'i. Mat
khac,
16,9 gam
X tac

dung
vai
dung dich
NaHCOs
du,
thu
dugc
V
lit
khi CO2
(dktc). Gia trj
aia V la
• - • .•
p
A. 5,60.
q
B. 8,40.
-
C. 4,48.
n
D.
7,84.
\
Khi
cho
Y
tac
dyng voi dung djch KOH:
1
mol -COOH

->
1
mol
-COONa
->
Khoi lugng tang
22
gam.
Theobai:
0,25
mol tang
22,4-16,9
=
5,5
gam
ncooH
=0,25
mol.
Khi
cho
Y tac
dyng vai dung djch NaHCOa:
nco2
=
"cooH
=
0/25
mol
^
Vco2

=
0/25.22,4
=
5,6 lit
^
->
Dap an
A.
Cty
TNHH
MTV DVVH
Khang
Vigt
Vi
dv
8:
Hon
hgp
X
gom
hai
amino axit
(mach
ho,
moi amino axit
deu
chiia
mgt nhom
chiic
-NH2

va
mgt
nhom
chuc
-COOH).
Cho
16,4
gam
X tac
dyng
vua du vai
dung dich KOH, thu
dugc
24
gam
muo'i. M|t
khac,
16,4
gam
X tac
dung vua du
V
lit dung djch HCl 2M. Gia tri ciia
V la
A.
0,1.
B.0,5. C.0,2.
D.0,3.
f.
Loigidi:

8""?' ""U:
i' ^
, ,
Khi
cho
X tac
dung vai dung dich KOH:
{i >
y,i
i
,/ <,
1
mol -COOH
>
1
mol
-COOK
khoi lugng tang
38
gam.
Theobai:
0,2
mol
<-
tang24-16,4
=
7,6
gam
nx
=

ncooH
=
0,2
mol. Khi cho
X tac
dung vai dung djch HCl:
nHCl=nNH2
=
"x ^
nHci=0,2mol
^
VHCI=0,2
lit.
-> Dap an C.
^ ,,,,, , ,
Vi
dy
9:
Cho 12
gam hon hgp
X
gom glyxin
va
etylamin
tac
dyng vua du voi
dung dich HCl, thu
dugc
dung dich
Y

chiia
19,3
gam muo'i. Cho
Y tac
dung
voi
dung dich NaOH du, thu
dugc
dung dich
chua
m
gam muo'i. Gia trj
cua
m
la
A.
9,60. B. 15,45. C. 21,40. D. 19,10.
Loigidi:
.
Glyxin
(H2N-CH2-COOH)
=
a
mol
va
etylamin
(C2H5NH2)
=
b
mol.

Khi
cho X tac
dung voi dung dich HCl:
^
1
mol -NH2
>
1
mol
-NH3CI
->
khoi lugng tang
36,5
gam.
Theobai:
0,2
mol tang
19,3-12
=
7,3
gam
->
a + b = 0,2
mol.
Mat
khac:
75a
+
45b
=

12
^
a
=
0,1 mol;
b
=
0,1 mol.
Khi
cho
Y +
NaOH du, thu
dugc
cac
muoi:
H2N-CH2-COONa
(0,1
mol)
va
NaCl
(0,2
mol).
Cac phan ung hoa hgc:
CIH3N-CH2-COOH
+
2NaOH
>
H2N-CH2-COONa
+
NaCl

Mol:
0,1 0,1 0,1
C2H5NH3CI
+
NaOH
>
C2H5NH2
+
NaCl
Mol:
ai 0,1 0,1 -
Vay:
m
=
97.0,1
+
58,5.0,2
=
21,40
gam
^
Dap
an C.
Luu
y:
Neu quen
tinh
khoi lugng NaCl
cac
em

se
chgn
nham phuong
an
A!
25
Ca'm
nang
6n luygn thi dgi hgc 18
chuySn
dg H6a hoc - Nguyin VSn Hii
4.
PHL/ONG
PHAP BAO
TOAN
ELECTRON
7
a. Npi dung
4v
si-lk-l
ymi
t
^'bdn vym
*•
,1
Tong so' mol electron
cac
chat khu nhuong = Tong so' mol electron
cac
cha't

oxi hoa nhan.
;
f-Aji,
b.
Cach
ap
di^ng
Cac
em can
xac
djnh dung
va
day du
cac
chat
khii
va
chat oxi hoa ciing nhu
su
bien doi trang thai oxi hoa cua chiing.
Viet
cac qua
trinh
oxi hoa
(nhuong
electron)
va qua
trinh
khu
(nhan

J,
electron)
de xac
dinh so'mol electron
trao
doi roi
ap
djmg dinh luat bao toan
electron.
, ;
VI
DU MAU
Vi
du
1
(CD-11): Hoa
tan
hoan toan
13
gam Zn trong dung dich
HNOs
loang,
du
thu dup-c dung dich
X
chua
m gam
muoi
va 0,448 lit
khi

Ni
(dktc). Gia
tricuamla
A.
18,90
gam. B.
37,80
gam. C.
28,35
gam. D.
39,80
gam.
Lot
gidi:
13
0 448
nzn=—=
0,2
mol;
nN2
= i:7T=
"lol.
Bai
toan nay
cac
em
c6
the giai khi viet phuong trinh phan ung.
6
day cac em

se
dugc huong dan giai
theo
phuong phap bao toan electron.
Chatkhu':
Zn - 2e
> Zn*^ —>
ne
nhuong
=
2nzn
=
0,4
mol.
Chat
oxi hoa:
2N^5 + lOe > N2 ^
nenh,in
=
lOn^^
=
0,2
mol.
"
Nhu
v$y so mol
electron
trao
doi
chua bang nhau—> "chua

on". O day,
'
mot san pham khu da dugc "gia'u di", do la su
tao
thanh muoi
NH4NO3:
-i
Chat
oxi hoa:
2N^5 + se >
NH4NO3
'
Mol:
0,2 -> 0,025
Vay:
m
=
mzn(N03)2
+
mNH4N03
=
0,2.189
+
0,025.80
=
39,8
gam.
—> Dap an D.
Luu
y: 1- Bai

toan
nay c6 the
giai nhanh
hon
bang each
ap
dung ngay
phuong trinh bao toan electron: 2 n^g = 3
n-^Q
+ 8
n[,jH4N03
•
2-
Ba
kim loai kha manh (Mg, Al, Zn)
tac
dung voi axit
HNO3,
c6 the t^io
thanh muoi
NH4NO3!
^
.
Vi
dy 2:
Hoa
tan
hoan toan h6n hop gom
0,03
mol FeS2

va a
mol CU2S vao axit
HNO3
(vira dii),
thu
dugc dung djch
X
(chi chua hai muoi sunfat)
va V lit
khi
duy nha't NO (dktc). Gia trj ciia V la
A.
1,12.
B.5,60.
C.2,24.
D.
4,48.
Cty
TNHH
MTV DVVH
Khang
Vi$t
Lai
gidi:
'Sh^n
xet: Cac em luu
y la
dung
dich
X

chi
chua
hai
muoi
sunfat
sau
cac
phan
ung, S nam he't
6
dang
goc
sunfat.
,
Ta
CO
cac
so
do
chuyen
hoa:
:
•
'
FeS2
>
-Fe2(S04)3
Cu2S
>
2CuS04

2
\
Mol:
0,03 0,015 Mol: a -> 2a
Bao toan nguyen
to
S,
ta c6:
2
npeSj
+
ncu2S
= "504
^ '
0,03.2 + a
=
0,015.3 + 2a ^ a
=
0,015. • ^^''^
Cac
phan
ung
khu:
FeS2-15e
>
Fe*^
+ 25*^
Cu2S-10e
>
2Cu^2

+ 5-6
Bao toan
electron:
3n^o =
15npesj
+10ncujS
"NO
=
0,2
mol
V^o
=
4,48
lit.
c
-^DapanD.
.oorfv
'^^WM-
Vi
dv 3
(B-08): Cho
2,16
gam Mg tac
dung
voi
dung
dich
HNO3
(du). Sau khi
phan

ung xay
ra
hoan
toan thu
dugc
0,896
lit
khi
NO
(6
dktc)
va
dung
dich
Y
chua
m
gam
muoi.
Gia tri cua
m la
A.
8,88. B. 13,92.
C.
6,52.
D.
13,32.
Lot
gidi:
"Me=-^^

=0,09
mol;
nNo=
^^^=0,04
mol.
„
^
24 22,4
md'hu't-
Chatkhu:
Mg - 2e
> Mg^2
f
Chat
oxi hoa:
N^^ + 3e ^ NO; 2N^5 + se >
NH4NO3
Bao toan
electron:
2
n^g
= 3 n^o + 8
nNH4N03
"
2.0,09-3.0,04 ,
nNH4N03
=
^=0,0075mol.
^
->

m=
mMg(N03)2
+
mNH4N03
=
0,09.142
+
0,0075.80
=
13,92
gam.
->DapanB.
^ ^ ^ : ,
Vi d\
4:
Hoa tan
hoan
toan
hon
hgp gom
9,75
gam Zn va
2,7
gam Al vao
200
ml
dung
dich
X
chua

dong
thoi
HNO3IM
va
H2SO4
1,5M.
Sau
khi
phan
ung
xay
ra
hoan
toan thu
dugc
khi
NO (san
pham
khu
duy nha't)
va
dung
dich
Y (chi
gom
cac
muoi).
Khoi
lugng
muoi

c6
trong
Y
la:
A.
41,25
gam. B.
53,65
gam. C.
44,05
gam.
D.
49,65
gam.
Lot
gidi:
9
75
'
n2n=-^—=
0,15
mol; nAl =
0,1
mol.
: • •
Ca'm
nang
On
luy$n
thi dgi hgc 18 chuy§n de Hoa hqc -

Nguyin
Van H&i
Nhan xet: Khi axit HNO3 c6 mat dong thai voi axit H2SO4 loang thi lirgng
trong dung dich la do 2 axit phan li ra -> giai
theo
phuong trinh ion.
=
'^HNOg
+ 2nH2S04 = 0,2 + 2.0,3 = 0,8 mol , ,^, ^
Trong
X:
Phuong trinh ion rut gon:
3Zn
+ 8H* + 2NO; >
3Zn2+
+ 2NO +
4H2O
Mol:
ai5 0,4-^ 0,1 s?,
ifr^f-8
Al + 4H* + NO~ > AP + NO +
2H2O
Mol:
0,1^ 0,4^ ai •
:Mig(m
ufifki-^iiO
—> H+vaNOj tham gia phan
ling
het. "+g': 4 - <K'!
->

Khoi
luong muoi trong Y = 9,75 + 2,7 +
0,3.96
= 41,25 gam Dap an A.
Vi
du 5 (B-07): Nung m gam hot sat trong oxi, thu
dugc
3 gam hon hg-p chat
ran X. Hoa tan het X trong dung dich
HNO3
du,
thoat
ra 0,56 lit NO (san
pham khu duy nha't 6 dktc). Gia tri cua m la:
A.
2,52.
B.2,22.
C. 2,62. D. 2,32.
Laigidi: •
nNO=—=0,025 mol. .88;;-',A
22,4
Bao toan khoi lugng:
m.Q^
= mx- mpg= 3-m.
Nhan xet: Neu dua
theo
phuong trinh phan ung, bai giai se rat dai va kho giai.
Cach
1: 6 day, cac em can su dung so do phan ung:
Fe (1) -^1°^ X (2) ,

Fe^3 (3)
Bao toan electron:
3.npg
= 4nQ^ + 3nj^Q
^
3.E. = 1:21 +
0,075
-> m = 2,52 gam : .^v
—> Dap an A.
Cach
2: Qui doi X thanh Fe (a mol) va O (b mol). Ta c6: 56a + 16b = 3.
Qua
trinh oxi hoa: Fe - 3e ^ Fe*^
Qua
trinh khu: O + 2e -> O-^; N*^ + 3e -> NO
Bao toan electron: 3a = 2b +
0,075
—> a =
0,045;
b = 0,03.
-> m =
0,045.56
= 2,52 gam.
Vi
d\ 6: Hoa tan hoan toan 5,84 gam hon hop gom
Fe304
va FeS2 trong dung
dich
axit
HNO3

(dac, du), thu dugc
3,808
lit khi
NO2
(dktc) va dung djch X.
Cho
X tac dung voi dung djch
Ba(OH)2
du, Ipc ket tiia va nung trong khong
khi
den kho'i lugng khong doi thu dugc m gam chat ran. Gia tri cua m la
A.
8,66.
B.9,68.
C.
15,86.
D.
10,24.
7S
Cty
TNHH
MTV
DVVH
Khang
Vigt
Led giai:
3,808
1
Ggisomol:Fe304(amol)vaFeS2(bmol).
.

Ta
c6:
232a
+ 120b = 5,84. .U. '
Cac
phan ung khu: ^ ^
Fe304
- le > 3Fe*3 "f ^ J-j^n;}».in - ' , ;
FeS2
- 15e > Fe- + 2S- ^, ,^ ^
jo ','-
Bao toan electron:
nN02=
"Fe304 + 15nFeS2 a + 15b = 0,17 ^ a = 0,02; b =
0,01.^
'
Ta
CO
cac so do chuyen hoa:
Fe304
, 3Fe(N03)3 "^'^^""'^ >
3Fe(OH)3
1,5
Fe203
Mol:
0,02 0,03
FeS2
> Fe(N03)3 +
2H2SO4
)

Fe(OH)3
+
2BaS04
0,5Fe2O3+2BaSO4
Mol:
0,01 ->
0,005
0,02
-> m =
160.0,035
+
0,02.233
= 10,24 -> Dap an D.
Vi
du 7: Cho 12,45 gam hSn hgp X gom Fe, Mg, Zn vao dung dich
HNO3
du,
thu dugc dung dich Y (khong chua
NH4NO3)
va hSn hgp khi Z gom 0,2 mol
NO
va 0,1 mol
NO2.
Co c^n dung dich Y thu dugc m gam muoi khan. Gia
tri
cua m la
A.
31,05.
B.
43,45.

C.
55,85.
D.
62,05.
Lcn
gtat:
Nhan xet: Day la bai toan c6 nhieu chat khu (3 kim loai) va tao ra 2 san
pham khi nen can ap dung dinh luat bao toan electron.
Chat
khu: Fe - 3e > Fe-; Mg - 2e > Mg^^; Zn - 2e >Zn^^.
Chat
oxi hoa: N- + 3e > NO; N- + le > NO2.
Bao toan electron: netraod6i = 3njsjo + l-nN02 =
^''^"^°^-
.
Den
day, cac em can luu y la khi cho kim loai tac dung voi
HNOs,
so mol
go'c NO
3
nam trong muoi dugc tinh nhu sau: ^
n,
(muoi)
= netraod6i = 0,7mol. <»i (' 1
jf'ju'/j
N03
Bao toan khoi lugng:
m
=

mpe,Mg,
zn+m^^oa = ^^'^^ ^ ^ ^^'^^
—> Dap an C. (v
.•••L
29
C^m
nang
6n
luy^n
thi
d<ii
hpc 18
chuyen
ai H6a
hpc
- IMguygn
VSn
HSi
Vi
8
(CD-12):
Hoa tan
hoan
toan
8,9
gam hon hg-p gom Mg
va
Zn
bang
lugng

vua du
500ml
dung dich
HNOa
IM. Sau khi
cac
phan ling
ket
thiic,
thu dugc
1,008 lit
khi
N2O
(dktc)
duy
nhat
va
dung
d|ch
X
chiia
m
gam
muoi. Gia
tri
ciia
m
la
, ,1
A.

31,22. B. 34,10.
C.
33,70.

D.
34,32.
'
Loigidi:
nN20=
^^=0,045
mol;nHN03
=
0/5
mol.
i» ,
Mg,Zn
>
Muoi nitrat
+
N2O
' * '>•
Chat oxi hoa:
2N*^ + Be > N2O "
va
CO
the xay ra
qua trinh:
2N*^ + Be >
NH4NO3
(a

mol)
Khi
cho
kim loai
+
HNO3:
r
*
i^jsjQ-(muoi)
=
ng
trao
doi
=
8
njs420
+
8
nNH4N03
~
(0/36
+
8a) mol.
Bao
toan
nguyen
to N:
nHNOs
=
"NO3

^
^"^^20
+
2nNH4N03
-> 0,5
=
0,36
+
8a
+
2.0,045
+
2a
a =
0,005
mol.
'
'
Bao
toan
khoi luong:
m
=
mMg, Zn+ m^^^^
+
mNH4N03
,^ ^
=
8,9
+

(0,36+B.0,005).62
+
0,005.80
=
34,1 gam
f
Dap anB.
Nhan
xet:
Can luu
y
Mg
tac
dung vai HNO3
c6
the tao ra
muoi NH4NO3.
Vi
dyt
9:
Dan luong khi CO
di qua hon
hgp
gom
CuO
va
Fe203
nung nong,
sau
mpt

thoi gian
thu
dug-c
cha't
ran
X
va
khi
Y.
Cho
Y
hap thy
hoan
toan
vao dung dich Ba(OH)2
du, thu
dugc
29,55 gam ket
hia. Chat
ran
X
phan
ung
voi
dung dich HNO3
du thu
dugc
V
lit
khi

NO (san
pham khu
duy
,s,i
nhat,
a
dktc). Gia
tri cua
V
la
A.
2,24.
B.4,48.
C.
6,72.
D.
3,36.
/J
<
Lffigiai:
"BaC03
= 0,15 mol
6 day,
cac em can su
dyng
so do
phan ung:
CuO,
Fe203
> X )

Muoi nitrat
+
NO
k>r>
n
CO2
+
Ba(OH)2
>
BaCOs
+
H2O
Mol:
0,15 0,15 '
Nhan
xet:
Khi
cho hSn hgp
oxit
tac
dung vdi khi CO thi:
n,, trao doi
=
IIXQQ
. Ma n^o = ^C02 ~ 0/15
mol
->
ng
trao
doi

= 0,3
mol.
Bao
toan
electron:
ngtraodoi
=
3nNo
^
T^NO
=0,1
mol.
VNO
=2,24
lit Dap
an A.
Cty
TNHH
MTV DVVH
Khang
Vijt
Vi
dV
10
(B-10):
Nung
2,23 gam hon hgp
X
gom cac
kim loai

Fe,
Al, Zn,
Mg
trong
khi oxi, sau mot
thoi gian
thu
dugc
2,71
gam hon hgp
Y.
Hoa tan
hoan
toan
Y vao
dung dich HNO3
(du), thu
dugc
0,672
lit
khi NO (san
pham khu duy nhat,
a
dktc).
So
mol HNO3
da
phan ling
la
1,;, v

A.
0,12.
B.
0,14.
C.
0,16.
.
j-'
D-
0/18.
;,; {
Lodgidi:
' W
:,,0,f
rrr'''.':
Bao
toan
khoi lugng: mo2
=
2,71
-
2,23
=
0,48 gam
rni-ii
ricirt
•
-> no2
=
0/015

mol
->
no= 0,03
mol
^
nQ.2 ^^^.jj=
0,03
mol.
each
1:
Su dung
so
do:
X > Y
^"^"^
>
Muoi nitrat
+
NO
Ltm
y:
Y
chua
ca
kim loai (con du)
va
oxit.
+ Khi cho kim loai
+
HNOs:

T^^Q-
("""OO
=
ng
trao
doi
=
3 n^o
= 0,09
mol.
+ Khi
cho
oxit
+
HNO3:
1
goc
O^^
trong
oxit
se
bi
thay
the
bang
2
go'c
3i'
NOo
de tao

muoi nitrat,
do
do:
n =
2n
, =
0,06
mol.
Bao
toan
nguyen
to
N: nnwos
=
"NO"
"
^'^^
"
^'"^^
—>
Dap an D.
Cach
2:
Qui doi
Y
thanh
X
va O (0,03
mol).
Bao

toan
electron:
ng(x)=2nQ+3nr^o
=0,15
mol
->
n^^. =0,15
mol.
Bao
toan
N: nnNOs
=
^^Q'^
"
^'^^
~^ ^' ''
Vi
d^ 11
(B-12):
Cho
29
gam
hon
hgp
Al, Cu
va Ag tac
dung vua
du
vai
950ml

dung dich HNO31,5M,
thu
dugc dung dich chua
m
gam
muoi
va
5,6
lit
khi
X (dktc)
gom
NO
va
N2O. Ti khoi
caa
X
so
vai H2
la 16,4.
Gia
tri
ciia
m la
A.
98,20. B. 97,20.
C.
98,75.
D.
91,00.

Lai
gidi:
6 bai nay,
truoc
he't
cac em can
tim
so
mol moi khi
trong
X
de thu
dugc ke't
qua: nNo=
0,20
mol;
n^^o=
0,05
mol.
Al,Cu,Ag
)
Muoi nitrat
+
NO
+
N2O
Chat oxi hoa:
N^=^ + 3e >
NO;
2N*^ + 8e > N2O

va
CO
the xay ra
qua trinh: 2N*5
+ Se >
NH4NO3
(a
mol)
,
Khi
cho
kim loai
+
HNO3:
"NO3
^
""^'"°
^ ^ ^ ^
"N2O
^ 8
riNH4N03
= (1 + 8a)
mol.
,, ^ j
Bao
toan
nguyen
to
N: nHN03
^^oi ^^'^

^
"N2O
2nNH4N03
'
-> 0,95.1,5
= 1 +
8a
+
0,2
+
2.0,05
+
2a
-> a =
0,0125
mol.
31
elm
nang On luyjn thi
dgi
hgc 18 chuySn
dg
H6a hpc
-
Nguyjn Van HSl
Bao toan khoi
luqmg:
m = m^i, cu, Ag+
"1,^0^
+

mNH4N03
=
29
+
(1+8.0,0125).62
+
0,0125.80
= 98,2 gam.
Dap
an A.
*'
Nhan xet: Can nhan ra bai toan da "giau di" san pham NH4NO3.
Vi
dv 12:
Cho hon hop khi
X
gom CI2 va O2 tac dung vua du vol hon
hgp
bpt
gom
10,8
gam Al
va 2,4
gam Mg, thu dugc
40,9
gam hon h(?p chat ran
Y.
Phan
tram the tich cua khi CI2 trong
X

la
A.
80%.
B.40%.
C.50%.
, .
D. 60%.
, Laigidi:
- •',
n^,=
M=a4mol;nM,=
^=0,lmol.
X
gnyb .ig
:f d.i5
27
^ 24
M&ii)
Igol^M m
mifl:4
y
i[ mil
Nhan xet: Day la bai toan hai chat khu
(2
kim loai) va hai chat oxi hoa
(2
phi
kim)
nen can ap dung dinh luat bao toan electron.
Chat

khu:
Al - 3e >
Al*^
Mg - 2e >
Mg*^
Mol:
0,4-^1,2
Mol: 0,1-> 0,2 *
f'^^'^
Chat
oxi hoa:
O2 + 4e > ICf^ CI2 + 2e > lOr
Mol:
X 4x Mol: y —> 2y
Bao toan electron:
4x
+ 2y =
1,2
+ 0,2 = 1,4 mol.
Bao toan kho'i luong: mx +
Tn^i
+
mj^g
= my
yt
->
32x + 71y + 10,8 + 2,4 = 40,9 -> 32x + 71y = 27,7.
->
X
= 0,2 mol; y = 0,3 mol

ImLrx.
• Q3
iff,:
^
%Vn
= '
.100% = 60%
->
Dap an D.
m/
CI2
0,2 + 0,3
^
Vi
dv 13:
Nung hon hop
X
gom
a
mol Fe va
0,03
mol Cu trong khong khi mot
thoi gian, thu dugc 12,64 gam chat ran Y. Hoa tan hoan toan
Y
bang dung
dich
HNO3 loang (du), thu dugc
0,896
lit khi NO (san pham khu duy nhat
6

dktc). Gia tri ciia
a
la
A.
0,08.
B.0,10.
C.0,12. D. 0,14.
r
Laigidi:
nNO=
^;r—= 0,04 mol.
22,4
;
Bao toan kho'i lugng:
TTIQ^
= my
-
mx
=
12,64
-
56a -1,92 = 10,72
-
56a.
Nhan xet: Neu dya
theo
phuong trinh phan ung se rat dai va kho giai.
Cach
1:
6

day, cac em can su diing so do phan ung:
Fe,
Cu
(1) —Y (2)
"""^^^
)
Fe^^
Cu^^
(3)
Cty
TNHH
MTV DWH Khang Vijt
Xet su
trao
doi electron
a
cac giai doan:
(3): Fe -3e > Fe*^ ->
nenhirang
= 3 npg = 3a s^;;n u,
Cu
-2e
>•
Cu*^
~^
nenhir6ng =
2ncu
=0,06
'"•
" ' '•'

^
, _^ .
10,72
-
56a
, „
(1) _^ (2): O2 +4e > IQ-^
nenh^n=4no2
= = 1,34-7a
(2) -* (3):
N*"'
+3e > NO
nenhSn=3nNo
=0,12.
Bao toan electron:
3a
+ 0,0"6 = 1,34
-
7a + 0,12
-> a
= 0,14 mol.
. ; ,
Dap
an D.
'.f.
Cach
2:
Qui doi
Y
thanh: Fe (a mol); Cu (0,03 mol) va

O
(b mol).
, „
lit,
;
Bao toan khoi lugng: 56a + 16b =
12,64-0,03.64
=
10,72.
ysb n
Bao toan electron: 3a +
2.a03
= 2b +
ai2 ^ a
=
ai4;
b =
ai8 \
].
(w^
-
,
-^DapanD.
.;„,, iiO
ftoi vMl'o^'i':
5
PHl/ONGPHAPTRUNGHOADIEN
;gnu nerfq
6u'>a
.

a. Npidung
Dung
dich cac chat dien li luon luon trung hoa ve di^n.
Q
si*: q&CJ
<
b. Bieuthiic
-'••^h
^ fi
.(OMI)
f
Tong
so'mol di^n tich duong = Tong so mol di^n tich am:
5^! ( ;v.
Lieu y: De tinh so'mol di?n tich, cac em la'y so mol ion
x
di^n tich ion do.
VIDVMAU
Vi
dv 1:
Dung dich
X
chua cac ion: Fe^* (0,1 mol), AP" (0,2 mol), Ch
(x
mol)
va
S04~ (y mol). Co c^n dung djch
X
thu dugc
46,9

gam chat ran khan Y. Gia
tri
cua
X
va
y
Ian lugt la
v •
nr,^ipf\n\~n
f:,
A.
0,2
va 0,3 .
B. 0,1
va 0,2.
C.
0,4 va 0,4.
D.
0,3 va 0,3.
>
Laigidi:
3
orbs
Nhan xet: Dung dich da cho chua
4
loai ion, trong do
2
loai ion chvra bie't
so
mol,

do vay can lap dugc
2
phuong trinh dai so de tim so mol cua chung.
+
Ap dyng dinh luat trung hoa
di?n:
Tong
so mol di^n tich duong = Tong so mol di^n tich am
'^^
-> 0,1.2 + 0,2.3 = x.l + y.2
^ X
+ 2y = 0,8 mol.
n €0,0 H
+
Ap dung bao toan khoi lugng: my
=
mpg2+
+ ^^3+
"^ci"
""^
"^sol"
'
56.0,1
+
27.0,2
+
35,5x
+ 96y = 46,9
35,5x
+ 96y = 35,9.

^
X
= 0,2 mol;
y
= 0,3 mol.
. ^, , ; ,
Dap
an A.
-'"JOb
+ >
33
dm
nang
On luygn
thi
dgi hgc
18
chuy6n
66
H6a hgc -
Nguygn
Van
Hai
Luu
y:
Diem mau
cho't
d
bai nay
la cac

em dua
ra
phuang
trinh
d^ii so' cua djnh
lu^t
trung hoa dien
tich.:t.'e'-f ,ji*i„trt,»sj:fr
'^4f^-
:>''J*!;/i .J.K
•:i.«jSJi':'iie|'
Vi
dy 2
(B-12):,
Mpt dung dich gom:
Na* (0,01
mol);
Ca^* (0,02
mol); HCO,
(0,02 mol) va ion
X (a
mol). Ion
X
va gia tn cua
a la
•••
A.
OH- va 0,03.
B. Ch
va 0,01.

C.
COs^- va 0,03. D. NO^ va
0,03.
Lai
gidi:
Goi dien tich ion
X la
-n. "•
<f
J>0
+
eC rrfor;
' ., ,('
+ Ap dyng dinh luat trung hoa di?n:
' ' ' ' 't;'^
1.0,01
+
2.0,02
=
1.0,02
+
n.a n.a
=
0,03.
fJom
«>
irtmrU
Y
rSffr
Den day

c6 2
phuang
an
thoa
man
la A
va D. Vay chpn ion
X la
ion OH" hay
ion
NO
3?
+ Nhan thay ion OH" khong
the
ton tai cung ion HCO
3
trong dung dich ban
daudocophanung:
,
i ,-'
A6i-U)MU*I
OH-
+
HCO;
>C0^- + H20.
'-IdU'dfi
—> Dap
an
D.
jt^n

ife graiQ
Vi
d^ 3
(B-IG):
Dung dich
X
chira
cac
ion:
Ca^
Na*, HCO,
va
CI"
(0,1
mol).
Cho
1/2
dung dich
X
phan
ling
voi dung dich NaOH du, thu
dugc
2 gam
ket tua.
Cho 1/2
dung dich
X con
lai phan
ling

voi dung dich Ca(OH)2 du,
thu
duoc
3
gam
ket
tua. Mat
khac,
neu dun soi den c^n dung dich
X
thi thu
dugc
m
gam
chat
ran khan. Gia trj cua
m la
A.
9,21.
; V
B.7,47.
C.
9,26. D. 8,79.
kUA^gi
fjv
\ Lai
gidi:
Itfbl'
' Gpi
so

mol trong
1/2
dung djch
X:
Ca^*
(a
mol); Na*
(b
mol), HCO
3
(c mol)
va CI-(0,05 mol).
+ Ap dyng dinh luat
triing
hoa di^n:
2a + b = c + 0,05.
Nhqn
xet:
Khi cho
Ca(OH)2
du vao 1/2 dung dich
X,
toan bp
goc
HCO
3
se
di
* vao ket tua:
OH-

+
HCO:
>
CO,^'
+
H2O.
Ca2>
+ CO^- >
CaCOs
„
>^;TOMb
rbil
Mol:
0,03
<r-
0,03 -> c
= 0,03 mol.
Khi
NaOH du vao
1/2
dung dich
X:
OH-
+
HCO
3"
>
CO
3"
+

H2O
Mol:
0,03 -> 0,03
Ca2*
+
CO3"
>
CaCa
Mol:
0,02 0,02 <- 0,02 ->
a = 0,02 mol.
Cty
TNHH
MTV
DVVH
Khang Vigt
Tirdo—>
b = 0,04
mol.
,
>,r
lim
y:
Khi dun soi den can dung dich
X, goc
HCO
3
bi
phan huy:
,

2HCOi
—^
CO^"
+
CO2
+
H2O.
Do v%y,
1/2
cha't
ran khan thu dupe
chua
cac
ion: Ca^*
(0,02
mol); Na*
(0,04
mol); CO
3'
(0,015
mol); CI-=
0,05
mol.
*^ '• ' '
m
=
2(0,02.40
+
0,04.23
+

0,015.60
+
35,5.0,05)
= 8,79
gam Dap
an B.
Vi
dy
4:
Mpt
coc
nuoc
chua:
a
mol Ca^^
b
mol Mg^^
va c
mol HCO3
.
Cho toi
thieu
V
lit dung dich Ca(OH)2
x
mol/1 vao
coc de
lam giam tong nong dp ion
kim
loai trong

coc
xuong muc nho
nhat.
Bieu
thuc
tinh
V
theo
a, b, x la
2a+
b a + b ^,,_a + 2b ^ ^ _ i±b '
AV=^^.
B.V^
—. C.V = ^^. D.V =
X
2x
Lin
gidi:
•
"mtlJl,
+ Ap dung dinh luat trung hoa dien:
2a + 2b = c. '~ '
+
Cac
phan
ling
hoa hpc khi cho
V
lit dung dich Ca(OH)2
x

mol/1 vao
coc:
Ca(OH)2
+
Ca(HC03)2
>
2CaC03
+
2H2O
U,,J,,W
Mol:
a <- a
••ttVqed
Ca(OH)2
+
Mg(HC03)2
>
MgCOs
+CaC03
+
2H20^'"
•'
Mol:
b <- b
Uiidnj-ir.,
'
a + b
"'"niih.
ri>\
;iM'«vJ>'>i>t

oifo
\-
—>Vx
= a
+ b—>
V =
—> Dap an
B. , , ,,
,,,,„,„,,
- - i •
Vi
dv
5:
Dung dich
Y c6
chvia:
K^ (0,01
mol), Fe^^ (0,02 mol), NO
3
(0,04
mol)
va SO 4~
(x
mol).
Co
can
Y
thu dupe
m
gam muoi khan. Gia trj cua

m la
A.
3,98.
B.6,87.
C.
5,43. D.
4,78.
Lai
gidi:
O bai nay, dung dich
da
cho
chua
4
lo?ii ion, trong
do
ion SO
4"
chua
biet
so
mol,
do vay
can
lap dupe 1 phuong
trinh
dai
so de
tim
so

mol nay.
+ Ap dyng dinh luat trung hoa di^n:
* i
^
Tong
so'mol
di^n tich duong
=
Tong
so'mol
dien tich
am
0,01.1
+
0,02.3
=
0,04.1
+
x.2^^,
x
=
0,015
mol.
m'^'^'^'^^^'^'
+
Bao
toan khoi lupng:
m =
m^^^
+ m^^^ + m^^^.

+
m^^2.
f,
=
39.0,01
+
56.0,02
+
62.0,04
+
96.0,015
= 5,43
gam.
.,.^j,
^
-» Dap an A.
Lieu
y: tai
nay
cac em can ap
dung djnh luat trung hoa di?n
de
tim dupe
so
mol
cua goc
sunfat.
"
t,Oi:<;
r 'i,„i

•
. 35
Ca'm nang
On
luyjn thi dgi hgc 18
chuySn
dS
H6a
hpc
- Nguygn Van
Hi'i
Vi
dv 6:
Dung djch
X c6
chua: Fe^*
(0,05
mol),
Na* (0,07
mol),
CI"
(0,03
mol)
va
SO
4"
. Cho
dung dich Ba(OH)2
du vao
dung dich

X thu
dugc
ket
tiia
Y.
Nung
Y a
nhi^t dg
cao
ngoai khong khi den khol lugng khong doi thu
dugc
^,
m
gam
chat
tin Z
(coi BaS04 khong hi nhi?t phan). Gia tri cua
m la
A.
19,91. B. 18,11.
C.
24,31.
D.
20,31.
Lai nidi:
*
^
Ap dung djnh luat trung hoa dien:
2^
n^,

- 2^
nj,
2.
0,05 +1.0,07
=
0,03.1
+
2n ^ n,.
=
0,07mol.
Cac
so
do phan ung:
SO
4"
"""'^ ) BaS04 > BaS04 y
Mol:
. 0,07 , ° •' 0,07 0,07 ^ j,,
Fe^-
)Fe(OH)2
>
Fe(OH)31
Fe203
Mol:
0,025 • • • 0,0125 ^
Vay
m =
233.0,07
+
160.0,025

=
20,31
gam.
—>•
Dap an D.
•
•
Vi
7:
Chia
dung dich
Y
chua
cac
ion: Mg2%
NH
J,
SO
4",
CI"
thanh hai phan
bang nhau.
+
Phan
1
cho
tac
dung voi dung dich
NaOH
du, dun nong, thu

dugc
0,58
gam
ket hia va
0,672
lit
khi
(dktc).
.y
+
Phan
2 tac
dyng voi dung djch BaCl2 du, thu
dugc
4,66
gam ket tiia.
Tong
kho'i lugng
cac
chat
tan trong
Y
la
A.
3,055
gam. B.
6,11
gam. C.
5,35
gam. D.

7,05
gam.
'
Lot
gtat:
NMn
xet: Bai nay
cac
em giai dua
theo
cac
phuong trinh ion.
o
Phan
1
+ dung djch
NaOH
du:
jj
Mg2^
+ 20H" >
Mg(OH)2i
o'ijhqil
NH^
+OH-
>NH3t+H20
^
->%2. =
nMg(OH),=
^=a01mol.

4
'^"^ 22,4
van^.
= nNH3=-rr-=
0,03
mol.
-A
Phan
2
+ dung djch BaCh du:
Ba^^
+ sol" ,
BaS04;
f '
-Aq«Q.^-'
^
'h
'"^nt^h
'qsi^^^IKi!
-yin
••{in
itff
->;;
:
~^
"Ba2+
=
"BaS04
= ^ = 0/02 mol. jf;|„yi
j^:.,

,j
Cty
TNHH
MTV DWH
Khang
Vi$t
Ap
dung djnh luat trung hoa
di^n:
0,01.2
+
0,03.1
=
0,02.2
+ n^^.
.1
-> n^j_ = 0,01 mol.
,'^f^•v:iJiJr^;,'^^ii:
^ ^':^^Otti >
+
Bao
toan
khoi lugng:
-
Khoi
lugng cac
chat
tan trong
Y
=

^^^^2+
+
'"NH^
"^cr
"^sol"
'
•'•
'
=
2.(0,01.24
+
0,03.18
+
0,01.35,5
+
0,02.96)
=
6,11 gam
_> Dap an B
Lim
y: Bai nay
cac
em
de
chgn nham dap
an A (3,055
gam)
do
chi tinh khoi
lugng

chat
tan trong mgt nua dung dich Y.
, , ,,
Vi
8:
Dung dich
Y c6
chua dong thoi
cac
ion: Ba^*; Ca^*;
CI"
(0,01
mol)
va
NO
J
(0,03
mol). Cho V ml dung djch Na2C03 IM vao
Y
de thu
dugc
ket tiia
Ion
nhat. Gia trj nho nhat ciia V la
A.
30. B.20. C.40.
D.
70.
LaigidU
iftAl

Ggi
so'mol ciia
cac
ion
:
Ba^*
=
X
;
Ca^*
= y.
-M.J'jj
\t
Ap
dyng djnh luat trung hoa di?n:
2x
+
2y
=
0,01.1+
0,03.1
=
0,04
mol.
Cac phuong trinh phan ung
: .
jii
0
Ba2*
+ CO3" >

BaCOsi
H^^-,
BV
p:^;-:
Ca-
+ COr
CaC03l
^ '
:
Xn^^^2-
=
X
+
y
=
0,02
mol.
,
3
nNa2C03
= 0,02 mol V= = 0,02 lit = 20 ml -> Dap an B.,
Vi
dv
9
(A-10): Dung dich
X c6
chua: Na^
(0,07
mol);
SO^"

(0,02
mol) va
OH"
(x
mol). Dung djch
Y c6
chua
CIO
4,
NO
3
va
H""
(y
mol);
tong
so
mol CIO4
va
NO3
la
0,04.
Trgn
X
va
Y
dugc
100ml
dung dich Z.
Gia

tri pH ciia
Z
la
A.
1.
B.
12.
C.
13. D.2.
Votigidi:
., ,
+
Ap dung dinh luat trung hoa
dif
n
vai dung djch X:
,
1.0,07
=
2.0,02+l.n
_ -» n _= 0,03
mol.
"
+
3.,'
•
OH
OH i ? -tit
•*•
Ap dung djnh luat trung hoa dien vai dung dich Y:

,^
(inil);
^•"ciOi
-^l-^NOS
=^-V V
=0'04mol.
• ^
Trgn
X
vai Y:
H* +
OH-
>
H2O
-> H^du
=
0,01 mol
-»
IH^]=
^
=
0,1
=
10"^ ->.pH =
l
-> Dap an A. rf.^
o: -
37
Ca'm
nang

6n
luygn
thi dgi hgc 18
chuy§n
dg H6a hgc -
Nguygn
van
H&\
Vi
dv 10
(CD-07):
Dung dich
Z
chiia:
Cu^*
(0,02
mol),
K""
(0,03
mol), Ch
(x mol)
va
SO4"
(y
mol). Tong kho'i lugng
cac
muo'i
tan c6
trong
Z la 5,435

gam. Gig
tri
cua x va y
Ian lugt
la
A.
0,02 va 0,05. B. 0,05 va 0,01.
C.
0,01 va 0,03.
D.
0,03 va 0,02.
LOT
gidi:
•
O
bai
nay,
dung dich
Z
chiia
4
loai ion,
trong
do 2
loai
ion
chua biet so'mol,
do
vay can lap
dugc

2
phuang trinh
dai so de tim
so' mol
cua
chung.
+
Ap
dung dinh luat trung
hoa
dien:
^t.
0,02.2 + 0,03.1 = x.l + y.2
X
+ 2y = 0,07
mol.
" ''^
W
Baotoankhoi lugng: mz=m^ 2. +
m .
+ m^,. +
m
'N^ anuQ
:8
i#b
>;
Bill
;H^i,^>vvi.
^" ^ ^'
-Kilofn'eOiO)

'rOVi
.
^ 64.0,02 + 39.0,03 + 35,5x + 96y = 5,435 S,,
35,5x + 96y = 2,985 ^ x = 0,03
mol;
y = 0,02
mol.
,
—> Dap
an
D.
Vi
dvi 11
(CD-08):
Chia
dung
dkh X
chiia
cac ion: Fe^^
SO|~, NH4,
NO3
thanh hai phan bang nhau.
"t* , ' f • ,-,4
v'
c ! * .1;^
,r|/
-
Phan
1 tac
dung

voi
dung dich NaOH
du, dun
nong
thu
dugc
0,672 lit
khi
(dktc)
va 1,07 gam ket
tiia;
, ,.
;„) i^^:;; •
:';©aji-4):;::.,3-plIiJ's
-
Phan
2 tac
dung voi dung djch BaCh du,
thu
dugc
4,66 gam ket
tua.
^
Co
can X thu
dugc
m gam
muo'i khan. Gia
tri cua m la
.

A.
3,52
gam.
B. 7,04
gam. C.
8,52
gam. D.
4,26
gam.
Lai
gidi:
Nhan
xet:
bai
nay cac em
giai dua
theo
cac
phuong trinh ion.
1*'
.
Phan
1: +
Bung dich NaOH du:
,
.'rin»*Wi
t.
Fe3*
+ 30H' >
Fe(OH)3l

,
t'C^
.
•
6':>
V
fti\b
nnw
NH;
+OH* >NH3T+H20
1'07
, 0,672 ,
-> "Fe3^
=
"MOH)3
= = ^'^^
"^"l-
"NH^
=
''NHg
= ^ = 0'03 mol.
Phan
2: +
dung dich BaCh du:
Ba^* +
SO
4"
>
BaSOU
i.:

jA
•
4
66
Ap
dung dinh luat trung
hoa
di?n:
0,01.3 + 0,03.1 = 0,02.2 +
n^^^.
.1 ^
->
n _ = 0,02
mol.
*' '
+
Bao
toan
khoi lugng:
m =
m^
3+
+
+
"^j>jo~
""^
"^so^'
=
2.(0,01.56 + 0,03.18 + 0,02.62 + 0,02.96) = 8,52 gam
Dap

an
C.
,
J:^.Asxu'^'^^^'^^^:
i^^^y
•
CtyTI\ih.i
1TV DWH Khang Vi^t
6
PHI/ONGPHAPDI/6NGCHEO
V6i
hSrhgp
bat
ki g6m hai
chat
X
va Y, khi biet
dugc
gia tri khoi lugng mol
trung bmh(M
)tasec6tile:
,
ny
M
-
MY
M-Mx
AMN
i
»!

^
>
I )
,
,,
,>.'i,
••
b. So
do hoa
X:
Y:
M,
1=0
n>
n,
lAMv
AM,
vi
oy
MAU
Vi
dxf.
1
(CD-07): Trong
tu
nhien, nguyen
to
dong
c6 hai
dong

vi la
29
Cu
va
29 Cu.
Nguyen
tu
khoi trung binh
cua
dong
la 63,54.
Thanh phan phan
tram
tong
so
nguyen
tu
ciia dong vj
29
Cu
la
A.
27%.
B.
50%.
C.
54%. '
D.
73%. n :
Lot

gidi:
" - 'o
Cach
1:
Ggi
so
nguyen
hi
29
Cu
la a va
2^
Cu
la b.
Ap
dung
cong
thuc
cua
phuong
phap
duong
cheo,
ta
c6:'
65 - 63,54
I
63-63,54
Dap
an D.

1,46
0,54
73
27
»%§Cu
= 73%. a
Cach
2:
Nhan
thay
Acu
= 63,54 <
63 +
65
dong
vi
29
Cu chiem
uu the hon
->
%
29 Cu
> 50%
Dap
an D
(Cac dap
an
khac deu
<
50%).

'
Vi
d^;i
2
(CD-07):
Cho 4,48 lit
khi CO
(dktc)
tu tir di qua ong
sii nung nong
dung
8 gam mgt
oxit
sat den
khi phan ling
xay ra
hoan
toan,
thu
dugc
hon
hgp khi
X c6 ti
khoi
so voi
hidro bang
20.
Cong thiic ciia
oxit
sat va

phan
tram
the
tich ciia khi
CO2
trong
X la
A.FeO;75%.
B.
FezOs;
75%.'
C. FeaOs;
65%.
D.Fe304;75%.
Lai
gidi:
4
48
Theo bai:
M = 20.2 = 40 va
nx =
nco=
TTT
=
0'2
mol
. , , ,
Ap
dving cong thiic ciia phuong phap duang
cheo,

ta c6: ' ''
"CO
=
nC02
44
-40
28
-40
1
3
0,05
0,15
%Vco2
=75%
chgn
B
hoac
D.
39
Ca'm
nang 6n luy$n thi dgi hpc 18
chuyfin
dg H6a
hpc
- Nguygn Van
Hi\
Taco:
no(oxit)=
"cOz = 0'15 mol ->
8-0,15.16

56
=
0,1
mol.
ion
_^ :
no=
0,1
:
0,15
=
2:3 ->
Fe203
Dap
an
B.
\ i '
fin.?'"/-:
Vi
du
3: Cho mot manh
Cu tac
dung vai dung dich HNOs du, thu duoc dung
dich
X va 0,896
lit (dktc) hon
hop
khi
Y
gom NO

va
NO2. Ti
khoi
ciia
Y so
voi
hidro bang
19. Co
can X thu dugc
m
gammuoi khan. Gia tri cua
m la
A.
5,64
gam. B.
7,52
gam. C.
9,4
gam. D.
15,04
gam.
'
"
Lcngidi:
0,896 ^. rr •
nY
=
22,4
=
0,04

mol. Theo bai:
My = 19.2
=
38.
Ap
dung cong thuc cua phuang
phap
duong cheo,
ta c6:
HNO
^ 46-38 _ 1 _ 0,02
nN02
30-38 ~1 ~0,02
Ap
dyng
bao
toan electron: 2 ncu = 3 n^o +
1
•
"NOZ
3.0,02 +
1.0,02
=
- V .
nNO=
0'02 mol; n^oj = 0,02 mol.
-«j
ijxj |V
^
"Cu

=
-=0,04 mol
->
ncu(N03)2 =
nCu
=
0.04mol,:
m
=
0,04.188 = 7,52
gam
->
Dap
an
B.
Vi
dy 4:
Hoa
tan
hoan toan
m
gam Al bang dung djch HNO3 loang,
thu
dugc
dung dich X (khong chiia NH4NO3)
va 5,6
lit (dktc) hon hgp khi Y gom
NO
va N2O.
Ti

khoi
hoi
ciia
Y so vai
hidro bang
19,2. So
mol axit HNOa
da
tham gia phan ung
la
A.
1,2.
B.
1,5. C. 1,7.
D.
2,0.
Loigidi:
5
6 — "
•
—
= 0,25
mol. Theo bai:
MY
= 19,2.2
=
38,4.
22,4 ; .„
,;i.,v<
—

!,(,./«
nY
=
Ap
dung cong thuc
ciia
phuong
phap
duong cheo,
ta
c6:
"N2O
^
30 - 38,4'
44 - 38,4
.3
'2
0,15
0,10
nN20=
0,15
mol; nMo=
0,10
mol.
Bao toan electron: 3
n^i
=
8.
n^^o + 3 n^o
8.0,15

+
3.0,1
"Al
=
=
0,5
mol
^
nAi(N03)3 =
^Al
=
0,5
mol.
Bao toan nguyen to
N:
nHNOa
=
3nAi(N03)3
+
2n
^20+ "NO
'
" " = 3.0,5
+
2.0,15
+
0,1 = 1,9
mol
^
Dap

an
D.
Vi
dy 5
(A-10):
Hon
hgp
khi
X
gom
N2 va
H2
c6 ti
khoi
so vai
heli
b^ng
1,8.
Dun
nong
X
mot
thoi
gian trong binh kin (c6
Fe
lam xuc tac),
thu
dugc hon
hgp khi
Y c6 ti

khoi
so
voi
heli
bang
2.
Hif
u
sua't
ciia
phan
ung
tong
hgp
NH3
la
A.
25%.
, r •'
B. 50%. C. 36%. D. 40%.
Cty
TNHH
MTV DVVH
Khang
Vigt
=
-
Ggi: n^^^ =
1
mol;

nH2 = 4
mol.
4
i
^
Lai
gidi:
iv
ri
I
^
n6'>
Theo bai:
Mx =
1,8.4
=
7,2;
MY
= 2.4 =
8 . i !
i:M
:
Ap
dyng cong thuc
ciia
phuong
phap
duong
cheo
voi X,

ta
c6:
"N;
_ 2-7,2
28-7,2
->
mx =
28.1
+ 2.4 = 36 gam mv = mx = 36 gam.
Phan
ling
xay
ra
nhu
sau:
N2
+ 3H2 ^=:±
2NH3
3x
-> 2x
X
Phan ung:
—
36
Ta
c6:
MY =
x
=
8

UY
=
(5
-
2x) mol
x
= 0,25 mol
^ H
= ^100% = 25%
.
5-2x
->
Dap
an
A.
Vi
dv 6
(A-12): Hon
hgp X
gom H2
va
C2H4
c6 ti
khoi
so
voi H2
la 7,5.
Dan
X
qua

Ni
nung nong,
thu
dugc
hon hgp Y c6 ti
khoi
so vai
H2
la 12,5.
Hi^u
suat
ciia
phan ung hidro hoa
la
A.
70%.
B.80%.
C.60%.
i^-'i
D.
50%.
Lai
gidi:
^ '•
Theo bai:
Mx = 7,5.2 =
15;
^_ | _
MY
=12,5.2

= 25.
S
lor~dS[
Ap
dung cong
thiic
ciia
phuang
phap
duong
cheo
voi X,
ta
c6:
rn,
2- 15
28 15
Ggi:
nc2H4
=
^ "H2
=
^
"C2H4
^
"H2
->
mx = 28.1 + 2.1 =
30
gam mv = mx = 30 gam.

Phan ung xay
ra
nhu
sau:
C2H4
+ H2
X
X
di
+
^
C2H6
i"'y
•.
' •
>
X ->
nY
=
(2-x)mol
Phan
ling:
Ta
c6:
MY
- = 25 ^ x = 0,8
mol
-> H
=
100%

= 80%.
2-x
1
Dap
an
B.
Vi
d^ 7
(CD-09):
Hon hgp X gom H2
va
C2H4
c6
ti
khoi
so
voi
heli
la
3,75. Dan
X
qua
Ni nung nong,
thu
dugc
hon hgp Y c6 ti
khoi
so
vai
heli

la 5.
Hi?u
sua't
ciia
phan
ling
hidro
hoa la
A.
20%. B.25%.
C.50%.
D.
40%.
Lai
gidi:
Theo bai:
Mx = 3,75.4 =
15;
• -
MY
=5.4=20. ;• ;r .
dm
nang
On
Iuy0n
thi
dgl
hpc
18
chuyfin

ai
H6a
hqc
-
Nguyin
van Hit
Ap
dyng cong thuc
ciia
phuang
phap
duong
cheo
vai X, ta c6:
Goi:
ric2H4
=
1
iriol;
riH2
= ^
"^ol-
;lado':.;riT.
nub IA
"C2H4
2-15 ^1
~
28-15 1
->
mx = 28.1 +

2.1
= 30 gam mv = mx = 30 gam. 1
Phan
ling
xay ra nhu sau: -1 |t
C2H4
+ H2
C2H5
- i^i^ 1 a
Phan
ling:
x x -> x ^ nv = (2 - x) mol "v/iH
Taco: My =^^ = 20
2-x
x
= 0,5mol^ H 100% = 50%.
—>
Dap an C.
Vi
du 8: Hon hgp X gom
H2
va
C2H2
c6 ti
kho'i
so vai hidro la 5. Dan X qua Ni
nung nong, thu dugc hon hgp Y c6 ti
kho'i
so vai hidro la 10.
Phan

tram the
X
Hr
tich
khi
H2
da tham gia phan
ling
hidro hoa la y
qt>rt
QQ;
'} t/b
JV
o^,j}
A. 60%. B. 75%. C. 50%. " D. 85%. ^
Lai
sidi:
Theo bai: Mx = 5.2 = 10; My = 10.2 = 20. ,; .
Ap
dung cong thuc cua phuong
phap
duong
cheo
vod X, ta c6:
"C2H2
^
nH2
2-10
26-10
=

- ^ Ggi:
nc2H2
=lmol;
nnj =2 mol.
->
mx = 26.1 + 2.2 = 30 gam mv = mx = 30 gam. )
mrfj
Cac phan ung xay ra nhu sau:
C2H2
C2H4
H2
2H2
C2H4
.0
C2H6
Nhan xet: So mol khi giam = so mol
H2
tham gia phan ung.
Taco:
MY
=20->
nY=
—= 1,5
20
So mol khi
giam=nY
-nx =3-1,5 = 1,5 mol
Dap an B.
H
= —.100% = 75%

Vi
dy 9: Hon hgp X gom hidro va vinylaxetilen c6 ti
khoi
so vai hidro la 6.
Dan
X qua Ni nung nong, thu dugc hon hgp Y c6 ti
khoi
so vai hidro la 10.
Phan
tram the
tich
khi
H2
da tham gia phan ung hidro hoa la
A.
60%. B.85%. C.50%. D. 75%.
Lcn
gtat:
Theo bai: Mx = 6.2 = 12; My = 10.2 = 20.
Ap
dung cong thuc cua phuong
phap
duong
cheo
vai X, ta c6:
Cty
TNHH
MTV DWH
Khang
Vigt

2-
-12
52
-12
=
- ^ Goi:
nc4H4
=lmol;
nH2 =4 mol.
nC4H4_^
_^
mx = 52.1 + 2.4 = 60 gam -> mv = mx = 60 gam. ^ ^
N/iflM
xet: So mol khi giam = so mol
H2
tham gia phan ung.
Taco: My =20 "^=- = 3 » ' ' J:
So mol khi giam =
nY-nx
= 6- 3 = 3 mol
H
= 100% = 75% -» Dap an D.
\1
(•
{)
ft
7. PHl/ONG PHAP
TRUNG
BINH
a. Npidung

Trong
mgt hon hgp nhieu chat, c6 the bieu dien mot dai lugng nao do cua
cac chat thong qua mot dai lugng chung, dai di?n cho ca hon hgp, ggi la dai
kmng
trung
binh.
' ^ ' '
b.
Bieu
thuc va van dung m^M:^'r.:hH4£vim
X
=
-i^^^
(X - dai lugng dang xet; n-so'mol; i - so thu tu chat).
+
Khoi
luang phan tu trung
binh:
M = ^^^^
-
"CO2
' ^,Dm+-M„
+
So' nguyen tu cacbon trung
binh:
C =
_
2nH,o :e£.o :.,.,„ 5,o
+
So nguyen tu hidro trung

binh:^H
= ——
VIDUMAU
Vi
du 1
(CD-11):
De hoa tan hoan toan 6,4 gam hon hgp X gom
kirn
loai
R (chi c6
hoa tri 2) va
oxit
ciia
no can vua du 400ml dung dich HCl
IM.
Kim
lo^i
R la
A.
Ba. B. Be. C.
Mg.
D. Ca. 163
:i.
Laigiai: -A.
Cac phuong
trinh
phan ung: '^'^'
•
R
+ 2HC1 >

RCI2
+ H2 RO + 2HC1 >
RCI2
+ H2O
Nhan
thay: nx =
-j
UHCI = 0,2 mol.
Mx=—=32R<32<RO
-> R<32<R + 16
16<R<32 R = 24
(Mg)->
Dap an C.
An
Ca'm
nang
6n
luygn
thi d^i hpc 18
chuySn
6i H6a hgc -
Nguygn
Van
Hai
Vi
d\ 2: Cho 1,67 gam hon hgip gom hai kim loai (6 2 chu ky lien tiep thupc
nhom IIA) tac dung het voi dung dich HCl (du),
thoat
ra
0,672

lit khi H2
(dktc). Hai kim loai do la
A.
BevaMg.
B.MgvaCa.
CSrvaBa.
D.
Cava
Sr.
nH2=
^^=0,03mol.
,V
^^'^ _ _ t ! >fr
^
-
Phuang
trinh
phan ung: M + 2HC1 > MCI2 + H2 ,
Nhan thay:
nj;^
= n^j = 0,03 mol. i.
-> ^ ^^^"^
55,67
Hai kim loai la Ca va Sr -> Dap an D.
Vi
dxjL 3
(A-10):
Cho 7,1 gam hon hop gom mot kim loai kiem X va mpt kim loai
if^l kiem tho Y tac dung he't vai lugng du dung dich HCl loang, thu dupe 5,6 lit
khi

H2 (dktc). Kim loai X, Y la
A.
Kali
va bari. B.
Liti
va beri. C.
Natri
va
magie.
D.
Kali
va canxi.
Lcri
gidi:
nH2
= a25mol. .V,;
v;.'^NI
Cach
1: Cac phuong
trinh
phan ung:
2X + 2HC1 > 2XC1 + H2 Y + 2HC1 > YCI2 + H2
Mol:
a 0,5a Mol: b b
Taco
: a5a + b = 0,25 -> a + 2b = 0,5.
Mat
khac:
a + b < a + 2b = 0,5 <
2(a+b)

-» 0,25 < a+b < 0,5
71-71 _
.••ulnAv,,.^,,'.
„
-> — < M < — > U< M <28.
0,5 0,25
Lap
luan:
Mx
hoac
MY
< M < 28 -» Loai A va D vi hai kim loai deu > 28.
Mx
hoac
MY
> M > 14 -> Loai B vi hai kim loai deu < 14.
b'j -> Dap an C.
Cach
2: Gpi
cong
thiic chung aia hai kim loai la M, hoa
tri
chung la n.
Ta c6: n^ = 2nj^^ = 0,5 mol. Bao toan
electron:
ng
(j^j = ng= 0,5 mol.
^
n.nM
=

0,5mol->
-^ = 0,5 ^ M =
14,2n.
M
Mat
khac:
1 < n < 2 14,2 < M < 28,4. Lap luan nhu tren -» Dap an C.
Vi
A\y
4
(CD-12):
Hoa tan
hoan
toan 1,1 gam hSn hpp gom mpt kim loai kiem X
va mpt kim loai kiem tho Y (Mx <
MY)
trong dung dich HCl du, thu dupe
1,12 lit khi H2 (dktc). Kim loai X la
A.
Li. B.Na. C. Rb. D. K.
Cty
TNHH
MTV DWH
Khang
Vi§t
Lot
gidi:
Gpi
cong
thiic chung ciia hai kim loai la M, hoa tri chung la n.

* *
-
Ta c6: n^ =
2n^^^
= 0,1 mol. Bao toan
electron:
n^ " "e" 0,1 mol.
_>
n.nM=0/l
mol-^ = 0,1 ^M =
lln.
Matkhac:l<n<2 -> ll<M<22.Vay Mx<M<22 Loai B, C, D.
Dap an A.
Vi
du 5
(B-08):
Cho 1,9 gam hon hpp X gom muoi
cacbonat
va
hidrocacbonat
ciia
kim
loai kiem M tac dung he't voi dung dich HCl (du), sinh ra
0,448
lit khi (a
dktc). Kim loai M la
A.Na.
B. K. C.Rb. D.Li. ,|,,/
^oigidi:
nco2

= 0,02 mol. Cac phuong
trinh
phan ung: ^j., j^- ^ .
M2CO3 + 2HC1 > 2MC1 + CO2 + H2O
MHCO3
+ HCl > MCI + CO2 + H2O k'mS.Dw W
Nh^n
thay : nx =
nco2
= 0,02 mol.
_^ Mx = — = 95 ^
MHCO3
< 95 < M2CO3.
0,02
^ M + 61< 95 < 2M + 60 17,5 < M <34 ^ M la Na ^ Dap an A.
Vi
d^ 6: Cho 1,7 gam hon hop gom kim loai X (nhom IIA) va Zn tac dung voi
dung dich HCl du, sinh ra
0,672
lit khi H2 (dktc). Mat
khac,
khi cho 1,9 gam
X tac dung voi axit H2SO4 loang, du thi the tich khi hidro sinh ra
chua
deh
1,12 lit (dktc). Kim loai X la
A.Ba. B.Ca. C. Sr. D. Mg. «) ;
Loigidi:
,
Cac phuong

trinh
phan ung:
X + 2HC1 > XCI2 + H2 Zn + 2HC1 > ZnCh + H2
Nhan thay: nx + nzn = nn, = 0,03 mol
M
=
AL=
56,67
Mx<
56,67
< Mzn ->
XlaMghoacCa.
. ,
0,03 . et-' •
Mat
khac:
Mx > 38 -> X la Ca -> Dap an B. 0 = d +
B,
m
0,05
Vi
dii 7: Dot
chay
hoan
toan 6,72 lit (dktc) hon hpp M gom hai
hidrocacbon
X
va Y (Mx <
MY),
thu dupe 11,2 lit khi CO2 (dktc) va 10,8 gam H2O. Cong thiic

phan tu cua X la
.C2H4.
B.C2H2.
C.C2H6. •Q™D.CH4. '
4";
Ca'm
nang On luygn thi dgi
h9C
18 chuygn H6a hpc -
Mguyin
Van Hii
Lai
giai:
11,2
22,4
"^€02
"
;rr7"
0,5
mol
^ nc
=
0,5
mol.
> :
•!
.M, /
10
8
"H20=

-T^=
mol ->
nH = 2.0,6 =
1,2
mol.
lo
Bao toan
khoi
lugng:
mM
=
mc + mn =
0,5.12
+
1,2.1
= 7,2
gam.
^ M =
-^=24.
ViMx<MY
^
Mx<24-> XlaCH4
'
0,3
' m
qit*
*
->DapanD.
.t
i*-')ti>f.

yfc
Vi
d\
8
(B-10):
Hon hop khi
X
gom mot ankan
va
mot anken.
Ti
khoi
cua
X so
voi
H2
bang
11,25.
Dot chay hoan toan
4,48
lit
X,
thu dugc
6,72
lit
CO2
(cac
the
tich
khi do

6
dktc). Cong thuc ciia ankan va anken Ian
lugt
la
A.CH4vaC2H4.
^^'V B.
C2H6
va
C2H4.
C.
CH4
va
C3H6.
<i»»' '^^''i^JD.
CH4
va
C4H8.
Lai
giai:
4
48 6 72
=
-zr =
0,2
mol;
nco2
=
^r—=
0,3 mol
-»

nc =
0,3
mol.
22,4
^22,4 , , ,.
M
=•,,(')
>n'
m
•
vsrfi
nv,r\v
Theo
bai: Mx
=11,25.2
= 22,5
X
chua mot hidrocacbon
c6
phan tvr
khoi
nho
hon
22,5 ->
Do
la
metan
(CH4)
Loai
B.

Ta
c6:
mx
=
Mx . nx =
22,5.0,2
=
4,5
gam.
J, Bao
toan
khoi
lirgng:
mx
=
mc
+
mH
I,
mH =
4,5 -
0,3.12
= 0,9
gam
->
nH
= 0,9
mol
'
''^

^ So
nguyen tu
H
trung
binh
trong
X
= ^=
^= 4,5
"X
0'2
;p|;y;
->LoaiA.
.u
• •
, v rr,,
.fO^.H ,
':,,tM\k
Thu50:50
^••"'^
it-Mi'^r:;
h':6
+
Ne'ulaC:CH4
=
amolvaC3H5 =
bmol.
Ta
c6:
nx

= a
+
b = 0,2 va
ncoj =
a
+ 3b
= 0,3 -» a =
0,15;
b = 0,05.
r-f
16.0,15
+
42.0,05
,
->
Mx = —
= 22,5
->
Thoa man
^
Dap an
C.
+
Neu la D:
CH4
= a mol va
C4H8
= b
mol.
' ''^

Ta
c6:
nx
= a
+
b = 0,2 va
nco2
= a
+
4b
=
0,3->a=—;b=
—
3
3
Mx
= 2 '-
0,2
Lo^i
phirong
an
D.
Mx
= 2— ^
^20,33
^ 22,5
46
Cty
TNHH
MlV DWH Khang

Vigt
Vi
dV
9
(B-08):
Dot chay hoan toan
1
lit hon hop khi gom C2H2
va
hidrocacbon
X sinh
ra 2
lit khi
CO2 va 2
lit hoi H2O
(cac the
tich
khi
va
hoi
do 6
ciing
dieu
kien
nhi^t
dg,
ap
suat). Cong thuc phan tu ciia
X la
A.C2H6.

,
B.C2H4. C.CH4. D.C3H8.
Lai
giai:
j.,,
Xheobai:
' ,:
"^*::i*>,^i:^-'O^H:,,>«:''-'''
-
+
So
nguyen
tu H
trung
binh
=
—^^^2-=
^
=
4
_> Hidrocacbon
X c6 so
nguyen
tu
H
Ion hon
4
(vi hon hgp chua C2H2
so
nguyen tu

H
nho hon 4).
Loai
B va C. : ; - i
+
So
nguyen tu
C
trung
binh
= ——
=
2
—> Hidrocacbon
X c6 so
nguyen
tu
C
bang
2
(vi hon hgp chua C2H2
so
nguyen tu
C
bang 2).
j j
,.
j
,
Dap an A.

Vi
dy 10:
Dot chay hoan toan
6
lit hon hgp
X
gom
2
anken
ke
tiep nhau
trong
day
dong
ding
can
vua
du 21
lit
O2 (cac the
tich
khi
do
trong
cung dieu
ki?n
nhi^t
do,
ap
suat).

Hidrat
hoa hoan toan
X
trong
dieu ki|n
thich
hgp
thu
dugc hon hgp ancol
Y,
trong
do
tong
khoi
lugng
cac
ancol
bac
mot gap
13/6 Ian
khoi
lugng
ancol
bac
hai. Phan
tram
khoi
lugng
ciia ancol
bac

mot
(c6
so
cacbon
Ian hon)
trong
Y la
A.
46,43%.
B.
10,88%.
C.
31,58%.
D.
7,89%,
'
Lai
giai:
Gpi
cong thuc chung ciia
2
anken
la
C-Hj
C;;H,-
+ —
O2
—^
nC02
+

nH20
n 2n 2
Lit:
6 9n
—
— 7
'
->
9
n =
21
^ n = >
Hai anken
la
C2H4
(a
mol)
va
CsHe
(b
mol).
3
Taco:
K = - ^ 2a +
3b^
7 ^
^^^b
->
Dat
a = 2

mol;
b =
1 mol.
3
a
+
b 3 _ , . . ^ „;>
Phan ung hoa hoc:
iul^ltXi
i
CH2=CH2
+
H2O —^^^^
CH3CH2OH
u'.
Mol:
2 2 i,:
C
Mol:
CH3-CH=CH2
+
H2O
>
CH3CH(OH)-CH3
s
dm
nang On luygn thi
dji
hqc 18
chuySn

dg H6a hpc - Nguygn
Van Hii
"
CH3-CH=CH2
+ H2O )
CH3CH2CH2OH
Mol:
1-x 1-x
46.2
+60(1-x)
13 no 1
Theobai:
- = > x =
0,8 mol.
|; w .1
1%
1 A
60x
6 " '
%mcH3CH2CH20H
=
^^7^-^00% = 7,89%
^
Dap an D.
Cach
2: De thay: % mcHgCHzOH
= "
60,53%.
'
. •

>
Mat khac,
theo
bai: %
mcH3CH(OH)CH3
=
-100%
=
31,58%.
^ %
mcH3CH2CH20H
=
100
-
60,53
-
31,58
=
7,89%
—> Dap an D.
' , , .,, .
Vi
dv
11: Hon hop
X
gom hai axit
cacboxylic
don chiic. Do't chay hoan toan
0,1 mol
X

can 0,26 mol O2, thu
duoc
CO2 va
0,2 mol H2O.
Cong
thiic hai
axit la
A.
CH3COOH
va
CH2=CHCOOH.
'
B.
CH2=CHCOOH
va
CH2=C(CH3)COOH.
X-
C.
HCOOH
va
C2H5COOH.
' "*
D.
CH3COOH
va
C2H5COOH.
'
Bao toan nguyen to O, ta c6: 2nx + 2 no2
=
2

nco2
+
"H2O
_^ ^111026^
, ,
Nhqn
xet: n^jo
<
^C02 ^'^P
^
^^^^ ^^^^ "^9* ^^^^ khong no
-»LoaiCvaD.
nP ^
«*) 1
-> So nguyen tu
C
trung binh
=—£22.=
2,6
->
X
chiia mgt axit c6 so nguyen
nx
ttr
C
nho han 2,6
—>
Loai B (Hai axit deu chua
so'cacbon
>

2,6)
.
-> Dap an
A.
Vi
d\ 12: Dot chay hoan toan hon hop
X
gom hai
este
can dung 14 lit khi
O2,
thu
dugc
12,32 lit khi
CO2 va
9,9 gam H2O. Neu cho
m
gam
X
tac dung
vira
du voi dung dich KOH, c6 can dung dich sau phan ung thi thu
dxxgc
muoi
khan cua mQt axit huu
co
va
m
gam hon hgip ancol
Y

la dong d5ng ke
tiep. Gia tri ciia
m
la
A. 8,5.
;« x
C.7,1.
D.
10,1.
48
Cty
TIMHH
MTV DVVH Khang
Vi$t
Loigiai:
fj
=
0,625
mol;
n =
O'^S mol; nH20= 0/55 mol.,,
,^ , ,
J^lhan
xet: ncoz
^
"H2O ^ac
este
deu no, dan chii-c
->
Cong

thuc cua
2
este
la C-Hj-Oj.
, , ^ _
^,
. ^ „ „
'&'f<:.ihtii
••
Bao
toan
nguyen
to
0x1: 2n ggte
+
2n
=
2n
+ "
H2O
= /. „ . ,
p
r ^
-
nco2
_
0,55
_ ,
7c
-ft

s;V!
mB<*(
d.e
m:,
rb'r:
n
=
T—-
-
A/3 •
Mat khac, khi cho
X
tac dung voi KOH, thu duQfc muoi ciia mgt axit
hihi
co
va hon hop ancol
Y
la dong dang ke tiep —>
2
este
trong
X
hon kem nhau
1
nguyen tu cacbon.
—>
Cong
thiic phan tii 2
este:
C2H4O2

va
CsHeCh.
- ,
.j.
j
->
Cong
thuc cau tao:
HCOOCH3
(a
mol) va
HCOOC2H5
(b mol).
,5.
Ta co:
a + b =
0,2 mol. Mat khac:
n =
2,75
=
^^—^
a +
b
-> 2a + 3b = 0,55
->
a = 0,05 mol; b = 0,15 mol.
51^* "
f
•
*

Phan ling hoa hpc:
HCOOCH3
+
NaOH
—^
HCOONa
+
CH3OH
'"
HCOOC2H5
+
NaOH
—^
HCOONa
+
C2H5OH
£ fv A
i&oJ
->
m =
0,05.32
+
0,15.46
=
8,5gam
. , fy 3
j§oJ
—> Dap an
A.
Vi

dvi 13
(A-10):
Cho hon
X
gom ancol metylic
va
hai axit
cacboxylic
(no, don
chiic, ke tiep nhau trong day dong
dang)
tac diang het voi Na, giai phong ra
6,72 lit khi
H2
(dktc). Neu dun nong
X (co
H2SO4
dac xiic tac) thi cac chat
trong hon hgp phan ling vtra dii voi nhau tao thanh 25 gam hon hgp
este
(gia thie't phan ling
este
hoa dat
hifu
suat 100%). Hai axit trong
X
la
A.
C3H7COOH
va

C4H9COOH.
B.
CH3COOH
va
C2H5COOH.
C.
C2H5COOH
va
C3H7COOH.
D.
HCOOH
va
CH3COOH.
Laigidi:
nH,=
—=a3mol.
'
22,4 _
Gpi
cong
thuc chung ciia 2 axit la
RCOOH.
4 v, ,.
Dat so mol:
CH3OH
=
a;
RCOOH
= b. Ta co:
a

+
b =
2 nHj
=
0,6 mol.
Phan ung
este
hoa:
RCOOH
+
CHaOH
;e=±
RCOOCH3
+ H2O
49