các dạng điển hình và phương pháp giải nhanh bài tập trắc nghiệm hóa học 12

NGUYEN
TUYEN
HA
CAC
DANG
fillN
HiNH
VA
PHJdNG
PHAP
GIAI
NHANH BAI
TAP
TRAC
NGH||M
I
J
i
1
BIEN
SOAN THEO CHUDNG
TRINH MCJI
DANH
CHO HQC
SINH
BAN CO
BAN
VA
NANG
CAO
ON

LUYEN
THI TU TAI, DAI HQC VA CAO DANG
rH(i
V!EN TINHBiNH
THU-V^
NHA XUAT
BAN
DAI HQC QUOC GIA TP.
HO
CHI MINH
CAC
DANG
OIEN
HINH
VA
PHl/dNG
PHAP
GIAI
NHANH
BAI
TAP
TRAC
NGHIEM
H6AHOCI2
Nguyen Tuyen Ha
NHA XUAT BAN
DAI
HOC
QUdc
GIA TP HO CHI MINH

Khu
pho
6,
phirdng
Linh
Trung, quan Thu Dutc, TP.HCM
So
3
Cong
tradng
Quoc
te, quan 3, TP ROM
DT:
38 239 172, 38 239 170
Fax:
38 239 172
-
Email:

Chiu
track nhiem xudt bdn
TS HUYNH BA LAN
T6 chiic bdn thao vd chiu track nkiem
vi
tac quyin
DOAN VAN KHANU
Bien tap
NGUYEN THI
NGOC
HAN

S^a bdn in
THAN
THI HONG
Trink
bay bia
. DIEM KHANH
TK.02.H(V)
494.2013/CXB/07-25
TK.H.453-13(T)
DHQG.HCM-13
In 1.000
cuon,
khd 16
x
24cm.
S6'
dang
ky ke
hogch
xuat
bin:
494-2013/CXB/07-25/DHQGTPHCM.
Quyet
dinh
xuSt
bhn so:
333/QD-DHQGTPHCM
ngay
21
thang

6
nam 2013 cua
Nha
xuat
bdn
DHQGTPHCM.
In t?i
Cong
ty In
Song
Nguy6n,
n^p
lau
chilu
quy IV
nSm 2013.
LClI
N(&I DAU
De h6 trg
viec
hoc tap va on thi TU TAI
-
DAI HOC
-
CAO DANG
hang nam, toi viet
cuon
sach
n^y
theo

tinh
than giiip
cac
em trang bi
du kien thiJc de tham diT ki thi dat ket qua tot nhat.
Ve noi dung
cuon
sach
cd
3
phan:
+
Phan
I:
Cac
dang
bai tap dp dung
cong
thiJc
tinh
nhanh
+
Phan
II:
Cac phiicfng
phap
giai bai tap trac nghiem
+
Phan
III:

Gidi thieu 05
de
thi thuf Tu tai va 06 de thi thuf Dai hoc
de cac em
thijf
siJc
minh.
Cac
de
thi thuf mang
tinh
he thong day du cau hoi li thuyet va hki
tap. Sau do cac em tham
khao
ddp an
de
rut
kinh
nghiem trong ki thi
sSp tdi nham dat ket qua nhuf
mong
doi. Chiing toi hi
vong
cuon
sdch
nay
la
tai lieu tham
khao
hCJu ich giup

cac em
ren luyen
ki
nSng,
nang
cao
kien thiJc cua minh
va
dat ket qua tot nhat trong ki thi
TU
TAI
-
DAI HOC
-
CAO DANG s^p tdi. Mac du rat co g^ng de bign
soan
nhuhg vdi thcfi gian c6 han
chac
ch^n van cbn nhufng khiem khuyet.
Tac gia rat
mong
nhan
dacfc
sir gop
y
xay diTng cua
dong
nghi?p,
doc gid
gan xa de nhufng Ian sau tai ban

se
duac
hokn
thien hdn.
Xin
tran
trong cam cfn.
Tac
gid
Nguyen Tuyen Hd
T
CAC
D4NG
BAI AP
D^JNG
CACH
TiNH
NHANH
Cach
tinh nhanh so
dong
phan ciia:
-
Ancol no, dctn
chiiCc
(C„H2n+20):
2n-2
(1<
n < 6)
-

Andehit dcfn
chvCc,
no (CnH2„0):
2n-3
(2< n <
7)
-
Axit cacboxylic dcfn chiic, no (C„H2n02):
2n-3
(2< n <
7)
-
Este
no, dcfn chtic (C„H2„02):
2n-2
(1 <
n <
5)
-
Ete darn chiic, no
(C„H2n+'20):
(n-l)(n-2)
2
(2<
n <
6)
-
Xeton dam chiic, no (C„H2„0):
(n-2)(n-3)
2

(2< n <
7)
-
Amin dctn chiic, no
(CnHzn+sN):
2n-l
(n<
5)
Dang
1: KIM
LOAI
TAG
DUNG
VOl
AXIT
A.
HCI, H2SO4
(loang)
Doi vdi hat loai
axit
tren
thi chi phdn ling voi nhUng kim loai diing
trade
H
trong
day
hoat
d6ng
hoa hoc.
K

Ca Na Mg Al Zn Cr Fe Ni Sn Pb H Cu
Hg
Ag Au
Phucfng trinh
tong
quat:
(Vdi M la kim loai)
2M
+
nH2S04
(loang)
-> M2(S04)n +
nHzl
ta luon c6
2M
+ 2nHCl ^ 2MC1„+ nHg J n„.
=
2. n^^
De
tinh khoi
luong
muoi thu
diTcfc
thi
•k Dung dich H2SO4:
m^^^i
sunfat
=
nihSn
h(7p

kim
loai
+
96 n,i^
*
Dung dich HCI:
nimudi
dorua
=
nihSn
hgp kim
loai
+ 71 nj,^
BAI TAP AP
DUNG
Bai
1: Cho l,04g hon hop hai kim loai tan hoan toan trong dung dich
H2SO4
loang du thodt ra
0,672
Ht khi H2 (dktc). Khoi luong hon hop
muoi sunfat khan thu
di/oc
la:
A.
3,92g
B. l,96g
C.3,52g
D.5,88g
HUdng

dan gidi
2 kim loai +
H2SO4I
hh muoi sunfat + H2
n„ ||f
=
0,03(mol)
Ap dung
COng
thufC: mmutfi sunfat =
mhSn
hap kim lo?i + 96 n„^
=
1,04 +
0,03.96
= 3,92 (gam)
Chon dap an A.
Bai
2: Hoa tan het 3,53 gam hon hop A gom ba kim loai Mg, Al va Fe
trong
dung dich HCl, c6
2,352
lit khi hidro thoat ra (dktc) va thu
daoc
dung dich D. Co can dung dich D, thu
difdc
m gam h6n hop muoi
khan.
Tri so ciia m la:
A.

12,405
gam B.
10,985
gam C.
11,195
gam D.
7,2575
gam
HU&ng
d&n gidi
nH |^ =
0,105(mol)
Ap dung
cong
thuTc:
mmuei doma = nihSn iwp kim loai +71 n„^
=
3,53 + 0,105 . 71=
10,985
gam
Chon ddp 6n B.
Bai
3: Hoa tan hoan to^n 2,17 gam hon hap 3 kim loai A, B, C trong
dung dich HCl da thu
diTOc
2,24 lit khi H2 (dktc) va m gam muoi. Gia
tri
cua m 1^.
A. 9,27 gam. B. 5,72 gam. C. 6,85 gam. D. 6,48 gam.
Hiidng

dan gidi
n„ =?^ = 0,l (mol)
22,4
m„us-iciorua
= nihSn h;>p kim lo^i + 71 n,,^ = 2,17 + 0,1 . 71 = 9,27 gam
Chon dap an A
Bai
4: Bok tan 1,19 gam hSn hop A gom Al, Zn b&ng dung dich HCl viTa
du
thu
difoc
dung dich X vk V lit khi Y (dktc). Co can dung dich X
di/oc
4,03 gam muoi khan. Gia tri cua V 1^. ,
A.
0,224
lit.
B.
0,448
lit.
C.
0,896
lit.
D. 1,792
lit.
Hii&ng
dan gidi
Ap dung
cong
thiJc:

m^ua-i
ciorua = nihSn hap kim loai + 71
n^^
4,03-1,19
r^r^A ^ ^^
nji^
= ^^^p—= 0,04
(mol)
^ V„^ = 0,04 . 22,4 =
0,896
lit
Chon dap an C.
Bai
5: Cho 5,35 gam h6n hop X gom Mg, Fe, Al vao 250ml dung dich Y
gom
H2SO4
0,5M va HCl IM thu
duoc
3,921it khi H2 (dktc) va dung
dich A. Co can dung dich A trong dieu kien khong c6 khong khi, thu
diTOc
m gam chat rSn khan. Gia tri cua m la
A.
20,900
gam. B.
26,225
gam. C.
26,375
gam. D.
28,600

gam.
Hitdng
dan gidi
nn^so,
=
0,25.0,5
= 0,125 mol =^ n^,, = 0,125 . 2 = 0,25 mol
HHCI
=
0,25.1
= 0,25 mol n„. = 0,25 mol
=> tong so mol H* = 0,25 + 0,25 = 0,5 mol
q
no
n„,
=1^
= 0,175 (mol)
Ta c6: 2H^ ^ H2
Ban dau: 0,5 mol
PU:
0,35 <- 0,175 mol
Sau pif: 0,15 mol
Vi
axit dir nen kho'i
\\iang
rin b^ng tong kho'i lifcfng kim loai va cdc
ion
CO
trong dung dich.
mr^nkhan = 5,35 +

(0,125.96)
+
(0,25.35,5)
=
26,225
gam
Chon dap an B
Bai
6: (DH khoi A.2010): Cho m gam hon hap bot X gom ba kim loai
Zn,
Cr, Sn c6 so mol b^ng nhau tac dung het vdi
lirotng
diT dung dich
HCl
loang, nong thu dugc dung dich Y khi Ha- Co can dung dich
Y
thu dtfdc 8,98 gam muoi khan. Neu cho m gam hon hop X tac
dung hoan toan vdi O2 (du) de tao hon hop 3 oxit thi the
tich
khi
(dktc) phan
ijrng
la
A. 2,016
lit.
B.
0,672
lit.
C. 1,344 Ht. D. 1,008 lit
Hii&ng

dan gidi
3 kim loai
tren
khi phan
ilng
vdi HCl loang nong deu hi oxi hoa
th^nh
so oxi hod +2. (X + HCl thu
di/Oc
muoi: ZnClz, CrClz, SnClg.)
- Con khi tac dung O2, Zn tao +2, Cr tao +3, Sn tao +4. (X + O2 thu
daoc
cac oxit: ZnO, CraOa,
Sn02)
- Goi so mol moi kim loai la x (mol) thi:
Vi
cac kim loai c6 so mol
bSng
nhau nen cAc muoi c6 so mol b^ng
nhau
va bang x
136x + 123x + 190x = 8,98 =>x = 0,02;
1
3
"oj =
2''^4^^'''^
2,25x
=
0,045
mol

=> Vo^ =
0,045.22,4
= 1,008 lit
Chon dap an D
Bai
7: Cho 11 gam hon hop Al va Fe tac dung het v6i dung dich HCl
thu
dirge 8,96 lit H2 (dktc). Phan
tram
khoi lugng cua Fe trong hon
hop la.
A.
49,09%.
B. 50,91%. C. 40,91%. D.
59,09%.
Hiiofng
dan gidi
Liiu
y: Kim loai tac dung vdi mot chat nao do ma sinh ra thi ti
le mol giufa kim loai va H2 luon b^ng hod tri kim log,i chia 2
So do Al -> - H2
2
X mol —> — X mol
2
Fe -4 H2
y mol y mol
"H,
=|g =
0,4(mol)
Ta lap va giai he phuong

trinh
0,1.56
27x + 56y =11
3
-X + y = 0,4
X = 0,2
y = 0,1
Fe
=
11
X
100% = 50,91%
Chon dap an B
Bai
8: (De KA.2007): Cho m gam hon hop Mg, Al vao 250ml dung dich
chufa hon hop HCl IM va
H2SO4
0,5M thu
dugc
5,32 lit khi H2 (dktc) v^
dung dich Y. Coi the
tich
dung dich khong doi. Dung dich Y c6 pH 1^
A. 1. B. 6. C. 2. D. 7.
Hiidng
dan gidi
n„^o, = 0.25 . 0,5 = 0,125 mol => n,j. = 0,125 . 2 = 0,25 mol
nHci
= 0,25 . 1 = 0,25 mol => n„. = 0,25 mol
=> tdng so mol H^ = 0,25 + 0,25 = 0,5 mol

5,32 ^ nor,,-/ — .V „ .
=
0,2375(mol)
22,4
Ta c6: 2H*
Ban dau: 0,5 mol
Pu:
0,475
<-
Sau pir:
0,025mol
Sau phan iJng axit diT H*
=1
H2
0,2375
mol
I
0,025
0,25
=
0,1M
pH
= -log [H^^
Chon dap an A
Bai
9: (De KB.2007): Cho 1,67 gam hon hop 2 kim loai d 2 chu ky ke
tiep
nhau thuoc nhom IIA tac dung het vdi dung dich HCl dii, thoat
ra
0,672

lit khi H2 (dktc). Hai kim loai do la (Be = 9,Ca = 40,Mg = 24,
^ Sr = 87,Ba = 137)
A. Mg va Ca.
B.
Ca va Sr.
C. Sr va Ba.
D.
Be va Mg.
Hiictng
dan
gidi
Dat 2 kim loai 1^ A
Ti
le mol gifla kim loai va hidro b^ng ho^ tri kim loai chia 2 n6n ti le
mol la 1: 1
=
MZ2^ 0,03 (mol)
22,4
A Ha
0,03mol 0,03mol
MA =^ =
55,67
^ Ca va Sr.
Chon dap an B
Bai
10: Cho
3,87gam
h6n hop X g6m Mg va Al v^o 250ml dung dich X
gom HCl IM va
H2SO4

0,5M thu dMc dung dich B va
4,368
Ht
H2(dktc). Phan trSm khoi laong Mg va Al trong X tucfng ufng la
A. 37,21% Mg va
62,79%
Al. B.
62,79%
Mg va 37,21% Al.
C.
45,24%
Mg va
54,76%
Al. D.
54,76%
Mg va
45,24%
Al.
B^i
nay lam tifcfng tif bai 5
Bai
11: Hoa tan m(g) h6n hop Zn va Fe can vCra du 11 dung dich HCl
3,65M (d = l,19g/ml) thu
dugc
mot chat khi va
1250g
dung dich D.
Vay m c6 gia
tri:
A.

65,63
(g) B. 61,63 (g) C.
63,65
(g) D. 63,61 (g)
HUdng
dan
gidi
m,,„c,
=
1000.1,19
= 1190
(g);n„c,
=
3,65.1
= 3,65 (mol)
Zn + 2HC1 ^
ZnCl2
+ H2
Fe + 2HC1 ^
FeCl2
+ H2
Ap dung DLBTKL, ta c6:
nihh.zn.Fe)
+
"^ddiici
=
"^udo
+
"^n,
=

1250 + 2(^) - 1190 ^
63,65(g)
Chon dap an C.
B. KIM LOAI TAG DyNG VC3l
HNO3,
H2SO4
(d|c)
Axit
H2SO4
(dSc) + kim loai = muoi sunfat + san pham khuf + H2O
San
phdm khii c6 thi Id SO2 (nhan 2e), H2S (nhan 8e),S (nhan 6e)
* mmuoi sunfat = "1
kin,
loai
+ molgpk.so 6 nhan. i .96
axit
HNO3
+ kim loai = muoi
nitrat
+ san pham
khijf
+ H2O
San
phdm khd c6 the la NO2 (nhan le), NO (nhan 3e), N2O (nhan
8e), N2 (nhan lOe),
NH4NO3
(nhan 8e).
* m muoi
nitrat

= HI kim
loai
+
molgpk.so
e nhan.62
Luu y: Al, Fe, Cr khong phan
ting
vcd
HNO3
dSc nguoi va
H2SO4
dac nguoi.
BAI TAP VAN
DUNG
Bai
1: Cho 1,86 gam hon hdp Al va Mg tac dung vdi dung dich
HNO3
loang dtf thi thu
di/cJc
560 ml lit khi N2O (dktc, san pham khuf duy
nhat) bay ra. Khoi liTOng muoi
nitrat
tao ra trong dung dich la:
A. 40,5 gam. B. 14,62 gam. C. 24,16 gam. D. 14,26 gam.
HtCclng dan
gidi
n^o =^^-0,025mol
N,o 22,4
Ap dung
cong

thuTc: m
musl
nitrat
= m
kim
loai
+ molspk-so e nhan.62
=
1,86 + 0,025.8.62 = 14,26 gam
Chon dap an D
Bai
2: Hoa tan hoan toan 9,94 gam hon hop X gom Al, Fe, Cu trong
lugng du dung dich
HNO3
thu
dugc
3,584 lit khi NO duy nhat (dktc).
Tong khoi lugng muoi khan tao thanh la:
A. 39,7gam B. 29,7gam C. 39,3gam D. 27,7gam
Hiidng
dan
gidi
n„„=||M.o,16n.o.
Ap dung
cong
thUc: mmuoi
nitrat
= m kim
loai
+ molspk.so e nhan.62

=
9,94 + 0,16.3.62 = 39,7 gam
Chon dap an A
Bai
3: Hoa tan 4,97 gam
hon
hap
Al,
Cu, Fe
trong dung dich HNOr,
loang
dii
thu dUcfc
1,792 Ut
khi NO(dktc).
Tong
khoi luong muoi khan
tao
thanh:
A.
19,85
gam
B. 26,5
gam
C. 39,7
gam
D. 40,2
gam
Bai
nay

gidi
tuang
tii bai 2.
Bai
4: Hoa
tan hoan toan
13,68
gam hon
hdp X
gom Al,
Cu, Fe
bSng
dung dich HNO3 loang, diX thu diroc
1,568 lit
khi N2O (dktc)
va
dung
dich
chuTa
m
gam muoi.
Gia
tri cua
m
la.
A.
48,40
gam.
B. 31,04
gam.

C. 57,08
gam.
D. 62,70
gam.
Bai
nay
gidi
tuang
tu bai 2.
Bai
5: (DH
khoi
A.2010):
Nung
2,23
gam hon
hop X
gom cac kim loai
Fe, Al,
Zn, Mg
trong oxi,
sau
mot thdi gian thu difdc
2,71
gam hon
hop
Y. Hoa
tan hoan toan
Y
vao dung dich HNO3 (da), thu dtfOc

0,672
Ht
khi
NO
(san pham khuf duy nhat,
d
dktc).
So
mol HNO3
da
phan
ufng
la
A.
0,12. B. 0,14. C. 0,16.
D.
0,18.
HiCdng
ddn gidi
Tom
t^t:
Fe, Al, Zn, Mg
+
O2
-> hh ; hh +
HNO3
->
muoi
+ NO
'

' ^'la^
0.672
(l)«5|^=0,03(mol)
Bao toan khoi liTOng:
mo
=
m,,
-
m„
= 2,71 - 2,23 = 0,48(g) ->
n^
- ^ = 0,03
(mol)
Ta
CO
bao toan nguyen
to N:
^ '^HNOj
~
^UNOj (pu O
trong
oxit)
^HNOj (pu oxi ho4
-
khut)
^NO
Mat khac:
HHNO^ (pu
o
t^ng oxiu

= ( cho H^O);
nHNo,
,pu
ox,
ho.
- uha, = 3.n^o ( Do N^^ —^ NO)
^^^iiNo,=
2no,oxit)
+
3nf^o+
n^o
=
^T^o(<«iM+
'^^m
=
2.0,03 + 4.0,03 = 0,18
mol
Chon
dap
an D
Bai
6: (BH
khoi
B.2008): Cho 2,16 gam Mg tAc
dung
vdi
dung dich
HNO3 (du).
Sau
khi cac phan iJng xay

ra
hoan toan, thu duoc
0,896 lit
NO
(d
dktc)
va
dung dich
X.
Khoi luong muoi khan thu diTOc khi lam
bay hoi dung dich
X
la
A. 8,88 gam
B.13,92
gam
C. 6,52
gam
D.13,32
gam
Hii&ng
dan gidi
n^g
= 246 : 24 = 0,09
(moDin^o
- 0,896 : 22,4 = 0,04
(mol)
Cdch
1:
3Mg

+
8HNO3
^
3Mg(N03)2
+ 2N0 +
4H2O
<
0,06 < 0,04
0,06 <
4Mg
+
IOHNO3
0,03 >
>
4Mg(N03)2
+
NH4NO3
+
3H2O
0,03 > 0,0075
Khoi
liTcfng muoi khan thu diioc khi lam bay hoi dung dich
X
\k
m
= 0,09. 148 + 0,0075.80 = 13,92
(gam)
^
chon
B.

Cdch
2:
Diia vao dinh luat bao toan electron
Mg
0,09
2e -
2.0,09
Mg^2
0,09
N
+5
N
+5
3e -
3.0,04
8e -
8.x
N^2(N0)
0,04
N-'
NH4NO3)
X
So
mol electron dLfOc bao to^n
2.0,09 = 3.0,04 + 8x -> X = 0,0075
(mol)
Khoi lirong muoi khan thu dtfoc khi lam bay hoi dung dich
X la
m
= 0,09. 148 + 0,0075. 80 = 13,92

(gam)
->
chon
B
Bai
7: Hoa
tan
23,4
gam
G
gom Al,
Fe, Cu
bang mot lirong
vCra
du dung
dich
H2SO4
d5c,
nong, thu di/dc
15,12 lit
khi SO2 (dktc)
va
dung dich
chuTa
m
gam muoi.
Gia
tri cua
m
la.

A.
153,0
gam.
B. 95,8
gam.
C. 88,2
gam.
D. 75,8
gam.
Hii&ng
ddn gidi
15,12
i^so, =
22,4
=
0,675mol
nimuol
simfat
= HI kjn,
loai
+
molgpk.SO
6
nhan.
—
.96
2
=
23,4 + 0,675.2. \6 = 88,2 gam
Chon

dap
an C. |
Bai 8: Kok tan ho^n to^n m gam hon X gom Al, Fe, Cu vko dung dich
HNO3
dSc nong du, thu duoc dung dich Y chufa
39,99
gam muoi va
7,168 Ht khi
NO2
(dktc). Gia tri cua m la.
A.
20,15 gam. B.
30,07
gam. C.
32,28
gam. D. 19,84 gam.
Hiicfng
ddn
gidi
HMO
=^^^ =
0,32mol
"""^ 22,4
Ap
dung cong
thiJc:
m raM
nitrat
= m kim loai + molspk.so e
nhan.62

=^ m kim loai = m muoi
nitrat
" molspk-SO G
nhan.62
=
39,99
-
0,32.1.62
= 20,15 gam
Chon dap an A
Bai 9: (DHKB.
2007)
ThUc hien hai thi nghiem:
1.
Cho 3,84 gam Cu phan iJng vdi 80 ml dung dich
HNO3
IM thodt ra
Vi
lit NO.
2. Cho 3,84 gam Cu phan
ling
vdi 80 ml dung dich chufa
HNO3
IM va
H2SO4
0,5 M thoat ra V2 Ht NO.Biet NO la san pham
khijf
duy
nhat, cac the tich khi do d cCing dieu ki$n.
Quan he giOfa Vi va

V2
la
A. V2
= Vi. B.
V2
= 2Vi. C.
V2
= 2,5Vi. D.
V2
=
l,5Vi.
Hiictng
dan
gidi
TNI:
nc„=^ = 0,06 mol
b4
n„,
=0,08
mol
n^^_
= 0,08 mol
nHN03
=0.08
mol ,
3Cu + 8H" +
2NO3-
> 3Cu^^ +
2N0T
+

4H2O
Dau
bai: 0,06 0,08 0,08 -> phan iJng het
Phan iJng: 0,03 <- 0,08 -> 0,02 ^ 0,02 mol
=> Vi tirang ufng vdi 0,02 mol NO.
TN2:
ncu = 0,06 mol; nn^o, = 0,08 mol; nn^so. = 0,04 mol.
=> Tong
n^^,
= 0,16 mol; n^^. = 0,08 mol.
3Cu + 8H* + 2NO3"
>3Cu^*
+ 2NO^ +
4H2O
Dau
bai: 0,06 0,16 0,08 -> Cu va
H"^
phan iJng het
Phan iJng: 0,06 ^ 0,16 ^ 0,04 -> 0,04 mol
=>
V2 ti/ong
urng vdi 0,04 mol NO.
Nhir
vay
V2
= 2Vi. Chon dap an B.
Bai 10: (DH
khoi
B-2009):
Khi hoa tan hoan toan 0,02 mol Au bang

ntfdc
cudng
toan thi so mol HCl
phan
ilng
va so mol NO (san pham
khuf
duy
nhat)
tao
thanh
Ian
liToft
la
A.
0,03 va 0,01 B. 0,06 va 0,02
C. 0,03 va 0,02 D. 0,06 va 0,01
Hiicfng
ddn
gidi
• Ni/dc CLfdng toan la ti le 3 : 1 giufa HCl va
HNO3
Au
+ 3HC1 +
HNO3
>
AUCI3
+ NO +
2H2O
0,02 -> 0,06 -> 0,02

Chon dap an B
Bai 11: Hoa tan 3 gam hon
hgfp
A gom kim loai R hoa tri 1 va kim loai
M
hoa tri 2 vifa du vao dung
dich
chda
HNO3
va
H2SO4
va dun
nong,
thu
dtroc
2,94 gam hon hap khi B gom NO2 va
SO2.
The tich cua h5n
hop khi B la 1,344 lit
(dktc).
Khoi
luang
muoi
khan
thu
diroc
la.
A.
6,36g.
B.

7,06g.
C.
10,56g.
D.
12,26g.
HUdng
dan
gidi
^ 1^344 ^ oGmol dSt
NO2
x mol,
SO2
ymol
22,4
r46x
+ 64y = 2,94 fx = 0,05
x + y = 0,06 [y = 0,01
nimuai
= 3 + (0,05 .62 +
0,01 2
.96) = 7,06 gam
2
•
Dap an B.
Bai 12: Cho 8,3 gam h5n hcfp Al va Fe tac dung vdi dung dich
HNO3
loang dtr thi thu dircfc 45,5 gam muoi nitrat khan. The tich khi NO
(dktc,
san pham khuf duy nhat) thoat ra la:
!

A.
4,48 lit. B. 6,72 lit. C. 2,24 lit. D. 3,36 lit.
Hiicmg
dan
giai
Ap dung
cong
thiJc:
m
musi
nitrat
=
ni
,0,1 +
molspk-so
e
nhan.62
^
DNO
=
"-^-"^
~ = = 0,2 mol
so e
nhan.62
3.62
=>
=
0,2.22,4
= 4,48
lit

Chon
dap an A
Bai
13: Hoa
tan hoan toan
1,23 gam hon hop X gom Cu va Al vao
dung
dich
HNO3
dSc,
nong thu
dixoc
1,344
lit khi
NO2
(san
pham khuf
duy
nhat,
d
dktc). Phan trSm
ve
khoi liiong
cua
Cu trong hon
hop X la
A.
21,95%.
B.
78,05%.

C.
68,05%.
D.
29,15%.
Hiidng
ddn
giai
n
NO,
= MM =
0,06mol
(Dat Cu
X
mol, Al
y mol)
22,4
Cu
X
Al •
y
. Cu^^
+ 2e
2x
Al'^
+ 3e
3y
N"^
+ 3e ->
0,18
0,06

Ta
c6:
64x + 27y
=
l,23
fx = 0,015
|2x +
3y-0,06
[y = 0,01
%Cu=M15:6i.78,05%
1,23
Chon
dap an B
Bai
14: Hoa
tan
15
gam
hon hop X gom
hai kim loai Mg
va
Al
vao
dung
dich
Y gom
HNO3
va
H2SO4
dSc

thu diioc
0,1 mol moi
khi SO2,
NO,
NO2, N2O.
Phan tr5m khoi Itfong cua Al
va
Mg trong
X
Ian lacft
la
A. 63%
va
37%.
B.
36%
va
64%.
C. 50%
va
50%. D. 46%
va 54%.
HU&ng
dan
gidi
Dat HMg
= X
mol; DAI
= y
mol.

Ta c6:
Qud
trinh
oxi ho^: Mg
->
Mg^*
+ 2e Al
x
2x y
=> Tong
so
mol
e
nhudng bing
(2x + 3y).
Al'"
+ 3e
3y
Qua
trinh
khii:
+ 3e ->
N*=^
2N*^
+ 2 x 4e 2N*^
0,3
0,1 0,8 0,2
N*^
+ le N"" S"^ + 2e S*^
0,1

0,1 0,2 0,1
=> Tdng
so
mol
e
nhan bkng
1,4
mol.
24x + 27y
= 15 fx = 0,4 mol
Theo
dinh luat
bao
toan electron:
97
n 9
.%A1
=
^^^.100%
=
36%.
2x +
3y = l,4 [y = 0,2 mol
15
%Mg
=
100%
-
36%
= 64%.

Bai
15:
Cho
2,8 gam hon hop bot
kim loai
bac va
dong
tac
dung v6i dung
dich
HNO3
dSc, dtr
thi
thu
duoc
0,896
lit
khi NO2
duy
nhat
(d
dktc).
Thanh phan phan tram cua
bac va
dong
trong h6n
hop
Ian liTOt la:
A. 73%
;

27%.
B.
77,14%
;
22,86%
C.
50%; 50%.
D. 44% ; 56%
Tuang
tu bai 14.
Dang
2:
OXIT
KIM
LOAI
TAG
DUNG
VOI
AXIT
HQ, H2SO4
[bang)
Phi/dng
trinh tdng
quat
MzOn
+
2nHCl
2MCln+
nHzO
M20„

+
nH2S04
(loang)
^
M2(S04)„
+
nHzO
Cach
tmh
nhanh
cho
trie nghi?m
* Doi vdi axit H2SO4
(loang)
Khoi
li/ong muoi thu
dugc
la:
m^^ai
^unfat
=
mhSn
hap
oxukim
loai
+
n,[^so,
-80
Doi vdi axitHCl
•

Khoi
luong muoi thu
duoc
la:
m,„„a-i
ciorua
=
nihon
hop oxit
kim
loai
+
27,5
n,,^;,
BAl
TAP
VAN
DgNG
Bai
1:
Cho
50g hon hop hot
oxit kim loai
gom
ZnO,
FeO,
Fe203,
Fe304,
MgO
tac

dung het
vdi
200ml dung dich HCl 4M (vCra
du) thu
duoc
dung dich
X.
Luong muoi
c6
trong dung dich
X
bSng
A. 79,2g
B.
78,4g
C. 72g D.
72,9g
HUcfng
ddn
gidi
Ap
dung COng thufC: nimudi
cloma
= nihon
h<7p
oxit
kim loai + 27,5 n^jp,
= 50 +
(0,2.4.27,5)
= 72 gam

Chpn dap an C.
Bai 2: Hoa tan
hoan
toan 2,81 gam hon hop A gom
FezOg,
MgO, ZnO
bkng
300ml
dung dich
H2SO4
0,1M
(vCra
du). Co can can than dung
dich thu diioc sau phan
ufng
thi thu difoc lirong muoi
sunfat
khan la:
A.
3,81 gam B. 4,81 gam C. 5,21 gam D. 4,8 gam
HU&ng
dan
gidi
Ap
dung
COng
thufC:
mmum
sunfat
=

nihSn
hop
oxit
kim loai + J^HaSOi -^^
= 2,81 +
(0,3.0,1.80)
= 5,21 gam
Chon dap an C.
Bai 3: De tac dung
vCra
du vdi
7,68g
hon hop gom FeO,
Fe304,
FegOs can
dung 260 ml dung
dich
HCl IM. Dung
dich
thu
diTOc
cho tac dyng vdi
NaOH du, ket tua thu Auac
mang
nung trong khong khi den khoi
iLfOng
khong doi
duoc
m gam chat r^n. Gia tri cua m la:
A.6g B. 7g C.8g D.9g

Hii&ng
dan
gidi
oxit kim loai + HCl > muoi
clorua
+
H2O
nnc.
= =
0,26.1
=0,26
mol
2H^ + O'-
>H20
0,26 >
0,13mol
5 6
=>
mpe
= 7,68 -
0,13.16
= 5,6 gam => ^fe == 0>l"iol
So do hop
thiJc:
2Fe ^ FeaOg
0,1 -> 0,05 mol
=> mp^^o^ =
160.0,05
= 8 gam
Chon dap an C.

Bai 4: Oxi hoa
cham
m gam Fe
ngoai
khong khi sau mot
thcJi
gian thu
duoc
12 gam hon hop X
g6m(Fe,
FeO, FezOg,
Fe304).
De hoa tan het
X, can vCra du 300 ml dung dich HCl IM, dong
thdi
giai phdng
0,672
Ht
khi (dktc). Tinh m?
A.10,08
B.8,96
C.9,84
D.10,64
n„^ = ' =
0,03mol
.nHci
= 0,3.1 = 0,3 mol
Taco:
2H^ > Hg
0,06

<-0,03
2H^+ O'- > H2O
0,24
->0,12mol
m
= mx-
mo(oxit)
= 12 -
0,12.16
= 10,08 gam
Chon dap an A.
Bai 5: Hoa tan
hoan
toan 2,8 gam hon hop FeO,
Fe203
va
Fe304
can vCra
du
V ml dung dich HCl IM, thu
di/oc
dung dich X. Cho tii tii dung dich
NaOH dir vao dung dich X thu dugfc ket tua Y. Nung Y trong khong
khi
den khoi
liiOng
khong doi thu dirge 3 gam
chat
r^n. Tinh V?
A.

87,5 ml B.125 ml
C.62,5
ml D.175 ml
Hii&ng
dan
gidi
-4
chat
rin la
Fe203
nFe,03
=Y|^=
0,01875
mol
So do hop thufc: 2Fe
Fe203
0,0375
<-
0,01875
=>
mo,ox,t)
= 2,8 -
(0,0375.56)
= 0,7 gam
no,oxit)=^-0,04375mol
2H^
+ O'- > H2O
0,0875
<-
0,04375

=> V = 87,5 ml
Chon dap an A.
Bai 6: Hoa tan
hoan
toan 5,4 gam mot oxit sSt vao dung dich
HNO3
dii
thu
dagc
1,456 lit hon hop NO va
NO2
(dktc va khong con san pham
khuf nao
khac).
Sau phan ufng khoi
lUOng
dung dich tang len 2,49 gam
so vdi ban dau.
Cong
thufc cua oxit s^t va so mol
HNO3
phan ufng la:
A.
FeO va 0,74 mol B.
Fe304
va 0,29 mol
C. FeO va 0,29 mol D.
Fe304
va 0,75 mol
Hiidng ddn giai

mkhi
= 5,4 - 2,49 = 2,91 gam (NO x mol va NO2 y mol).
1,456
Lap he:
x + y =
22,4
30x + 46y = 2,91
=
0,065
Fe -> Fe + 3e
a ^ 3a
NO3' + 3e ^ NO
0,015 <-
0,005
2-
X =
0,005
mol
y = 0,06 mol
O + 2e ^ O
b ^ 2b
NO3- + le ^ NO2
0,06 <- 0,06
56a + 16b = 5,4 , , ^ ,
Tac6:<^
=> {a = b =
0,075
mol
[3a-2b
= 0,06 + 0,015 ^

J ^ MZ5 ^ 1 oxit sat la FeO
no
0,075
1
FeO
Fe(N03)3
0,075 0,075
BTNT NitO: njjfHNo,) - 3nN(Fe(N03)3 +
^N(NO)
+
^NCNO^)
=
3.0,075
+
0,005
-t- 0,06 = 0,29 mol
n„N03
=
nj,,HN03)
^ ^imo, = 0.29 mol.
Chon
dap an C.
Dang
3:
KHLf
CAC
GXIT
KIM
LOAI
BANG

(CO. C. Ha. Al)
A.
LfTHUYET
I.
Phxftfng trinh phan i^ng
tong
quat:
Dieu kien: (M Id kim
lo<^i
diKng sau Al trong day di?n hod)
- Khi khuf oxit kim loai
bSng
cac
chat
khuf CO (H2) thi CO (H2) lay oxi
ctja oxit kim loai ra khoi oxit.
* Doi vdi trudng hop
chat
khuf la CO, H2 thi
Thuc
chat
phan
ufng khuf cac oxit tren la
CO +
0(trong
oxit) > CO2
H2 +
0(trong
oxit) > H2O
Nen ta luon c6:

no,t„,„g„,it)
=
"H,
= n„^o
>
nctrongoxit) = ^co = ^co,
B. BAI TAP AP
DUNG
Bai
1:
Khijf
hoan
toan 17,6g hon hap gom Fe, FeO,
FegOs,
can 4,48 lit
H2
(dktc). Khoi
luong
sat thu
di/oc
la:
A. 14,5 g B. 15,5 g C. 14,4 g D. 16,5 g
Hitdng ddn gidi
4 48
no(trongoxit)
= "^11, = »H,0 = = 0.2
HlOl
mo = 16 . 0,2 = 3,2 g
mpe = 17,6 - 3,2 = 14,4 g
Chon

dap dn C.
Bai
2: H6n hop A gom sSt va oxit sSt c6 khoi luofng 2,6 g . Cho khi CO di
qua A dun
nong,
khi di ra sau
phan
ufng
di/gfc
dan vao binh diing nirdfc voi
trong diT, thu
diTcfc
lOg ket tua trSng. Khoi Itfcfng sSt trong A la:
A. 1 g B. 1,1 g C. 1,2 g D. 2,1 g
Hi£&ng
ddn gidi
Ca(0H)2 + CO2 ->
CaCOsi
+ H2O
10
•
ncaco3
= ^00, = ^co = — = 0,1 mol
t •
J^odrongoxit)
= = n^o^ =0,1 mol
Khoi li/gfng sit trong h5n hap A la: mpe = 2,6 - 16 . 0,1 = 1 g.
Chon
dap an A.
Bai

3: Cho V lit
(dktc)
khi Hg di qua hot CuO dun
nong,
thu
duoc
32 g Cu.
Neu cho V lit H2 di qua hot FeO dun
nong
thi ItfOng Fe thu
duoc
la:
A. 24g B. 26g C. 28g
D.30g
HUitng dan gidi
32
= ncu = npe = — = 0,5 mol
D4
mpe = 56 . 0,5 = 28 g
Chon
dan an C.
Bai
4: Khuf ho^n toan 32g hon hop CuO va
FezOg
b^ng khi H2, thay tao
ra
9 g nude. Khoi lirong h6n hop kim loai thu
duoc
la:
A. 12 g B. 16g C. 24 g D. 26 g.

Hitdng
d&n gidi
9
nodrong oxit) = = n„^o = = "^'^
niO (trong oxit) = 16 . 0,5 = 8g
m
kim io,i = 32 - 8 = 24 g
Chon
dap an C.
Bai 5: Cho khi CO qua ong diTng a (g) hon hop gom CuO,
Fe.304,
FeO,
AI2O3 nung
nong.
Khi thoat ra
duoc
cho vao niidc vol trong du thay c6
30g ket tua trang. Sau phan ufng, chat rSn trong ong suf c6 khoi luong
202g.
Khoi lirong a (g) cua hon hop cac oxit ban dau la:
A.
200,8g
B.
216,8g
C.
206,8g
D.
103,4g.
HU&ng
dan gidi

• a = mchat r^n + ^lodrong oxit)
Ca(0H)2 + CO2 ^
CaCOgi
+ H2O
30
nc«co3
= "co, = "c;o = — = 0,3 mol
• nourongoxit) = ^co = ^00, = 0.3 "^ol;
ma
= 202 +
0,3.16
=
206,8
g '
Chon
dap an C.
Bai
6: (CD
-2009):
Khuf
hoan
toan mot oxit sat X d nhiet do cao can
viia
du V lit khi CO (d dktc), sau phan ufng thu
dugc
0,84 gam Fe va
0,02 mol khi CO2.
Cong
thufc cua X va gia tri V Ian luot la.
A. FeO va

0,224
B.
FeaOg
va
0,448
C. Fe304 va
0,448
D. Fe304 va
0,224.
HiCdng
dan gidi
0,84
56
x 0,015 3
no(irongoxit)
= "^co = "co, = 0,02 mol) npe = -b^ = 0,015 (mol)
Fe.Oy:
- = ^^^^ = - X la Fe304
' y 0,02 4
V =
0,02.22,4
=
0,448
(lit)
Bai
7: Nung
nong
hon hop X gom PbO v^ FeO vdi mot luong C vijfa du.
Sau khi phan ufng xay ra
hoan

toan. thu dUoc hon hop chat r^n Y va
khi
khong mau Z. Dem can hon hop rSn Y thay khoi liTOng giam 4,8g
so vdi hon hop X. Cho hon hop Y tac dung vdi dung dich HCl duf, thu
di/oc
chat khi A. Sue khi Z vao dung dieh
nude
voi trong dii
di/oc
ket
tua
trSng. The tich khi A (dktc) va khoi liTOng ket tua thu diroc la:
A. 6,72 lit va 15g B. 3,36 lit va 30g
C. 6,72 lit va 30g D. 3,36 lit va 15g.
Hitdng
dan gidi
2PbO
+
C—^2Pb
+ C02
2FeO
+
C—^2Fe
+ C02
Khoi
li/ong chat r^n Y giam so vdi h6n hop X 1^ khoi li/Ong O trong
oxit da bi C lay di tao CO2.
4 8
=> mo = 4,8g => no - -V = 0,3 (mol) ^
n^j,^

= n^hv = - 0,3 (mol)
lb
1
"co,
^no
=0,15(mol)
Pb + 2 HCl->
PbClz
+ H2
Fe + 2 HCl ^ Fe + H2
• n„^ =
n^hY
=0,3 (mol) ^ V,,^ =
0,3.22,4
= 6,72 (lit)
CO2
+ Ca(0H)2 ^ CaC03 + H2O
ncaco3
= nco, = 0,15 (mol) => mc^co, =
0,15.100
= 15(g)
Vay dap an diing la A.
Bai
8: Cho 0,3 mol Fe^Oy tham gia phan ufng nhiet nhom thay tao ra 0,4
mol AI2O3.
Cong
thufc oxit ski la:
A. FeO B. Fe203 C. Fe304
D. Khong xac dinh diTOc vi khong cho biet so mol Fe tao ra.
Hiidng

dan gidi
Al
lay di oxi cua Fe^Oy de tao ra AI2O3. Vi vay so mol nguyen tuf O
trong
AI2O3 va trong FoxOy phai
bang
nhau.
Do do: 0,3 y = 0,4 . 3 = 1,2 => y = 4 =^ Fe304
Chon
dap an C.
Bai
9: Dot
chay
khong ho^n toan 1 luong sSt da dung het 2,24 lit O2 d
dktc, thu
dtfoc
h6n hop A gom cac oxit sat va s^t du. Khuf
hoan
toan
A
bang
khi CO diS, khi di ra sau phan ling
duoc
dSn vao binh difng
ni/dc
voi trong dU . Khoi
liiOng
ket tua thu dUOc la:
A. 10 g B. 20g C. 30g D. 40 g
Hiidng

ddn giai
Ca(0H)2
+ CO2 ^
CaCOai
+ H2O
2 24
natrong
oxit) = nco = n^o^ =
n^^co^
= -2 = 0,2 mol
mc,co3 =
100-0.2
= 20g
Chon
dap An B.
Bai
10: De
khijT
hoan
toan h6n hop FeO va ZnO thanh
kirn
loai can 2,24
ht
H2 (ot dktc).Neu dem hon hop kim loai thu
di/ac
hoa tan
hoan
toan
vao axit HCl thi the tich khi H2 (dktc) thu
dUcJc

la:
A. 4,48 lit B. 1,12 lit
C.3,36
lit
D.2,24
lit
Hii&ng
dan giai
2.24 „ , ,
nhhoxit =
J^Hj
= n(hh kim
loai)
= . = u,i moi
Khi
hoa tan h6n hgp kim loai vao axit thi:
n„ = n hh kim
loai
= 0,lmol
VH^=
22,4.0,1
= 2,24 lit
Chon
dap an D.
Bai
11: Thoi mot luong khi CO dtf di qua ong dung hon hap hai oxit
Fe304 va CuO nung
nong
den khi phan ufng xay ra
hoan

toan thu
dugfc
2,32 g hon hop kim loai. Khi thoat ra
duac
dua vao binh dung dung
dich
Ca(0H)2
du thay c6 5g ket tua
tring.
Khoi lugfng h5n hap hai
oxit kim loai ban dau la:
A. 3,12g B. 3,21g C. 4g D. 4,2g
Hii&ng
dan giai
nO(trong oxit) = Hco = HCQ^ ^
"caCOa
=
0,05(mol)
moxitkimio^i = nikim io,i + m oxi = 2,32 +
0,05.16
= 3,12 (g)
Chon
dap dn A.
Bai
12: Khuf
39,2g
mot hon hop A gom
FegOa
va FeO b&ng khi CO thu
duoc

hon hop B gom FeO va Fe. B tan vUa du trong 2,5 lit dung dich
H2SO4 0,2M cho ra 4,48 lit khi (dktc). Tinh khoi luong
FezOg
va FeO
trong hon hop A.
A. 32g
FezOs;
7,2g FeO B. 16g Fe203;
23,2g
FeO
C. 18g
FeaOg;
21,2g FeO D. 20g Fe203; 19,2g FeO.
HUdng
dan giai
^ . 1 - , .
[FeoOg
:
x(mol)
Goi hon hap
A<^
^ ' ^ ^ 160x + 72y = 39,2
FeO
:
y(mol)
(1)
(2)
TCr (1), (2)
mp,^o^
=0,2.160

= 32 (g);
Hon hgp B + H2SO4: FeO + H2SO4 FeS04 + H2O
Fe + H2SO4 ^ FeS04 + H2
n„^so^
=0,2.2,5
= 0,5 (mol);
Ap dung DLBTNT Fe trong hai hon hgp A va B ta c6:
160x + 72y = 39,2
"2x + y = 0,5
x = 0,2
,y = o,i
mp^o
=
0,1.72
= 7,2 (g).
Vay dap an dung la A.
Bai
13: De khuf
hoan
toan 45 gam hon hgp gom CuO, FeO, Fe304, Fe va
MgO can dung vCra du 8,4 lit CO d (dktc). Khoi lugng
chat
rin thu
dugc
sau phan u"ng la:
A. 39g B. 38g C. 24g D. 42g
Hii&ng
ddn giai
Khoi
lugng

chat
r^n sau phan iJng la: 45 -
16 ^^
= 39g
Chon
dap an A.
Bai
14: Khuf
hoan
to^n 17,6 gam hon hgp X gom Fe, FeO, Fe203 can
2,24 lit CO (or dktc). Klio" lugng sSt thu
dugc
la
A. 5,6 gam. B. 6,72 gam. C. 16,0 gam. D. 8,0 gam.
Hii&ng
dan gidi
2 24
Khoi
luong chat ran sau phan ufng la: 17,6 - 16 ~- = 16,0 gam
Chon dap an C.
Bai
15: De
khil
hoan to^n 30 gam hon hop CuO, FeO, FezOg, Fe304,
MgO can dung 5,6 lit khi CO (6 dktc). Khoi lifong chat ran sau phan
ufng
la
A. 28 gam. B. 26 gam. C. 22 gam. D. 24 gam.
HUdng
dan gidi

5 6
Khoi
liiang
chat ran sau phan ufng la: 30 - 16-^^ = 26 gam
Chon dap an B.
Bai
16: Dan tCf t\i V lit khi CO (6 dktc) di qua mot ong sil
difng lUdng
dii
hon hop ran gom CuO,
FezOs
(d nhiet do cao). Sau khi cac phan ufng
xay ra hoan toan, thu
duoc
khi X. Din toan bo khi X d
tren
vao
lUOng
diT
dung
dich Ca(0H)2 thi tao thanh 4 gam ket tua. Gia tri cua V la
A. 1,120
lit.
B.
0,896
lit.
C.
0,448
lit.
D.

0,224
lit.
Hii&ng
dan gidi
nco = nco, =
nc«co,
= ^ = ^'^^
=> Vco = 0,04. 22,4 =
0,896
lit
Chon dap an B.
Bai
17: Thoi mot luong khi CO di qua ong suf
ditog
m gam hon hcfp
Fe304 va CuO nung nong thu
diTOc
2,32 gam hon hap ran. Toan bo
khi
thoat ra cho hap thu het vao binh dung dung dich Ca(0H)2 du thu
dtfoc
5 gam ket tua. Gia tri cua m la:
A. 3,22 gam. B. 3,12 gam. C. 4,0 gam. D. 4,2 gam.
Hii&ng
dan gidi
m hSnhop = 2,32 + 16. = 3,12 gam
Chon dap an B.
Bai
18 Cho dong khi CO du di qua hon hap (X) chufa 31,9 gam gom
AI2O3,

ZnO, FeO va CaO thi thu
difcfc
28,7 gam hon hgp chat ran (Y).
Cho toan bo h6n hop chat ran (Y) tac dung vdi dung dich HCl dif thu
duoc
V lit H2 (dkc). Gia tri V la
A. 5.60
lit.
B. 4,48
lit.
C. 6,72
lit.
D. 2,24 lit.
Hii&ng
dan gidi
Khoi
luong nguyen tuf oxi = do giam cua chat r^n
m
= 31,9 - 28,7 = 3,2 g
3 2
So mol nguyen tuf O = ^ = 0,2 mol= so mol CO
16
CO
CO2 + 2e
(ZnO + 2e ^ Zn ^ Zn^* + 2e
FeO
+ 2e -> Fe ^
Fe^*
+ 2e)
2W + 2e -> H2

Vay so mol nguyen tuf O = so mol CO = so mol H2 = 0,2 mol
Chon dap an B.
C. BAI TAP VAN
DUNG
Bai
1: Cho luong khi CO di qua m gam Fe203 dun ndng, thu
di/oc
39,2
gam h6n hop gom bon chat ran la sat kim loai va ba oxit cua n6,
dong
thcfi
CO
hon hop khi thoat ra. Cho h6n hop khi nay hap thu vao
dung dich
nudc
voi trong c6 dii, thi thu
di/Oc
55 gam ket tua. Tri so
cua m la:
A. 48 gam B. 40 gam C. 64 gam
D.
Tat ca deu sai, vi se khong xac dinh
duoc.
Bai
2: Cho luong khi H2 c6 dif di qua ong suf c6 chufa 20 gam hon hop A
gom MgO va CuO nung nong. Sau khi phan iJng hoan toan, dem can
lai,
thay khoi liXOng chat ran giam 3,2 gam. Khoi liicfng moi cha't
trong
hon hop A la:

A. 2gam;
18gam
B. 4gam;
16gam
C. 6gam;
14gam
D. 8gam; 12gam.
Bai
3: Cho luong khi CO (du) di qua 9,1 gam hon hop gom CuO va AI2O3
nung
nong den khi phan ufng hoan toan, thu
dUdc
8,3 gam chat ran.
Khoi
luong CuO c6 trong hon hop ban dau la
A. 0,8 gam. B. 8,3 gam. C. 2,0 gam. D. 4,0 gam.
Dang 4: KIM LOAI TAG
DUNG
VOI
DUNG
DICH
MUOl
Ca'^ Na" Mg'^ Al'" Zn'* Cr'*
Fe^*
Ni'* Sn'* Pb'*
Fe^*
H* Cu'*
Fe'*
Ag* Au'*
__ •

Tinh
oxi hod ion kim log,i tang
K
Ca Na Mg Al Zn Cr Fe Ni Sn Pb Fe H Cu
Fe'*
Ag Au
. —— •
Tinh
khvC kim loi^i gidm
- Chieu phan ufng: Chat oxi hod m^nh + Chat
khii
mg,nh -> Chat
oxi hod yeu + Chat khvi yeu
PT:
Cu'* + Fe Fe'* + Cu
Dicing
bai tap nay can lUu y den quy tdc a
Ddu
a cang Ian khd ndng phdn ling xdy ra cdng manh
BAI TAP AP DUNG
Bai
1: Cho O,lmol Fe vao 500 ml dung
dich
AgNOs IM thi dung
dich
thu
difOc
chufa:
A.
AgNOg B. AgNOs va Fe(N03)2

C.
Fe(N03)3 D. AgNOa va Fe(N03)3
Hitcfng
dan gidi
"AgNo^ = 0.5-1 = 0,5 mol
Fe +
2AgN03
^ Fe(N03)2 + 2Ag
0.1 0.2 0.1 0.2
Fe(N03)2
+
AgNOa
Fe(N03)3
+ Ag
0,1 0,1
nAgNo,
= 0,5 - 0,3 = 0,2 mol dif
Vay muoi gom c6 Fe(N03)3 va AgNOs dii
Bai
2: (DH khoi B.2009): Cho 2,24 gam hot sSt vao 200 ml dung dich
chufa hon hop gom AgNOs 0,1M va Cu(N03)2 0,5M. Sau khi cac phan
ufng
xay ra hoan toan, thu
diTOc
dung dich X va m gam chat ran Y.
Gia tri ciia m la.
A.
2,80. B. 4,08. C. 2,16. D. 0,64.
Hiidng
dan gidi

nFe
= 0,04 mol, n^^. = 0,02 mol, n^^^. = 0,1 mol
Chat khuf manh nhat se tac dung vdi chat oxi hoa manh nhat
triTdc:
Fe + 2Ag*^Fe'* + 2Ag
0,01 0,02 0,02mol
Fe + Cu'* ^ Fe'* + Cu
0,03 0,03 0,03 mol
m
=
0,02.108
+
0,03.64
=
4,08g.
Chon dap an B.
Bai
3: (CD -2009): Cho m gam Mg vao dung dich chufa 0,12 mol FeCls.
Sau khi phan ufng xay ra hoan toan thu
dtfoc
3,36 gam chat rkn. Gia
tri
cua m la
A.
5,04 B. 4,32 C. 2,88 D. 2,16.
HiCdng
dan gidi
npe = ^ = 0,06 mol
56
Mg

+
2FeCl3
MgCl2
+
2FeCl2
0,06 -> 0,12 ^ 0,12
Mg
+
FeClz
^
MgCl2
+ Fe
0,06 0,06 <- 0,06 (mol)
m
= (0,06 +
0,06).24
= 2,88 (g). Chon dap an C.
Bai
4: (DH khoi A.2010): Cho 19,3 gam hon hop bot Zn va Cu c6 ti le mol
tuong
ijfng
la 1: 2 vao dung dich chufa 0,2 mol
Fe2(S04)3.
Sau khi cac phan
ufng
xay ra hoan toan, thu
duoc
m gam kim loai. Gia tri cua m la
A.
6,40g

B.
16,53g
C.
12,00g
D.
12,80g.
Hii&ng
dan
gidi
Goi X la so mol cua Zn thi so mol cua Cu la 2x
65x +
64.2x
= 19,3
^ X = 0,1 (nzn = 0,1 mol ; ncu = 0,2 mol)
n^,^,.
= 0,2.2 = 0,4 mol
Ta c6: 2Fe'^ + Zn > 2Fe^" + Zn^^
0,2 <- 0,1 mol
2Fe^^ + Cu > 2Fe^^ + Cu^^
0,2 0,1 mol
ncucondu
= 0,1 mol
Khoi
luong kim loai con lai la khoi ItfOng cua Cu:
0,1.64
= 6,4g
Chon dap an A.
Bai 5: (CD -
2009;
Cho mi gam Al vao 100 ml dung dich gom Cu(N03)2

0,3M va AgNOs 0,3M. Sau khi cac phan ufng xay ra
hoan
toan thi thu
dtfgc m2 gam
chat
rAn X. Neu cho m2 gam X tac dung vdi lugng du
dung dich HCl thi thu
duac
0,336
lit khi (d dktc). Gia tri cua mi va mg
Ian lUOt la
A.
8,10 va 5,43 B. 1,08 va 5,16
C. 0,54 va 5,16 D. 1,08 va 5,43.
Hi^&ng
dan
gidi
Theo
bai ra Al con
diX:
2A1 + 6HC1 ->
2AICI3
+
3H2
0,01 0,015
Al
Al^^
+ 3e Ag^ + le Ag Cu^^ + 2e ^ Cu
X 3x 0,03 0,03 0,03 0,03 0,06 0,03
Ap

dung dinh luat bao toan e: 3x = 0,03 + 0,06 => x = 0,03 mol.
z:>
Tong so mol Al =
0,04(mol)
mi =
27.0,04
= 1,08 (g)
=> m2 = mAi + mcu +
niAg
=
0,01.27
+
0,03.108
+
0,03.64
= 5,43 (g)
Chon dap an D
Bai 6: (DH
KA.2008):
Cho hon hdp bot gom 2,7 gam Al va 5,6 gam Fe
vao 550 ml dung dich AgNOa IM. Sau khi cac phan ufng xay ra
hoan
toan, thu dufcfc m gam
chat
r^n.
Gia tri cua m la
(biet
thuf tu trong
day the dien hoa:
Fe^VFe^'

diJng
trUdc AgVAg).
A.
64,8 B. 54,0 C. 59,4 D. 32,4
Hii&ng
dan
gidi
Al
+ 3Ag* ^ Al'^ + 3Ag
0,1 0,3 0,3 mol
Sau phan ufng vdi Al, Ag* c6n 0,55 - 0,3 = 0,25 mol dung phan ufng
vdi
Fe
Fe + 2Ag^ Fe^* + 2Ag
0,1 ^ 0,2 ^ 0,1 ^ 0,2
Dir
Ag^= 0,25 - 0,2 = 0,05 mol
Fe'" + Ag" ^ Fe^* + Ag
0,05 -> 0,05
Sau cac phan
ilng
chat
ran la Ag c6 so mol 0,3 + 0,2 + 0,05 = 0,55 mol.
Khoi
liigfng Ag =
0,55.108
= 59,4 gam
Chon dap an C.
Bai 7: Nhiing mot thanh kem va mot thanh sat vao cung mot dung dich
CUSO4.

Sau mot
thdi
gian lay hai thanh kim loai ra thay trong dung
dich con lai c6 nong do mol
ZnS04
hang
2,5 Ian nong do mol
FeS04.
Mat
khac,
khoi li/ofng dung dich giam 2,2g. I^oi Itrofng dong bam len
thanh kem va thanh sSt Ian lucft la:
A.
12,8g;
32g B. 64g;
25,6g
C. 32g; 12,8g D.
25,6g;
64g.
HiC&ng
dan
gidi
CM(ZnS04)
=
2,5CM(FeS04)
^
n^^^o,
= '^MF.SO,
Zn
+

CUSO4
ZnS04
+ Cui
2,5x 2,5x 2,5x 2,5x
Fe +
CUSO4
-^FeS04
+ Cui
X <— X <— x —> X
Do giam khoi lagfng cua dung dich la:
Am
=
mcu
(Mm)
- nizn
(tan)
"
"IPe
(tan)
o 2,2 =
64(2,5x
+ x)
-65.2,5x-
56x
Ax
= 0,4 (mol)
Vay mcu b^m len tiianh Zn = 64g; mcu bAm len
thanh
Fe =
25,6g.

Chon dap an B.
Bai 8: Nhiing mot thanh graphit dUdc phu mot Idp kim loai hoa tri II
vao dung dich
CUSO4
dtf. Sau phan ufng khoi iMng cua thanh graphit
giam di
0,24g.
Cung thanh graphit nay neu diTdc nhiing vao dung dich
AgNOa thi khi phan uTng xong khoi luang thanh graphit tang len
0,52g.
Kim loai hoa tri II la kim loai nao sau day:
A.
Pb B. Cd C. Fe D. Sn
Hiidng
dan
gidi
Goi kim loai c6 hoa tri II do la M c6 khoi lifcfng m(g)
M
+ Cu^^ ^ M'" + Cui
1 mol 1 mol —> giam M - 64 (g)
-^^(mol) 4-Amgian, = 0,24 (g)
M
- 64
M
+ 2 Ag^ -> 2 Agi
1
mol 2 mol
tSng
2.108 - M = 216 - M (g)
(mol) ^

Amtang
= 0,52 (g)
Vi
cung mot thanh graphit tham gia phan
iJng
nen:
_a2^ = _M2_^M = 112
M-64 216-M
Vay dap an
diing
la B: Cd.
Bai 9: Ngam mot thanh Cu trong dung dich c6
chiJa
0,04 mol AgNOa,
sau mot
thdi
gian lay thanh kim loai ra thay khoi
li/dng
tSng
hdn so
vdfi
luc dau la 2,28 gam. Coi toan bo kim loai sinh ra deu bam het vao
thanh Cu. So mol AgNOs con lai trong dung dich la
A.
0,01. B.
0,005.
C. 0,02. D.
0,015.
HUdng
dan

gidi
Cu +
2AgN03
-4 Cu(N03)2 + 2Ag
X 2x 2x
108.2X
- 64x = 2,28
X
=
0,015mol
n^^No^
da
=0,04
-
(0,015.2)
=0,01 mol
Chon dap an A.
Bai 10: Ngam mot vat
bSng
dong c6 khoi
li/cJng
15 gam trong 340 gam
dung dich AgNOs 6%. Sau mot
thdi
gian lay vat ra thay khoi luong
AgNOa trong dung dich giam 25%. Khoi lUcfng cua vat sau phan
iJng
la
A.
3,24 gam. B. 2,28 gam. C. 17,28 gam. D. 24,12 gam.

HiCctng
dan
giai
340.6
,
HAgNO,
(ban dSu, = ^^Q^QQ = 0.1^ mol;
25
nA.NO3<«=0,12.—
=0,03
mol.
Cu +
2AgN03
>
Cu(N03)2
+ 2Agi
0,015 <- 0,03 ^ 0,03 mol
nivat sau phan ilng = ^v&t ban diu + (b^m) ~ (tan)
= 15 +
(108.0,03)
-
(64.0,015)
= 17,28 gam.
Chon dap an C.
Bai 11: Hoa tan 3,23 gam hon hgtp gom
CuCl2
va
Cu(N03)2
vao niTdc
difcfc dung dich X. Nhung thanh kim loai Mg vao dung dich X den khi

dung dich mat mau xanh roi lay thanh Mg ra, can lai thay
tSng
them
0,8 gam. Khoi lU(?ng muoi tao ra trong dung dich la
A.
1,15 gam. B. 1,43 gam. C. 2,43 gam. D. 4,13 gam.
Hii&ng
dan
gidi
Mg
+
CuCl2
MgCla + Cu
X X
Mg
+
Cu(N03)2 ^Cu(N03)2
+ Cu
y y
Ta c6: (64 -
24).(x
+ y) = 0,8
=> 40(x + y) =0,8
=> X + y = 0,02
Khoi
lirong muoi tao thanh la 3,23 +
(0,02.24)
-
(0,02.64)
= 2,43 gam

Chon dap an C.
Bai 12: Ngii6i ta phu mot Idp bac tren mot vat b^ng dong c6 khoi lifOng
8,48 gam
bSng
each
ngam
vat do trong dung dich AgNOa. Sau mot
th6i gian \zy vat do ra khoi dung dich, rufa nhe, lam kho can
dufeJc
10
gam. Khoi luong Ag da phu tren be mSt cua vat la
A
.1,52 gam. B .2,16 gam. C. 1,08 gam. D. 3,2 gam.
Hii&ng
dan gidi
Cu +
2AgN03
-> Cu(N03)2 + 2Ag
X
2x 2x
Ta c6:
108.2x
- 64x = 10 - 8,48
X
= 0,01 mol
mAg
=
0,01.2.108
= 2,16 gam
Chon

dap an B.
Bai
13: Cho m gam hon
hcjp
hot Zn va Fe vao li/cfng du dung dich
CUSO4.
Sau khi ket thiic cac phan ufng, loc bo phan dung dich thu
duac
m gam bot r^n. Thanh phan %
theo
khoi li/ong cua Zn trong h6n
hop ban dau la
A.
90,27%.
B.
85,30%.
C.
82,20%.
D.
12,67%.
Hiidng
dan gidi
Zn + CUSO4 ->
ZnS04
+ Cui
X
X
Fe + CUSO4 ^
FeS04
+ Cui

y
y
Vi
m hon hop r^n dau b^ng m hon hdp r^n sau
Nen: 64x + 64y = 65x + 56y <=> x = 8y
% Zn = —^^-^y .100% =
90,27%
65.8y
+ 56y
Chon
dap An A.
Bai
14: Cho 1,12 gam bot Fe vh 0,24 gam bot Mg tac dung vdi 250 ml
dung dich CUSO4 xM, khuay nhe cho den khi dung dich mat mau
xanh nhan thay khoi liicfng kim loai sau phan ufng la 1,88 gam. Gia
tri
cua x la
A.
0,04M.
B.
0,06M.
C. O.IM. D.
0,025M.
Hii&ng
ddn gidi
nikim
io,i sau phan ling = 1,88 - 1,12 - 0,24 = 0,52 gam
=> Am
=0,01.(64-24)
= 0,4 gam

nicbniai
= 0,52 - 0,4 = 0,12
Mg + Cu^^ ^ Mg2" + Cu
0,01 0,01
Fe + CuS04 ^FeS04 + Cui
X
X
Ta c6: 64x - 56x = 0,12
X
= 0,015 mol
Z
^cuso,
=0,01 + 0,015 =
0,025
mol
_
0,025
Chon
dap an C.
Dang
5. CDs. SOa TAG
DUNG
VOl
DUNG
D|CH
KIEM
A.
KIM
LOAI
KIEM:

TOM TAT LI THUYET
1. Vi tri trong bang tuan hoan, cau hinh electron
Kim
loai kiem gom:
Liti
(Li),
Natri
(Na),
Kali
(K), Rubidi (Rb),
Xesi
(Cs), Franxi (Fr).
Thuoc
nhom lA
Cau
hinh electron: Li (Z=3) ls'2s* hay [He]2si
Na (Z=ll)
ls'2s'2p'3s*
hay
[Ne]3s'
K
(Z=19)
Is22s22p^3s^3p^4s' hay [Ar]4s'
Deu
CO
1
electron
d Idp
ngoai
cung

II.
Tinh
chat ho a hoc
Co tinh khut manh: M -> + e
1- Tac dung v6i phi kim:
Thi du: 4Na + 02-^ 2Na20
2Na + C]2^2NaCl
2. Tac dung vdi axit (HCl,
H2SO4
loang): tao muoi H2
Thi
du: 2Na + 2HC1 ^ 2NaCl + H2t
3.
Tac dung vdi n\i6c: tao dung dich kiem va H2
Thi du: 2Na + 2H2O 2NaOH + Hgt
III.
Dieu che:
1.
Nguyen
tSc:
khuf
ion kim loai kiem thanh nguyen tur.
2.
Phifcfng
phap:
dien phan nong
chay
muoi
halogen
hoSc

hidroxit cua
chung.
Thi
du: dieu che Na b^ng
each
dien phan
n6ng
chay
NaCl NaOH
PTDP:
2NaCl
—^^^^^
2Na + CI2
4NaOH ) 4Na + 2H2O + O2
B. MQT SO HOP
CHAT
QUAN
TRQNG
CUA KIM
LOAI
KIEM:
I.
Natri
hidroxit
- NaOH
+
Tac dung vdi axit: tao muoi va niTdc
Thi
du: NaOH + HCl ^ NaCl + H2O
+

Tac dung
vdfi
oxit axit: tao
mtioi
va
niidc
Thi
du: 2NaOH + CO2 -> Na2C03 + H2O
+
Tac dung vdi dung
dich
muoi:
Thi
du:
2NaOH
+ CUSO4 ^
Na2S04
+
Cu(0H)2i
n.
Natri
hidrocacbonat
- NaHCOg
1.
Phan
v(ng
phan
hiiy:
Thi
du:

2NaHC03
—^
Na2C03
+ COot + H2O
2. Tinh
litofng
tinh:
+
Tac dung vdi axit:
NaHCOs
+ HCl ^ NaCl + C02t + H2O
+
Tac dung vdi dung dich bazd: ?
NaHCOa + NaOH ^ Na2C03 + H2O
III.
Natri
cacbonat
- Na2C03 •*
+
Tac dung
vdi
dung
dich
axit manh:
Thi
du:
Na2C03
+ 2HC1 ^
2NaCl
+ CO2T + H2O

Muoi
cacbonat
cua kim loai kiem trong
nadc
cho moi triTdng kiem
IV.
Kali
nitrat:
KNO3
Tinh
chat:
c6 phan
ling
nhiet phan
Thi du:
2KNO3
2KNO2
+ O2
C. DANG TOAN CO2 (HOAC SO2) VAO DUNG DICH KIEM NaOH (HOAC
KOH)
Cac
pht^ofng
trinh
xay ra:
CO2 + NaOH NaHCOa (1)
CO2
+ 2NaOH ^ Na2C03 + H2O (2)
Thiic
chat
ta

dung
hai pt sau
CO2 + OH- ^ HCO3-
CO2 + 2
OH"
^ CO3'- + H2O
Ta can lap t
le
^NaOH
San phani
< 1
= 1
1 <2
= 2
> 2
• NaHCOa
• CO2 du
• NaHCOa
• NaHCOa
• NaaCOa
• NajCOi
• NajCOa
• NaOH dir
BAI TAP VAN
DUNG
Bai 1: Sue 2,24 lit khi CO2 vao
100ml
dung dich NaOH IM,
tinh
khoi

lUdng
muoi thu
duac.
HUdng
dan
gidi
^^^^ = 1 ehi tao muoi NaHCOg.
CO2
+ NaOH ^
NaHC03
0,1 moi -> 0,1 moi
^ii^ico,
=
0.1-84
= 8,4 gam.
Bai 2: Hap thu
hoan
toan 4,48 lit khi SO2 (d dktc) vao dung dich
chufa
16g NaOH thu
dugc
dung dich X. Khoi liicfng muoi tan thu
duac
trong
dung dich X la bao
nhieu?
A.20,8g
B.18,9g
C.23,0g
D.25,2g.

Hii&ng
dan
gidi
2NaOH+
SO2 ->
NazSOg
0,4 0,2 0,2
m
muoi
=
0,2.126
= 25,2 gam
Chon dap an D.
Bai
3: DSn 10 lit h5n hap khi gom N2 va Cd2 do or dktc sue vao 2 lit
dung dich Ca(OH)2 0,02M thu difgrc Ig ket tua.Tinh phan
tram
theo
the
tich
CO2 trong h5n hap khi.
A.2,24% va
15,68%
B. 2,24%
C.
15,68%
D. 2,24% va
7,84%.
Hitcfng
ddn giai

Triforng
hofp 1:
• nco^ =nc,co3 =Y^ = 0,01mol
nca.oH),
=
0,02.2
= 0,04mol
=Mi:^.100%
=
2,24%
CO.
-^Q
TrvLUns
hc/p 2: • n^o^ = -
n^^co^
=
2.0,04
- 0,01 = 0,07mol
^
0,07.22,4
^^g^^
Chon dap an A
Bai
4: Hap thu 3,36 lit SO2 (dktc) vac 0,5 lit h5n hgp gom NaOH 0,2M
va KOH 0,2M. Co can dung dich sau phan ufng thu
duc/c
khoi I'Jcfng
muoi
khan la.
A.

9,5gam
B. 13,5g C. 12,6g D. 18,3g
HU&ng
dan gidi
nwaOH
=
n^^.
= n^„_ -
0,5.0,2
= 0,1 mol
nKOH
= n^, =
n^j,,
=
0,5.0,2
= 0,1 mol
Zn^^.
= 0,1 + 0,1 = 0,2 mol
SO2
+ 0H >
HSO3
XXX
SO2
+ 20H ^ S0^-+ H2O
y 2y y
x + y = 0,15 fx = 0,1
"x + 2y = 0,2 |y = 0,05
Theo
dinh
luat

bao toan nguyen
tuT
(tdng khoi lugng ion trong muoi
bang
long khoi lugng muoi)
m^uo.
=
0,1.23
+
0,1.39
+ 0,1 .81 +
0,05.80
= 18,3 gam
Chon dap an D.
Bai
5: Cho 6,72 lit khi
CO2
(dktc) vao 380 ml dd NaOH IM, thu
duac
dd "
A.
Cho 100 ml dung dich Ba(0H)2 IM vao dung dich A
di/ac
m gam
ket
tua. Gia tri m bang:
A.
19,7g B.
15,76g
C. 59,lg

D.55,16g
Hiidng
dan gidi
"N^OH =
Hj^^,
= n^,,^ = 0,38 mol "^n^iom, = n^^^, =0,1 mol
En^j^.
= 0,38 + 0,2 = 0,58 mol nco, = - 0,3 mol
22,4
CO2
+ OH-
HCO3-
XXX
CO2
+ 20H- ->
CO3'-
+ H2O
y 2y y
Ta c6:
x + y = 0,3 fx = 0,02
x + 2y = 0,58 [y = 0,28
Ba'" + CO^- BaCOsi
0,1 0,28 0,1
Khoi
lirang
ket tua la:
mB^cOj
=
0,1.197
= 19,7 gam

Chon dap an A.
Bai
6: Dot
chay
hoan toan 0,1 mol etan roi hap thu toan bo san pham
chay
vao binh chufa 300 ml dd NaOH IM. Khoi liiong muoi thu difgrc
sau phan iJng?
A.
8,4g va 10,6g B. 84g va 106g
C.
0,84g
va l,06g D. 4,2g va 5,3g.
Hiidng
ddn gidi
Taco:
C2H6
^
2CO2
0,1 -> 0,2mol
CO2
+ NaOH -> NaHCOa
XX
X
COa +
2NaOH
^
NazCOg
+ H2O
y 2y y

rx + y-0,2 fx = 0,1
Ta c6: < => <
x + 2y = 0,3 ,.y = 0,l
V4y
-mN,HC03
=
0.1-84
= 8,4 gam
mNa,co,
-
0,1.106
= 10,6 gam
Chon
dap an A.
D.
DANG
TOAN
CO2
(HOAC
SO2) VAO
DUNG
DjCH
Ca(0H)2
hogc
Ba(0H)2.
KIM
LOAI
PHAN
NHOM
CHINH

NHOM
II
T6M TAT
LITHUYET
1. Vj tri
trong
bang
thong
tuan
hoan,
tinh
chat
v^t If
a) Vi tri
Kim
loai
phan
nhom
II gom:
Beri (Be);
Magie
(Mg); Canxi (Ca);
Stronti (Sr); Bari (Ba) va
Radi
(Ra)
Trong
cac chu ki cac
nguyen
to nay diing lien sau khi loai kiem.
b)

Tinh
chdt
vat li
- Nhiet do
nong
chay
nhiet do soi
thap
- La kim loai mem (mem hon
nhom)
- Khoi
li/Ong
rieng tiidng doi nho
2.
Tinh
chat
hoa hoc
Cac
nguyen
to
phan
nhom
chinh
nhom
II c6:
- 2
electron
hoa tri (s^)
- Co ban kinh
nguyen

tuf \6n
- La
nhOfng
chat
khuf
manh
M - 2e -> M^^
Trong
cac hap
chat
c^c
nguyen
to nay c6 so oxi hoa +2.
a) Tdc
dung
vai phi kim
- Vdi oxi khi dot
nong:
2M + O2 2M0 (M la
nguyen
tut kim
loai)
2Ca + 02^
2CaO
- Vdi CI2: M + CI2 ->
MCI2
Mg + CI2 ^
MgCl2
b) Tdc
dung

vai
axit
_ De dang
khilf
ion trong dung dich axit (HCl,
H2SO4
loang) thanh
H2 t\i do.
M
+
H2SO4
MSO4
+ Hzt
M
+ 2H^ M^* + H2t
- Doi vdi axit c6 tinh oxi hod manh nhiT
HNOs,
H2SO4.
->
N"'(N02),
N*^(NO),
N-^(NH4N03)
S^*^ ^
S^'(S02),
S-=^(H2S),
S°(S).
4M +
IOHNO3
4M(N03)2
+

NH4NO3
+
3H2O
c) Tdc
dung
vai H2O
Trong
H2O, Be khong phan urng, Mg khuf cham, cac kim loai con lai
khuf
manh.
M
+
2H2O
^
M(0H)2
+ H2t
Ca
+
2H2O
Ca(0H)2
+ H2T
d) Tdc
dung
vai
dung
dich
muoi
- Mg day cac kim loai hoat dong yeu hcfn ra khoi dung dich muoi
Mg +
CUSO4

^
MgS04
+ Cui
- Cac kim loai con lai tac dung vdi H2O trong dung dich
3.
Dieu
che
Dien
phan nong chay muoi halogenua cua chiing
MX2 Jig" phan n6ng chay ^ M + X2
X:
halogen
MQT SO HOP
CHAT
QUAN
TRONG
CUA
CANXI
1.
Canxi
oxit:
CaO
Caxi
oxit la oxit baza
- Tac dung manh liet vdi H2O tao baza
CaO
+ H2O ^
Ca(0H)2
- Tac dung vdi nhieu axit tao muoi tuang ufng
CaO

+ 2HC1 ^
CaCl2
+ H2O
- Tac dung vdi oxit axit tao muoi tuang
iJng
CaO
+ CO2 CaC03
- Canxi oxit dirge dieu che bkng phijang phap nhiet phan muoi cacbonat.
CaC03
—^CaO + CO2
2.
Canxi hidroxit:
Ca(OH)2
La
chat rSn it tan trong H2O
Dung dich
Ca(0H)2
c6
tinh
bazO yeu hon NaOH
- Tac dung vdi axit va oxit axit tao muoi tUofng ting
Ca(0H)2 + 2HC1 ^
CaCla
+
2H2O
Ca(0H)2 + CO2 ^
CaCOai
+ H2Q
Ca(0H)2 +
2CO2

^ Ca(HC03)2
,
CaCOH),
1 ^
Neu ti le mol < — tao muoi axit
CO2 2 •
Neu ti le mol
^^^^^^2
< ^ ^.^Q
jjjy5'j
trung
tinh
CO2
Neu ti le mol 9^^^k trong khoang i < ^^^^^ < 1 tao
d6ng
CO2 2
thdi
2 muoi
- Tac dung vdi dung dich muoi
Ca(0H)2
+ NaaCOs
CaCOai
+ 2NaOH
Ca'^
+ CO^- ^
CaCOai
3.
Canxi cacbonat CaCOa
Canxi
cacbonat la chat rSn mau trang khong tan trong H2O

CaCOs
la muoi cua axit yeu va khong ben
CaCOa
+ 2HC1
CaCl2
+ H2O +
COzt
CaCOa
+
2CH3COOH
->
Ca(CH3COO)2
+ H2O +
COat
ot nhiet do thap
CaCOa
tan dan trong H2O c6 CO2
CaCOa
+ H2O + CO2 -> Ca(HC03)2
4. Canxi sunfat: CaS04
CaS04
con goi la thach cao, m^u tr^ng, it tan trong H2O
CaS04.2H20:
thach cao song
2CaS04.H20: thach cao nung nho lufa
CaS04:
thach cao khan
Thach
cao nung duoc dung de due tiiOng.dung trong y hoc de bo bot
5.

Nifofc
ciJtng
1. Nx^oTc
cufng
Nadc
curng
la nUdc c6
chura
nhieu ion Ca^*, Mg^*. Nifdc khong chiJa
hoSc
chura
it nhOfng ion tren, goi la niTdc mem.
2.
Phan loai
nvCdc
cufng
Nude
curng
chia thanh 3 loai
1. Nifde ciJng tarn
thdi:
la nirdc curng c6
chura
ion
HCO3"
2.
NiXdc
curng
vinh
cijfu:

la niJdc
curng
c6 chufa ion CI"
hoSe
SO^
3.
Nifdc
curng
toan phan: La
niidc
cufng c6 chiJa dong thcfi cac ion
HCO3",
Cr
hoSe
SO^-
3.
Tac hai ciia nitofc ciifng
- Xa phong khong tan
- Vai soi mau muc nat
- Nau
thurc
an lau chin, giam mui vi
- Tao chat c&n trong noi hai lam lang phi nhien lieu
4. Cach lam mem ni^cfc
Nguyen
tSc: Lam giam nong do cac ion Ca^^ va Mg^* trong nufdc bang
each
chuyen nhij'ng ion tii do nay vao thanh phan chat khong tan.
Phirong phap: Phuong phap hoa hoc va phuong phap trao ddi ion.
a)

Phuang phap hod hoc
* Doi vdi ntrdc cufng tam
thdi.
Dun nong trUdc khi dung
Ca(HC0a)2
—^
CaCOai
+ H2O +
C02t
Loc bo chat khong tan, dirge nifdc mem
- Dung
Ca(0H)2
vto du de trung hoa
Ca(HC03)2
+
Ca(0H)2
2CaC0ai
+
2H2O
Loc bo chat khong tan dtrgc nude mem
* Doi vdi nude cufng vinh cufu va ni/dc cufng toan phan
Dung dung dich
Na2C0a,
Na3P04
CaS04
+
Na2C03
-> CaC03i +
Na2S04
Ca(HC03)2

+
Na2C03
^
CaCOgi
+
2NaHC03
Ca?*
+
CO3
^
CaC03>l
Mg'*
+
CO3
^" MgCOa
b) Phuang
phdp
trao
doi ion
Cho
nutdfc
cufng
di qua
chat trao
doi ion
(ionit)
chat
nay se hap thu
cac
ion Ca^^

va
Mg^*
the vao do la
ion Na\a duoc nude mem.
D^ng
bai tap
CO2
vao
dung dich Ca(OH)2 hoqc Ba(OH)2
thi:
Dang
1:
Biet
nc^o,,,,^
.ncaco,
"co,
^Ca(0H)2
THI
:
n^^o^
-
nc^co^
TH2
:
n^o^
-
2.nc^,o„)^
-''^c^co^
Dang
2:

Biet n^o^, nc^oH)^
tim
nc^co^
W2
o
^^CatOlllj
THI:
nco,
=
nc.co,
TH2
:
ncacos
-
•^^cMom,
~
"co^
Dang
3:
Biet nc^co^.
"00,
tim
nCaC03
_ ,
^
"Ca(OIl)o
~ •
Neu
n^o,
^

ric^co,
=^
^CMOMU
BAI TAP VAN
DUNG
Bai
1.
Hap thu toan
bo x
mol
CO2
vho
dung
dich
chiJa 0,03
mol Ca(0H)2
diroc
2
gam
ket
tua.
gia
tri
x?
A.
0,02
mol
va 0,04
mol
B. 0,02

mol
va 0,05
mol
C.
0,01
mol
va 0,03
mol
D. 0,03
mol
va 0,04
mol.
HiC&ng
dan
gidi
2
nr,ro
= = 0,02 mol
TrUang
hap 1:
rico,
=
ntettua
= 0,02
mol
TrUdng
hop 2: HCO, =
2nb«,„
-
ntettua

= 2.0,03 - 0,02 = 0,04
mol
Chon
dap
an A.
Bai
2: (DH A - 2008):
Hap thu hoan toan
4,48
lit khi
CO2
(d
dktc)
vao
500
ml
dung
dich
hon hop gom
NaOH
0,1M
va
Ba(0H)2 0,2M,
sinh
ra
m
gam
ket
tua.
Gia

tri
cua m la.
A.
19,70 g. B. 17,73 g. C. 9,85 g. D. 11,82 g.
Hiidng
dan
gidi
nNaoii
= "Na* =
"oH"
"
^'^^
^^^'^
"^u^iom,
= ^3, =0,1
mol
I
n^„.
= 0,05 + 0,2 =0,25
mol;
n^o, M| = 0,2
mol
CO2
+ OH" ^
HCO3"
XXX
CO2
+
20H-
^ CO3'" + H2O

y
2y y
x
+ y = 0,2
rx
= 0,15
"x
+ 2y = 0,25 [y = 0,05
Ba'"
+
CO3'-
^
BaCOai
0,1
0,05 0,05
Khoi
laong
ket
tua
la:
ra^^co,
=0,05 .197 = 9,85 gam
Chon
dap
an C.
Bai
3: (BH A - 2009): Cho 0,448 lit
khi
CO2
(d

dktc)
hap thu het vao
100
ml
dung
dich
chuTa hon
hop
NaOH
0,06M
va
Ba(0H)2 0,12M,
thu
duoc
m
gam
ket
tua.
Gia
tri
cua m la.
A.
1,182 g. B. 3,940 g. C. 1,970 g. D. 2,364 g.
Hit&ng
dan
gidi
nNaOH
=
n^^.
=

nQ„_
=0,006
mol;
HB^IOH),
=
n^^,,
=
0,012mol
In^jj
=0,006 +
0,012.2
= 0,03
mol;
nco, = =
0,02mol
C02 + Off ^ HCO3"
XXX
CO2 + 20H- -> COa^" + H2O
y 2y y
x + y = 0,02 fx = 0,01
' x + 2y = 0,03 ^ [y = 0,01
Ba^^
+ CO^- BaCOai
0,012 0,01 0,01
Khoi
laang ket tua la:
mg^co,
=0,01 .197 = 1,97 gam
Chon dap an C.
Bai

4: (DH A - 2007; Hap thu hoan toan
2,688
lit khi CO2 (or dktc) vao
2,5 lit dung dich Ba(0H)2 nong do a mol/1, thu difdc 15,76 gam ket
tua.
Gia tri cua a la.
A.
0,032
M. B.
0,048
M. C. 0,06 M. D. 0,04 M.
Hiiafng
d&n gidi
2,688
^,,„„,_ _15^
n^f> =
=
0,12mol;
ng^co,
= = 0,08mol
nco, >
neaco,
chi xay ra
tri/dng
hap 2
C„^,„„, =-^ = 0,04M
Chon dap an D.
Bai
5: Dan V ht CO2 (dkc) vao 300ml dung dich Ca(0H)2 0,5 M. Sau
phan

ling
duac
lOg ket tua. Gia tri cua V bSng:
A.
2,24 lit B. 3,36 lit C. 4,48 lit D. Ca A, C deu dung
Hildng
d&n gidi
TriCang
hap 1: n^o^ = nkettoa = 0,1 mol
Vco = 0,1 .22,4 = 2,24 lit
TriCdng
hap 2: n^^^ =
2nb,,o
- ntettua =
2.0,15
- 0,1 = 0,2 mol
Vco, =
0,2.22,4
= 4,48 lit
Chon dap an D.
Bai
6: Thoi V lit (dktc) CO2 vao 100 ml dung dich Ca(0H)2 IM, thu diroc
6 gam ket
tiia.
Loc bo ket tua lay dung dich thu
ducfc
dun nong lai c6
4gam ket
tiia
nufa. Gia tri V la:

A.3,136 lit B. 1,344 lit
C. 1,344 lit
hoSc
3,136 lit D. 3,36 lit
hoSc
1,12 lit.
Hit6ng
dan gidi
Ca(0H)2 + CO2 CaCOsi + H2O
0,06 <- 0,06
Ca(0H)2 +
2CO2
^ Ca(HC03)2
0,08 <- 0,04
Ca(HC03)2 —^ CaCOai + H2O + C02t
0,04 <- 0,04
Inco,
=0,08 + 0,06 = 0,14 mol =>
Vco,=
0,14 .22,4 = 3,136 lit
Chon dap an A.
Bai
7: Hap thu toan bo 0,3 mol CO2 vao dung dich chijfa 0,25 mol Ca(0H)2.
Khoi
lifofng dung dich
sau
phan uTng tang hay giam bao nhieu
gam?
A.
Tang 13,2gam B. Tang 20gam

C. Giam 16,8gam D Giam
6,8gam.
Hii&ng
dan gidi
Be
tinh khoi lixang chat sau phan i2ng tang hay giam a lay khoi liigng
vao trie khoi lilgng ra khdi dung dich niu > 0 la tang n^u < 0 la giam
Ta c6: n^o^ >
n^^^Q^^^^
nen chi xay ra TH2
J^COj
- 2nbaza ~ ^kgt
tua
nc,co,
=
2nc„oH,,
- ^co, =
2.0,25
- 0,3 = 0,2mol
Am
= mco, - ixic^co, =
0,3.44
-
0,2.100
= -6,8 gam
Chon dap an D.
Bai
8: Cho 5,6 ht hon hdp X gom Ng va CO2 (dktc) di cham qua 5 Kt
dung dich
Ca(0H)2

0,02M de phan
ilng
xay ra hoan toan thu
dUdc
5
gam ket tua.
Tinh
ti khoi hoi cua h6n hcfp X so vdi H2.
A. 18,8 B. 1,88 C. 37,6 D. 21
Hiidng
dan gidi
5
=
n^ = O.lmol
n
CRCOO
100
=
0,05mol
Truang
hap 1: n^o^ =
nkettua
= 0,05 mol
^
0,05.44
+
0,2.28
M
=
=

31,2 =^ d,/,,=^ = 15,6
n
KOH
- - ^on-
0,25 ' "'"^ 2
Khac dap an loai
Truang
hap 2: n^^^ =
2nb^,„
-
nkeuaa
= 2.0,1 - 0,05 = 0,15 mol
j^>,15.44.0,1.28
^ =3^ = 18,8
0,25 ""'"^ 2
Chon dap an A.
Bai
9: Sue CO2 vao 200 ml h5n hgfp dung dich gom KOH IM va Ba(0H)2
0,75M.
Sau khi khi hi hap thu hoan toan thay tao 23,6 gam ket tua.
Tinh
Vco^ da dung d (dktc).
A.
8,512 lit B.
2,688
lit C. 2,24 lit D. Ca A va B dung
HU&ng
dan gidi
=
0,2 mol;

ri^^mn,
=
^Ba^*
= ^'
1^"^°^
_ 23,6
'Oir
' '
-^BaCOj
~
Truang
hap 1: n^o^ = nkettta = 0,12 mol
Vco.=
0,12 .22,4 =
2,688
lit
Truang
hap 2: CO2 + OH" ->
HCO3"
XX
X
CO2 + 20H- ^
003'-+
H2O
y 2y y
y = 0,12 fx = 0,26
"x + 2y = 0,5 ^|y = 0,12
=> X "co, =0,12 + 0,26 = 0,38 mol
Vay Vco^
=0,38.22,4

= 8,512 lit
Chon dap an D.
In,,,,.
=0,2 +
0,15.2
= 0,5 mol;
197
=
0,12mol
ai
10: Sue 4,48 ht (dktc) CO2 vao 100ml hon hop dung dich gom KOH
IM
va Ba(OH)2 0,75M. Sau khi khi hi hap thu hoan toan thay tao m
gam ket tua.
Tinh
m.
A.
23,64g
B.
14,775g
C.
9,85g
D.
16,745g
Giai
tiicfng
tii
bai
tren
Bai

11: Thoi V ml (dktc) CO2 vao 300 ml dung dich Ca(0H)2 0,02M, thu
duac 0,2g ket tua. Gia tri V ml la:
A. 44,8
hoSc
89,6
B.44,8
hoSc
224 C. 224 D. 44,8
TMng
tuf cac bai
tren.
Dang
6.
MUQI
CACBONAT
TAG
DUNG
VO!
DUNG
D|CH
HCI, H2S04
loang
Nguyen tdc cua phuang phdp nay Id dua vdo sU tang, gidm khoi lugng
khi
chuyen tii chat nay sang chdt khac Ung vdi 1 mol chdt.
Pt:
XCO3
+ 2HC1 ->
XCI2
+ H2O +

CO2T
(1)
Y2(C03)3
+ 6HC1
2YCI3
+
3H2O
+3CO2T (2)
Cach
tinh
nhanh dung cho
tr&c
nghiem.
Theo
2 phuong
trinh
(1),(2) ta thay
ncf = 2n^^,_ (ma M^,. = 35,5;
M^^,.
= 60)
•^muo'i
clorua
~ ^ hh
rauoi
cacbonat -'^cOo
.(71-60)
Muoi
cacbonat
tac dung vdi dung dich
H2SO4

loang
Pt:
ACO3
+
H2SO4
(loang) ^
ASO4
+ CO2T + H2O
B2CO3
+
H2SO4
(loang) ^
A2SO4
+ CO2T + H2O
Cach
tinh
nhanh cho muoi sunfatv.
nirauo'i
sunfat —
rnmuo'i
cacbonat ^CO^
BAl TAP VAN
DUNG
Bai
1: Cho 115g hon hcrp gom
ACO3,
B2CO3,
R2CO3
tac dung het vdi dung
dich HCI thay thoat ra

0,4481
CO2 (dktc). Khoi liicfng muoi clorua tao ra
trong
dung dich la:
A.
115,22g
B.151,22g
C.
116,22g
D.
161,22g
HUcfng
dan gidi
Cach
giai 1:
ACI2
+ H^O + CO2 t
>
2BC1 + H2O + CO2 t
R2CO,
+ 2HC1 ^ 2RC1 + H.,0 + CO, t
ACO3
+ 2HC1 -
B2CO3
+ 2HC1