Lời giải đề thi học sinh giỏi hóa học 9

DE SO 1
DE THI HOC S1I\IH GiOl HOA HOC
LOfP
9, TP. HQ CHI MINH NAM HOC
1998
-
1999
Cdu
I.
Viet
3
phuang
trlnli
khdc nhau
de
dicu
die
muoi
ZnCl2.
Cdu
II
Viet
phuang trinh phdn ling
de
bieu dlen
chiiSi
bien
hoa sau:
FeCls
>
Fe(0H)2

^
FeSO,
>
Fe(N0s)2
FeCh
>
Fe(0H)3
>
FesOs
> Fe.
Cdu
III. Co 6 ong
nghiem dugc ddnh
so tii 1 den 6
chica
cdc
dung
dicli:
NaOH;
(NHJ.SO,;
NaoCOs; Ba(N03)2; PbiNO^)^; CaCl
Hay cho
bict
6'ng
mang
so ndo
dUng
chat
ndo?
Viet

phdn icng minh
hga.
Biet rdng:
a)
Dung dich
(2) cho ket
tila
trdng
vai cdc
dung dich
(1), (3), (4). ''•
b) Dung dich
(5) cho ket tua
trdng
vai cdc
dung dich
(1), (3), (4).
c)
Dung dich
(2)
khong
tgo ket
tila
vai
dung dich
(5).
d)
Dung dich
(1)
khong

tgo ket tua vai cdc
dung dich
(3), (4).
e) Dung dich
(G)
khong phdn
i(ng vai
dung dich
(5).
f) Dung dich
(5) bi
trung
hoa bdi
dung dich
HCl.
g) Dung dich
(3) tgo ket tua
trdng vai
HCl, khi dun
nong
ket tua nay se tan.
CduIV.
a)
Nong
do
dung dich
bdo hoa KCl d
40°C
la
28,57%.

Tinh
do tan cua
dung dich
KCl a
ciing nhiet
do.
»
)
Xdc
dinh lugng AgNOs tdch
ra khi Idm
Ignh
2500
gam
dung dich
AgNOs
bdo hoa a
60°C
xuong
10°C. Cho
biet
do tan cua
AgNOs
o
60°C
Id 525 gam, a 10°C Id 170 gam.
Cau
V. (A) la
dung dich
H2SO4;

(B) la
dung dich NaOH.
• Trgn
0,3 lit (B) vai 0,2 lit (A)
dugc
0,5 lit (C).
Lay
20 nd (C),
them
mot it quy tim vdo
thdy
c6 mdu
xanli.
Sau do
them
tic tit
dung dich
HCl
0,05M
tai khi quy tim ddi
thdnh
mdu tim
thdy
het 40
??il axit.
Trgn
0,2 lit (B) vai 0,3 lit (A)
dugc
0,5 lit (D).
Lay

20 ml
dung dich
(D),
them
mot it quy tim vdo
thdy
c6 mdu do.
Sau
do
them
tii
tii: dung dich NaOH
0,1M tai khi quy ddi
thdnh
mdu
tim
thdy
het 80 ml
dung dich NaOH.
Tim
nong
do
mol/l dung dich
(A) vd (B).
I
r(i
niAi
nc
TUI unr
CIMM

nifii
HHA
Hfin Q
Cdu
VI. Xdc
dinh cong
thiJtc
ciia
hai
oxit
sdt A vd B,
biet
r&iig:
•
23,2 gam (A) tan
vica
dil
trong
0,8 lit HCl IM.
•
32 gam (B) khi
k/iii
bdng
Ho tqo
thdnh
sdt vd 10,8 gam IhO.
L6\I
Cdu
I. * Zn + CI2 >
ZnCla

. ZnO
+
2HC1
>
ZnCl2
+
H2O
•
ZnS04
+ BaCl2 >
ZnCh
+
BaS04i
Cdu
11.
Fe +
2IIC1
> FeCl2 + H2t
FeCl2 +
2NaOH
>
Fe(0H)2^
+
2NaCl
(trdng
xanh)
Fe(0H)2
+
H2SO4
>

FeS04
+
2H2O
FeS04
+
Ba(N03)2
>
BaS04^
+
Fe(N03)2
2Fe
+ 3CI2 >
2FeCl3
FeCla
+
3NaOH
>
FeCOIDai
+ 3NaCl
(ndu
do) •
2Fe(OH)3
—>
Fe203
+ 3H2O
Fe203
+ 3C0 —> 2Fe + 3CO2T
Cdu
III.
Theo

cac
duf kien
de
bai neu ra,
cac lo
diing
cac
hoa chat sau:
f)
=>
Lo so 5
diing:
Na2C03
g)
Lo so 3
diing: Pb(N03)2
d)
Lo so 1
chufa: Ba(N03)2,
lo 4
chuTa
CaCl2.
a)
=> Lo 2
chijfa: (NH4)2S04.
e) =^
Lo 6
chijfa: NaOH.
Phan
irng:

(NH4)2S04
+
Ba(N03)2
>
BaS04i
+
2NH4NO3
(NH4)2S04
+
Pb(N03)2
>
PbS04i
+
2NH4NO3
(NH4)2S04
+ CaCl2 >
CaS04i
+
2NH4CI
'
Cac phan
ilng
con
lai
hoc
sinh
tii
vie't.
Cdu
rv.

a)
Goi
S la do
tan cua KCl
d
40°C.
Khoi
lUcfng dung dich thu dUgfc:
(S +
100) gam
Nong
do
phan
tram
cua chat tan trong dung dich
bao
hoa:
S
+ 100 S + 100
S
+ 100 - 3,5S » S = 40
(gam).
I
fi\l
n£
Tui
unr sluu
Rini
Hni Hnr.
a

'^)^
60"C, trong
525 + 100
bang
625
gam dung dich
c6 525
gam AgNOg
va
100 gam H2O.
Trong
2500
gam
dung dich
c6 x gam
AgNOa
va
y gam nxidc:
2500
X 525
X
=
625
=
2100 (gam)
y
=
2500
-
2100

=
400g
(nxidc)
Vay
&
60°C
trong
2500
gam
dung dich
c6 2100 gam
AgNOs
va
400 gam ni/dfc.
CJ 10°C,
CLf
100
gam ntfcJc h6a tan 170 gam AgNOa
400 gam niJcJc
hoa
tan
z
gam AgNOa
400x170
z
=
100
=
680
(gam) AgNOa.

Do
do
khoi liTgng AgNOa ket
tinh
khi lam
lanh:
2100
-
680
=
1420 (gam)
Cdu
V.
Phan ufng:
H2SO4
+
2NaOH
>
Na2S04
+
2H2O
(1)
Lan
thuf
nhat
quy
tim
c6
mau xanh chufng
to

diT NaOH. Them axit
HCl
de
trung
hoa NaOH dif:
HCl
+
NaOH
>
NaCl
+
H2O
Lan
2
lam quy tim
c6
mau
do
chufng
to
H2SO4
dir.
Them NaOH
de
trung
h5a
H2SO4
di/.
2NaOH
+

H2SO4
>
Na2S04
+
2H2O
Goi X,
y
lan lugt
la
nong
do
cua
H2SO4
va NaOH.
Theo
cac
phan ufng (1), (2),
(3) ta c6
phiiofng
trinh:
500
(2)
(3)
0,3y
- 2 X
0,2x
=
0,05
x
0,04

x
0,3x
-
0,2y
0,1 X 0,08
20
500
=
0,05
=
0,1
2
2 20
Giai
he
phuang
trinh
(a),
(b)
=>
x =
0,7M
va y =
1,1M.
Cdu
VI.
Goi
cong
thufc oxit
sat la

FexOy.
Fe^Oy
+
2yHCl
> x
FeClayx
+
yH20
0,4
(a)
(b)
(1)
(mol)
0,8
^^IfilAlflg
THI HOC
SINH
GIO
GIOI
HOA HOC
9
Ta
c6:
IIHCI
= 0,8 X 1 = 0,8 (mol)
Theo
de: 23,2
= —
(56x + 16y)
y

<=> 56x
=
42y
<=>
X
_ 3
fx
=
3
y
4
[y =
4
Vay oxit sSt (A):
Fe304
(sdt
tii
oxit)
•
FexOy
+
yH2
>
xFe
+
yH20
0,6
(mol)
0,6
(2)

Ta c6:
10,8
18
=
0,6 (mol)
x
= 2
TheodI: ^(56x + 16y] =32
« -
=
-=>•
y
^ ' y 3
[y =
3
=>
Cong
thufc oxit (B):
FeaOs-
DE
SO
2
DE
THI
CHOIV
flOl
TUYEN
KQC
SINH
GIOIHOA

HOC 9,
qUAAl
TAN
BINH
TP. HO CHJ
MINH
NAM HOC 1998
-
1S99
PHAN
A.
Li thuyet
Cdu
I. Chi dugc diing
thuoc
thii
de
nhgn
biet
cdc
muoi
sau:
NH4CI,
FeClo,
FeCls,
MgCl2, NaCl,
AICI3
(Gidi thich
vd
viet

phuang trinh phdn ling
neu c6).
Cdu
II. Viet phuang trinh phdn ling bieu dien cdc bien hoa sau:
a)
Fe
>
Fe2(S04)3
> Fe(0H)3 > FcsOa
>Fe
>
FeClz > FeCls
>FeCl2
> AgCl.
b) MgCOs > MgS04 > MgCl2 > Mg(0H)2 > MgO
>
MgCOs
> CO2 >
Ca(HC03)2
>
CaCOs
Cdu
III. Chi
dugc dung
guy
tini
vd
dung dich AgN03
c6 sdn, neu
each

phdn
biet
cdc dung dich: NaOH, NaCl, HCl, H2S,
H2SO4.
Cdu
IV.
Sdt
nguyen
chdt trong khong
khi thi
khong
bi
han
gl,
nhUng sdl
CO
tap
chdt
de
Idu
ngdy
trong khong
khi Igi
bi
han
gl. Hay
gidi thich
hien tugng nay.
a
Lfll

GIAI Oi
THI HOC
SINH GIOI
HOA HOC 9
PHAN
B.
Bai toan
Bai
1-
Co
mot hon hgp gom
Na2S04
vd
K2SO4
dugc trgn
Idn
theo
tl Ic
I
: 2 ve so
mol (1 mol
Na2S04 yd.2
mol
K2SO4).
Hoa tan hon hgp vdo
102 gam nUac
thi
thu dugc dung dich
A. Cho
1664 gam dung dich

BaCL
10% vdo dung dich
A. Lgc ket
tua,
them
H2SO4
du
vdo tiudc vUa Igc
tin
thdy
tgo ra
46,6 gam
ket
tua.
Xdc
dinh nong
do
phdn tram cua
NosSOj
vd
K2SO4
trong dung dich ddu.
•
Bai
2.
Tinh
CM
cua
dung dich
H2SO4

vd
NaOH,
biet
rdng
10 nd
dung
dich
H2SO4
tdc
dung vUa du vdi
30 ml
dung dich NaOH.
Neu lay 20 ud
dung dich
H2SO4
cho tdc
dung
vdi 2,5
gam CaCOs
thi
axit
con du vd
lugng
du
ndy tdc dung vita du vdi
10
nd NaOH.
LdlGIAI
PHAN
A.

Li
thuyet
Cdu
I.
-
Trich m6i
lo
mot it lam mt\ thuf.
- Cho dung dich NaOH dii Ian
liicft
vao
cac
mau
thif
tren va dun nhe;
Mau
thuf
nao
c6
khi
miii
khai bay ra la
NH4CI:
NaOH
+
NH4CI
—>
NaCl
+
NHgt

+
H^O
Mau tao ket
tiia
trSng xanh, hoa nau
do
trong khong khi
la
FeC]2:
2NaOH
+
FeCl2
>
Fe(0H)2i
+
2NaCl
trang xanh
,
2Fe(OH)2
+ -
O2
+
H2O
>
2Fe(OH)3
2
Mau
CO
ket
tija

nau
do la
FeCls:
^
3NaOH
+
FeCls
>
Fe(0H)3i
+
3NaCl
nau do
'
Mau tao ket tua keo trSng sau d6 tan trong NaOH dif la
AICI3:
3NaOH
+
AICI3
>
Al(0H)3t
+
3NaCl
keo trdng
A1(0H)3
+
NaOH
>
NaAlOa
+
2H2O

Mau tao ket tua trSng la
MgCl2:
2NaOH
+
MgCl2
^
Mg(0H)2
+
2NaCl
Mau khong
c6
hien tJong gi
la
NaCl.
Lfll
GIAI
THI
unr
emu
ni/\i
un»
unr a ^
Cdu
II.
a) 2Fe +
6H2SO4
damdac > 1^02(804)3 +
3SO2T
+ 6II2O
Fe2(S04)3 + 6NaOH > 2Fe(OII)3i + 3Na2S04

2Fe(OH)3 —>
FeaOs
+
3H2O
FeaOg + SCO —> 2Fe + SCOat
Fe
+ 2HC1 > FeCla + Hat
2FeCl2
+ CI2 >
2FeCl3
2FeCl3 + Fe > SFeC^
FeCla + 2AgN03 > Fe(N03)2 + 2AgCU
b) MgCOg +
H2SO4
loang >
MgSOi
+ CO^^t + HgO
MgS04 + BaCla >
MgCl2
+ BaS04i
MgCl2
+ 2NaOH > MgCOIDai + 2NaCl
Mg(0H)2 —> MgO + H2O
MgO + CO2 —^ MgCOg
MgCOa
>
MgO + COat
2CO2
+ Ca(0H)2 > Ca(HC03)2
Ca(HC03)2 —> CaC03i + H2O + C02t

hoac
Ca(HC03)2 +
2NaOII
> CaCOai + NaaCOg +
2H2O
hoac
Ca(HC03)2 + Ca(0H)2 2CaC03i +
2H2O
Cau
HI.
Trich
m6i lo mot it lam mau
thtf.
- Cho quy tim tam ni/dc Ian \\iat vao cac mau
thijf
tren:
• Mau
thuf
nao l^m quy tim hoa xanh la NaOH.
• Mau khong c6 hien
tiiong
la NaCl.
• Nhufng mau l^m quy tim hda do:
IICl,
H2S,
H2SO4
- Sau do cho dung dich AgNOa Ian
li/gt
vao cac mau lam quy tim hoa do.
• Mau nao tao ket tua

trSng
la HCl:
AgNOs + HCl > AgCli + HNO3
irdng
• MSU nao CO ke't tua den la H2S:
2AgN03 + H2S >
kgiSi
+
2HNO3
• Mdu khong c6 hien tuong gi la
H2SO4.
10
ifil
RI4I
fiF TH]
Hfir
SlUH
f;inr
unA unr n
Cciu
IV.
- sat nguyen chat khong bi ban gi vi n6
dugc
bao ve
bori
Idp
Fe203
ben trong khong khi a nhiet do thu6ng.
- Khi trong sat bi Ian tap chat, de lau ngay bi ban gi do xay ra su
an

mon kim
loai
tufc
la bie'n sat
thanh
hop chat cua sat.
Gidi
thidi:
Tren be mat kim
loai
c6 Idp
nUdrc
am, da hoa tan mot
luong
oxi nen chuyen Fe -> Fe^\
Va oxi hoa tan trong niidc theo qua
trinh:
O2 +
2H2O
>• 4011"
Sau do Fe^' ket hop
vdri
OH" > Fe(0H)2 (trdng xanh)
Mot
phan Fe(0H)2 bi oxi hoa tao Fe(0H)3 (nau do)
4Fe(0H)2 + O2 +
2H2O
-> 4Fe(OH)3
Do do sat bi han gi c6 mau nau do.
PHAN

B. Bdi toun
rr.
- C% X m,, 10 X 1664 n o / 1^
Bai
1. Ta co:
n3,,,^
= = = 0,8 (mol)
"Na^so,
•
"K^SO,
=1:2
Goi X la so mol
ciia
Na2S04 =>
Hj^^g^^
= 2x (mol)
Cac phan ufng xay ra:
BaCl2 + Na2S04 > BaS04i + 2NaCl (1)
(mol) X <- X -> X
BaCl2 +
K2SO4
> BaS04i + 2KC1 (2)
(mol) 2x 2x ^ 2x
Vi
khi them dung dich
H2SO4
vao
nLfdc
loc lai tao ket tua nen
trong

niidc loc con dtf BaCla.
BaCl2 +
H2SO4
> BaS04i + 2HC1 (3)
(mol) 0,2 <- 0,2
«o = ^ = 0,2 (mol)
B^so,/,3, 233
Theo de:
ng^ci,
= ^-^
^n^^l)
ng^^i./d)
+
"BaCi,/(2)
+
^Baa.,im
= ^'^
» X + 2x + 0,2 = 0,8 => X = 0,2
Khoi
lagng dung dich A
la:
nidungdich = mNa.so^ + "^K,SO, + "^H^O
« mdung dich = 0,2 X 142 + 0,4 x 174 + 102 = 200 (gam)
C%Na so = ^^^^^ X 100 = 14,2%
^^2^°"
200
1
74 X 0 4
Ml
GIAI

DE
THI
HOC
SINH
GIOI
HOA
HOC
9
11
Bai
2. Ta c6: n
CaCO,
2.5
100
=
0,025
(mol)
4. Tafco 3 6'ng nghi&m deu dung dung dich bari clorua, ngUdi ta cho
them
vao:
diig
l:~dung dich kali
cacbonat
Ong 2: dung dicli natri
cacbonat
Ong 3: dung dich bac nitrat
Sau
do cho
them
axit nitric vao ca ba ong nghiem. Em hay cho

biet
dng ndo con
chat
ket tua. Gidi thich vd
viet
phiXang trinh phdn icng.
5. Mot hon hap NaCl vd
MgCl2
them
nuac vao hon hap ta c6 dung dich A.
Them dung dich AgNOs vao dung dich A, khi phdn iing ket thuc logi
bo
chat
ket tua trdng, phdn dung dich con Igi la dung dich D.
Chia
B lam 2 phdn: a vd b.
Phdn
a: Sau khi c6 can tiep tuc dun nong thl diigc mot hon
Jigp
khi
C. Cho hon hap khi nay qua binh dUng KOH.
Phdn
b: Cho vdo lugng dU dung dich HCl thi thu dugc ket tua trdng D.
Viet cdc phuang trinh phdn dng.
6. Lam the ndo de phdn
biet
cdc Ig hoa
chat
diidi
day md khong dugc

dung theni hoa
chat
ndo khdc:
MgCl2,
H2SO4,
NaCl, CuSO^, NaOH.
7. Mot hon hap gom: CuO, FeO,
AI2O3.
Lam cdch ndo de tdch chiing ra
khoi nhau.
Cdu
II. Bai todn
Cho 26,Ig
MnOo
tdc dung vdi dung dich HCl c6 20 gam HCl. Cho hct
khi
do qua mot lit dung dich NaOH lodng dU.
a)
Lugng HCl nay c6 du de phdn dng het vdi
Mn02
khong?
b)
Tinh
nong do mol
11
cua
muoi
thu dUgc trong phdn I'ing giOa do vd NaOH.
c)
Nung qugng pyrit sdt 'de tgo ra SO2. Cho khi SO2 sue vdo dung dich

chda 2
muoi
tren.
Sau
do
them
vdo mot lugng du Ba(N03)2. Tim khoi lugng ket tua va
khoi lugng pyrit can dung. Biet rdng lugng SO2 tdc dung vUa du
dung dich muoi.
(Cho Mr, = 55; O = 16; H = 1; CI = 35,5; S = 32; Fe = 56; Ba = 137)
^^F:.
, 13
Phiin
ufng:
H2SO4
+ CaCOa -
(mol)
0,025
<-
0,025
H2SO4
+ 2NaOH
(mol) a -> 2a
Trong
20 ml dung dich
H2SO4
chufa
0,025
mol
H0SO4

10 ml dung dich
H2SO4
chtfa x mol
H2SO4
10 X
0,025
->
CaS04
+ COat + H2O
Na2S04 +
2H2O
X
=
20
=
0,0125
(mol)
Trong
10 ml dung dich NaOH chijfa 2a mol NaOH
30 ml dung dich NaOH chufa y mol NaOH
2a X 30
y =
=
6a (mol)
Na2S04
+
2H2O
10
H2SO4
+

2NaOH
(mol)
0,0125
->
0,025
Vi
phan ufng
trung
hoa nen 6a = ^' (mol)
6
"2SO4 " 0,01
0,025
=
1,25M
^6 2,5
0,03
M
OE SO 3
DE THI HOC
SINH
GiOl HOA HOC 9
QUAN
3 TP. HQ CHJ MINH NAM HOC 1998 - 1999
Cdu
1. Li thuyet
1. Tit 7 Ig hoa
chat
sau, em c6 the dieu chc nhUng
chat
khi

nuo?
Axit sunfuric; natri hidroxit;
amoni
nitrat; canxi
cacbonat;
natri
sunfit; sdt sunfua vd kim logi keni.
2. Ta
H2SO4
CO may cdch de dieu che
CaSOJ
3. Viet
cong
thUc vd ten ggi 3
muoi
diing trong nong nghiep (phdn dam,
phdn Idn vd plidn
kali).
Hay gidi thich tgi sao ngUdi ta khong trgn tro
bep vdi phdn dam de ban rugng?
19
Ldl
GIAI
Cdu
I. Li thuyet
1. Ta dieu che
duac
cac chat khi sau:
CaCOa
1000°C

hoac
Zn +
H2SO4
loang
Zn + 2H2S04dac
FeS +
H2SO4
—
-> CaO +
COat
>
ZnS04
+ H2t
—>
ZnS04
+ S02t + 2H2O
->
FeS04
+ H2St
hoac
NaOH + NH4NO3
Na2S03
+
H2SO4
NH4NO3 >
NaOt
+ 2H2O
NH4NO3
>
NaNOs

+ NHat + H2O
>
Na2S04
+
SOst
+ H2O
>^°°°^
)
N2t-H^02t
+ 2H20
2
2. Dieu che
CaS04
tif
H2SO4:
Cdch
1: Tac dung
vdri
canxi:
H2SO4
+ Ca
Cdch
2: Tac dung
vdri
CaO:
H2SO4
+ CaO
Cdch
3: Tac dung v6i
bazcJ

Ca(0H)2:
H2SO4
+ Ca(0H)2 >
CaS04
+ 2H2O
Cdch
4: Tac dung
vcri
muo'i cija canxi:
H2SO4
+
CaC03
>
CaS04
+ COat + H2O
Chu
y: Con nliieu cdch klidc, xin
nJiuang
ban doc!
3.
Cong
thufc ba muoi dung trong nong nghiep:
• CO(NH2)2 (46%
nita):
Ure (phan dam)
• Ca(H2P04)2:
Supe
phot phat kep (phan Ian)
[Ca(H2P04)2
>

CaS04
+ Hat
CaS04
+ II2O
hoac
supe
pho't phat dcfn
2CaS0,
•
KRO'i.
kali
nitrat
(phan kali)
hoac
KCl: kah clorua.
Phan dam de bi phan
hiiy
trong moi
trtfdng
kiem, vi tro bep
mang
tinh
kiem nen ngi/di ta khong
tron
phan dam vcfi tro bep.
4.
Ong 1: BaCl2 +
K2CO3
>
BaCOgi

+ 2KC1
2HNO3
+
BaCOg
> Ba(N03)2 + C02t + H2O
Ong
2:
BaCIa
+
Na^CO^
>
BaCOgi
+
2NaCl
2HNO3
+
BaCOa
->
Ba(N03)2 + C02t + H2O
dng
3: BaCla +
2AgN03
> Ba(N03)2 + 2AgCli
AgCl + HNO3
Trong 3 ong nghiem thi ta thay ket
tija
AgCl khong tan trong axit
mac dti la axit HNO3 vi the trong ong 3 con chat ket tua.
5. Phan ufng: NaCl + AgN03 > AgCU + NaNOg
MgCl2

+ 2AgN03 > 2AgCU + Mg(N03)2
Sau khi Ipc bo ket tua thi dung dich (B) gom: NaNOs, Mg(N03)2
va
AgNOs
dif.
Phdn
a: Khi dun nong dung dich B:
NaNOs —> NaNOa + -02t
2
2Mg(N03)2 —> 2MgO + 4N02t + O-^t
2AgN03 —> 2Ag + 2N02t + 02t
2NO2
+ 2K0H —>
KNO3
+
KNO2
+ H2O
Phdn
b: Khi cho HCl du vao dung dich B thu
diJOc
ket tua. Dieu
nay cho ta thay rang trong dung dich con
AgNOs
dif.
HCl
+
AgNOs
> AgCli + HNO3
(D)
6, Ta nhan thay: trong tat ca cac dung dich

tren
thi chi c6 mot dung
dich
CO
mau xanh la:
CUSO4,
cac dung dich con lai la khong mau.
Trich
moi lo mot it lam mau
thijf.
- Cho dung dich
CUSO4
Ian
lifot
vao cac mau
thuf
tren:
• Mau thCf cho ke't
tiia
mau xanh la NaOH:
CUSO4
+ 2NaOH > Cu(0H)2i +
Na2S04
• Cac mlu con lai khong c6 hien tu'ong.
- Dung NaOH lam thuoc
thiif,
cho Ian
liiot
vao cac mau ihxi con lai:
• Mau

thLf
CO ket tua mau trSng la MgClg:
2NaOH + MgCla > Mg(0H)2 + 2NaCl
• Mau
thii
tao dung dich trong suot va toa nhiet manh la
H2SO4.
2NaOH
+
H2SO4
>
Na2S04
+
2H2O
• Mau khong c6 hien tugng la NaCl.
7.
Tach
CuO, FeO, AI2O3 ra khoi
nhau:
CuO
AI2O3
FeO
NaOH
da
^NaAlOg
CuO_
FeO^
+C0,
+ HC1
du

-> Al(OH),
->A1,03
CuCl^
FeCl,,
[Cu(NH3)J(OH)2
+HC1
->CuCL
+NaOH du
+
NH.OH dil
-^Cu(OH),
Fe(OH),
cliaii
kliong
-^FeO
Phan
ufng:
• AI2O3
+
2NaOH
>
2NaA102
+
H2O
NaAlOz
+
CO2
+
2H2O
>

Al(0H)3i
+
NaHCOg
2A1(0H)3
-
CuO
+
2HC1
FeO
+
2HC1
-> AI2O3
+
3H2O
—>
CUCI2
+
H2O
->
FeCl2
+ H2O
>
Cu(0H)2i
+
2NH4CI.
CuClz
+
2NH4OH
—
Cu(0H)2

+
4NH3
>
[Cu(NH3)4](OH)2
(xanh tim, tan)
[Cu(NH3)4](OH)2
+
2HC1
>
CuCh
+
4NH3T
+
H2O
CUCI2
+
2NaOH
>
CuCOIDal
+
2NaCl
Cu(0H)2
>
CuO
+
H2O
FeCl2
+
2NaOH
Fe(0H)2

Cdu
II.
Bai
todn
Ta
c6:
n
chankhong
26,1
-> Fe(0H)2i
+
2NaCl
FeO
+
H2O
MnO.,
87
=
0,3 (mol): n^j^j
=
20
0,55 (mol)
MnOa
+
4HC1
(mol)
0,55
CI2
+
2NaOH

(mol)
0,1375
36,5
>
MnCla
+
Cl2t
+
2H2O
0,1375
-> NaCl
+
NaClO
+
H2O
0,1375
a)
Lifgng HCl nay c6 du
de
phan
lifng
het vcfi Mn02
khong?
TCr phan ufng (1) ta c6 ti le:
n
MiiO,
=
0,3>
n
-»CuO

(1)
(2)
HCl
0,1375
1
4 4
Vay
lirang
HCl nay khong dii
de
phan ufng het liicfng
Mn02
da cho
hay
IICl
phan
ijfng
het va
Mn02
con dii.
b) The
tich
dung
dich
thu dUcfc
chinh
1^ the
tich
cua NaOH tufc la:
Vduiigdich

= 1 (lit)
So'
mol cua NaCl trong phan ufng (2):
0,1375
(mol)
C^
= C„ = =
0,1375M
c)
Phan
ufng:
4FeS2
+ IIO2
>
2Fe203
+
8S02t
(i)
(mol)
^
0,1375
2
SO2
+
NaClO
+
2NaOH
>
Na2S04
+

NaCl
+
H2O
(2)
(mol)
0,1375
<-
0,1375
->
0,1375
Ba(N03)2
+
Na2S04
>
Ba^O^i
+
2NaN03
(3)
(mol)
0,1375
^
0,1375
Txi
(3)
^
ne^so,
=
^i^.,so,
=
0.1375

(mol)
mg^so^
=
0,1375
x 233
=
32,04
(gam)
TO
(1), (2)
:^
np^3
= ^ = (mol)
^so,
_
0,1375
2
" 2
0,1375
=>
mpes,
=
-^-r—
X 120
=
8,25 (gam).
DE
SO 4
OE THI HOC
SiNH

GIOI
HOA
HOC
9, CAP TP. HO
CHI
MINH
NAM
HOC
1999
-
ZOOO
Cdu
I.
Khi cho
kirn
loai
vdo
dung dich mud'i
c6 the xdy ra
nhitng phan
ling
gi? Cho
vi du
minh
hoa.
Cdu
II. Viet cdc phuang
trinh
phdn
vCng

theo
chuSi bien
hoa
sau:
FeCh
>
Fe(0H)3
>
FezOs
Fe
>
FesO^
<^
^
FeCl2
>
Fe(0H)2
> FeO
Cdu
III.
Mot nhd hoa hoc
dieu
che
dugc
3
mdu
kirn
loai gidng nhau
ve
dang

ben
ngoai (mdu sdc)
va da
tim dugc phuang phdp phdn
biet
nhanh
chong.
dng lay cdc
mdu
kim
log.i
cho tdc
dung vdi^axitjm_d^g dich
NaOH,
ket qua do dugc ghi
tron^^M^:^,^^"^
6SNH
THUAN
Lfll
RlAi
n£
Tui unn SIUH
Rini
HflA HtlR
9 — • _ . JO
17
Thuoc
thii
Kim
loai

I
Kim
loai
II
Kim
loai
III
Axit
HCl
-
+
+
Axit
HNO3
+
-
+
Dung
dich
NaOH
-
+
+
Trong do ddu (+) de chi trudng hap kim loai hoa tan, ddu (-) chi triiang
hap kim loai
khong
tdc dung vai dung dich kicm hay axit.
Hay xdc dinh kim loai nghien ciiu,
viet
phuang trinh phdn ling vd

gidi
thich vi sao kim loai
khong
tdc dung vai cdc
chat
dd cho.
Cdu
TV. Chi diing kim loai, hay nhgn
biet
cdc dung dich sau day:
HCl,
HNO3
ddc,
AgNOa,
KCl, KOH.
Vii't
cdc phuang trinh phdn dug xdy ra trong qua trinh
nhgn
biet.
Cdu
V. Hoa tan
oxit
cua kim
logi
hoa tri II trong mot lilgng
vita
dil dung
dich
H2SO4
20% thi thu

dicgc
dung
dich mudi c6
nSng
do 22,69c.
Xdc dinh kim
logi
do.
Cdu
VI. Co mot hon hap gom Na2S04 vd
K2SO4
dilgc trgn Idn
iheo
ti le
I
: 2 ve so mol. Hoa tan hon hap vdo 102 gam
nUac
thi thu
dilgc
dung
dich A. Cho 1664 gam dung dich
BaClz
10% vao dung dich A, xudt hien
ket tua. Loc bo ket tua,
them
H2SO4
dU vdo
nilac
Igc thi tlidy tgo ra
46,6 gam ket tua. Xdc dinh

nong
do phdn tram cua
A^a^SOj
vd
K2SO4
trong dung dich A ban ddu.
(Cho Na = 23; S = 32; K = 39; Ba = 137; CI = 35,5).
Ld\I
Cdu
I. Khi cho kim loai vao dung dich muoi c6 the xay ra:
- Neu kim loai cho vao khac kim loai trong muoi thi xay ra phan uTng
the - oxi hoa
khijf.
Fe + CUSO4 >
FeS04
+ Cu
- Neu kim loai
trung
v6i kim loai trong muoi thi xay ra phan
tfng
tiT
oxi hoa.
Fe + 2FeCl3 > SFeClg
Cdu
II. Phan ufng:
3Fe+
202 ^°
>
Fe304
Fe304 + 8HC1 > FeCla + 2FeCl3 +

4H2O
FeCl3 + 3NaOH > FeCOIDgi + 3NaCl
2Fe(OH)3 —> FezOs +
SHsO
FeCla
+ NaOH >
Fe(0H)2i
+ 2NaCl
Fe(0H)2
TT-^rt
> FeO + H2O
chan
khong
Cdu
III
Theo
de bai thi: Kim loai I: Ag (Bac)
Kim
loai
II:
Al (Nhom)
Kim
loai
III:
Zn (Kem)
• Ag khong phan uTng vdi HCl vi Ag dufng sau hidro trong day hoat
dong kim loai. Ag khong phan ufng vdi NaOPI vi khong phai la kim
loai
iLfdng
tinh.

• Al khong tac dung vdi
HNO3
vi Al bi thu dong trong moi trUcfng
HNO3
dac nguoi.
Cdu
IV. Trich moi lo mot It lam mau
thijf.
- Cho bot kim loai Cu dti Ian luat vao cac mau
thtf
tren:
+) Mau
thii
nao dung dich id khong mau chuyen sang xanh la AgNOs.
Cu + 2AgN03 >
Cu(N03)2
+ 2Ag
+) Mau
thuf
nao vifa tao dung dich mau xanh va c6 khi nau do bay ru
la
HNO3.
Cu
+
4HNO3
dac
> Cu(N03)2 + 2N02t +
2II2O
- Sau do cho dung dich viia thu
di/oc

a
tren
Ian
liicft
vao cac mau con lai
+) Mau
CO
ket tua mau xanh la KOH.
Cu(N03)2
+ 2K0H >
2KNO3
+ Cu(0H)2i •
+) Cac mau con lai khong c6 hien tugng.
- Cho dung dich KOH vao 2 mau con lai, mau nao c6 phan ufng toa
nhiet
la HCl
KOH
+ HCl > KCl + H2O
Mau
con lai la KCl.
Cdu
V. Goi kim loai hoa tri II la A va c6 a mol => oxit la: AO.
AO
+
H2SO4
>
ASO4
+ H2O
(mol) a ^ a a
Theode:

«,,„,^.=^
"^dd
Hi,S04
a(A +
96)100
C%ddASO, =—
™dd
AS04
Ma:
a(A + 96)xl00 a x 98 x 100
<=> • = a(A + 16) +
22,6 20
=> A = 24: Magie (Mg).
Cdu
VI. Ta c6: 03 „, = 10 x 1664 ^ ^
^^^'2 100 X 208
'^NajSO,,
• '^K2S04 =• 1 • 2
Goi
X la so' mol cua Na2S04 => n^^g^^ = 2x (mol)
Cac phan
ijfng:
BaCl2
+ Na2S04 > BaS04i + 2NaCl (1)
(mol) X <- X -> X
BaCls
+
K2SO4
> BaS04l + 2KC1 (2)
(mol) 2x <- 2x 2x

Klii
them dung dich
H2SO4
vao lo niidfc loc thi tao ket tua nCfa nen
trong nude loc con dii
BaCl2.
BaCl2
+
H2SO4
> BaS04i + 2HC1 (3)
(mol) 0,2 <- 0,2
nBaso,„3,
= H = 0,2 (mol)
Theo de: ng^Q^
ban
ddu
= 0,8 (mol)
TCr (1), (2), (3) => x + 2x + 0,2 = 0,8 =:> x = 0,2
Khoi
lugng dung dich (A) la: md.ngdich =
m^^^^^^
+
m^^^^^^
+ m^^^
=> mdungdich = 0,2 X 142 + 0,4 x 174 + 102 = 200 gam
142 X 0,2
200
174x0,4
C%Na
so = —^—— 100 = 14,2%

NaaSO^
200
C%„ „„ = ^ X 100 =
34,8%.
'^2^^i
200
DE SO 5
OE
THI
HOC
SINH
GlDl
HOA
HOC,
QUAN 9
(VONG
2)
TP.
HCM
NAM
HOC
19S9
-
2000
Cdu
I. Cho chuoi phuang trinh plidn ling sau:
+A, I, +B,
CaCOs
CaCOs
Tim

cdc
chat
A, B, C, X, Y, Z (Id cdc
chat
khdc nhau vd khdc CaCOs)-
Viet
cdc phuang trinh phdn ting.
Cdu
II. Dung mot kim loai de nhdn
biet
cdc lo dung dich sau:
FeCh;
FeCh; HCl; BaCh; (NH^2S04; AlCh;
NH4CI
Cdu
III. Cho Na vdo 2 dung dich mud'i Al2(S04)3 vd CuSOj thi thu
dagc
khi
A, dung dich B vd ket tua C. Nung ket tila
dugc
chat
rdn D. Cho H2
di
qua D nung
nong
dUgc
chat
rdn E. Hda tan E vdo dung dich HCl thi
thdy
E tan mot phdn.

Gidi
thich.
Viet
phuang trinh phdn iCng.
Cdu
IV.
a)
Cho 0,25 mol CuO tan het trong dung dich
II9SO4
20% dcm nung
nong
lugng vila dii, sau do Idm ngugi dung dich den 10°C.
Tinh
khoi
lugng tinh the CUSO4.5H2O tdch ra khoi dung dich. Biet do tan
CUSO4
a 10°C Id
17,4g.
b)
Clio
biet
do tan CaSO^ Id 0,2g (a
20°C)
vd khoi lugng rieng cua dung
dich
bdo hda coi bdng Iglml.
•
Tinh
do tan cua CaSO^
theo

nong
do mol.
• Khi trgn 50 ml dung dich CaCl2 0,012M vai 150 ml dung dich
NaoSO^ 0,004M a 20°C thi c6 ket tila xudt hien
khong?
Cdu
V.
a)
Hda tan hodn todn 1 hidroxit cua kim logi M bdng mot lugng vita dii
dung dich HCl 10%. Sau phdn ling thu
dugc
dung dich A. Them vdo
dung dich A mot lugng vUa dii dung dich AgNOs 20% thu
dugc
dung
dich
muoi
c6
nong
do
8,965%.
Xdc dinli
cong
thvCc
hidroxit tren.
b)
Khi phdn tich 2
oxit
vd 2 hidroxit tUong
iCng

cila ciuig mot
nguyen
to
hda hoc
dugc
so lieu sau: ti so thdnh phdn % ve khoi lugng ciia 0x1
20
trong 2
oxit
do bdng —. ,
2T
Tl
so thdnh phdn % ve khoi lugng ciia nhdm Iiidroxit trorig hai
107
hidroxit do bdng . Hay xdc dinh
nguyen
to' do.
135
Cdu
VI. Cho X, Y Id hai dung dich HCl c6
nong
do khdc
nhau.
Lay V ml
dung dich X tdc dung vai-AgNOs du tao thdnh
35,876g
ket tila. De trung
hda V ml dung dich Y can 500 ml dung dich NaOH 0,3M.
a)
Khi trgn V (lit) dung dich X vol V (lit) dung dich Y thu dUgc 2 (lit^

dung dich Z.
Tinh
CM dung dich Z.
b)
Neu lay 100 nd dung dich X vd lay 100 nd dung dich Y cho tdc dung
het vdi kim logi Fe thi lugng hidro
thodt
ra trong hai trUdng hgp lech
iihau
0,448
(lit) (dktc).
Tinh
CM dung dich X, Y.
LCII
Riai
nc
Tu,
„„o unA unr a 21
Cdu
VII. Trgn Idn 10 ml dung dich HCl vai 20 ml dung dich
HNO3
vd
20 nd dung dich
H2SO4
thu dugc dung dich A, pha
them
II2O
vdo dung
dich
A de diCgc dung dich B c6 the tich gap doi. Trung hoa 25 nd dung

dich
B can 8 ml dung dich NaOH 8% (D = l,25g/nd). Bern c6 can dung
dich
tqo thdnh dugc
l,365g
muoi
khan. Neu cho 40 tnl dung dich B tdc
dung vai mot liCgng dU dung dich BaCl-z thi thu dugc
0,932
ket
tiia.
a)
Tinli
CM dung dich axit ban ddu.
b) Dung dich C chi'ia hon hap NaOH 0,8M vd Ba(0H)2 0,2M. Can bao
nhieu ml dung dich C de trung hoa het 50 ml dung dich B?
Cho: H= 1,N= 14, O = 16, Na = 23, S ^ 32; CI = 35,5, Co = 40, Fe = 56,
Cu = 64, Ag = 108, Ba = 137
LCfl
GIAI
Cdu
I. ThiTc hien chuoi phan ufng:
CaCOg
CaO
lAl
CO,,
+
NH011
->Ca{OH),
(Bl

-^Na^CO,
t-HCl
->CaCl,
CaCO;,
+CO.,*U.,0
->NaHCO.,
Phan
ijfng:
CaC03
CaO +
II2O
iooo"c
CaO +
CO2T
(A) (X)
Ca(0H)2
(B)
>
CaCl2
+
2H2O
Ca(0H)2
+ 2HC1 —
CaCl2
+
NaaCOa
>
CaCO^i
+ 2NaCl
CO2 + 2NaOH >

Na2C03
+ H2O
(Y)
NaaCOs
+ CO2 +
II2O
>
2NaHC03
(Z)
-> CaCOyi + 2NaN03 + CO.T +
II2O
2NaIiC03 +
Ca(N03)2
—
Cdu
II. Trlch moi dung dich mot it lam mau thuf.
Cho
kirn
loai Ba vao cac mau
ihii
tren,
dau tiea c6 phan ufng:
Ba +
2H2O
Ba(0H)2
+ H2T
Mau nao cho ket
tiia
trfing
xanh la FeCl2:

Ba(01i)2
+ FeCL — >
FeCOIDai
+
2BaCl2
22
I
rii
piAi
nc
Tui
unr
Qtwu
nni unA unr Q
M&u
nao cho ket tua nau do la FeCla:
3Ba(OH)2
+
2FeCl3
> 2Fe(OH)3i +
3BaCl2
Mau
CO
ket
tiia
va khi
miii
khai bay ra la (NH4)2S04:
Ba(0H)2
+

(NH4)2S04
>
BaS04^
+
2NH3T
+
2H2O
MSU
cho ket
tiia
keo trang la
AICI3:
3Ba(OH)2
+
2AICI3
> 2Al(OH)3i +
SBaCla
Mau chi
CO
khi
mtii
khai bay ra la
NH4CI:
Ba(0H)2
+
2NH4CI
>
BaCl2
+
2NH3t

+
2H2O
Mau cho dung dich trong
suo't
va toa nhiet la HCl:
Ba(0H)2
+ 2HC1 >
BaCl2
+
2H2O
Con lai la BaCla.
'Cdu
III. Phan ufng:
Na + 1120
-> NaOH + ^Hat
6NaOH +
Al2(S04)3
2NaOH +
CUSO4
—
>
2Al(OH)3i
+
3Na2S04
-> Cu(0H)2i +
Na2S04
2A1(0H)3
-
Cu(0H)2
—

AI2O3
+ H2
CuO + H2 -
->
AI2O3
+
3H2O
-> CuO + H2O
^—> Cu + H2O
Vi
E tan 1 phan nen E chufa: CuO, Cu,
AI2O3
AI2O3
+ 6HC1
CuO + 2HC1 -
->
2AICI3
+
3H2O
-> CuCla + H2O
CuO
+
H2SO4
—
(mol) 0,25 -> 0,25
Khoi
iLfoag dung dich
H2SO4
20%:
0, 25 X 98 X 100%

>
CUSO4
+ H2O
0,25
20
=
122,5 (gam)
Khoi
lugng
CUSO4:
0,25 x 160 = 40 (gam)
Khoi
lu'gng dung dich sau phan ufng:
'^duiig
dich
sau
phan
iMig —
niCuO
+
'^^ddn^SO^
=
0,25 X 80 + 122,5 = 142,5 (gam)
Khi
ha
nhiet
do:
•
CUSO4
+ 5H2O —>

CUSO4.5H2O
(mol)
a -> a
Kho'i
liicrng
CUSO4
c6n lai
trong
dung
dich:
40 - 160a (gam)
Khoi
liiong
dung
dich
con
lai:
142,5 - 250a (gam)
.
40-160a _ 17,4
142,5 - 250a " 100 + 17,4
<r>
4696 - 18784a = 2479,5 - 4350a
« 2216,5 = 14434a a = 0,154 (mol)
Vay
khoi
lifgng
CUSO4.5H2O
ket
tinh:

250a = 250 x 0,154 = 38,5 (gam)
h)*a 20°C: 100 gam H2O + 0,2g CaS04 -> 100,2 gam
dung
dich
CaSO,
CO
d =
Ig/ml
Vay
thi
tich
dung
dich
CaS04 Ik:
V^^^^^
= = 100,2 (ml)
Suy
ra: C^. = x —-— = 0,015M
Mc,so,
136 0,1002
Khi
tron:
CaCla + Na2S04 > CaS04 +
2NaCl
(mol)
0,0006 0,0006
"caci,
= X 0,012 = 0,0006 (mol)
^Na^so^
= O'^'^ ^ 0'004 = 0,0006 (mol)

d
2Q°C
trong
1 lit
dung
dich
CaS04 bao hoa c6 0,015 mol CaS04
trong
0,2 lit
dung
dich
CaS04 bao hoa c6 0,003 mol CaS04
Vi
0,6.10"^ < 3.10"^ nen
khong
c6
hien
ti/gng
ket
tiia.
Cdu
V.
a)
Goi cong
thufc
cua
hidroxit
kim
loai
M:

M(OH)a
M(OH)a
+
aHCl
>
MCla
+ aHaO (I)
MCla
+ aAgNOg >
aAgCll
+
M(N03)a
(2)
Klio'i
lugng
dung
dich
trong
(1)
=
M + 17a + 36,5a x 10 = M + 382a (gam)
.
I rli nil nc
TUI
unn ciuu nini unA unp n
Khoi
lucfng
dung
dich
trong

(2):
mdu„gdich2 = ra^i^ +
ra^^j^^o,
= + 382a + 170a : 20% - 143,5a
=
M + 1232a - 143,5a (gam)
Theo de, nong do cua
dung
dich
mudi
thu
dirge
sau
phan
ufng
(2) la:
M
+ 62a
C%
M(N0,),
'3'a
8,965% =
M
+1232a-143,5a
M
+ 62a
X
100%
M+1232a-143,5a
=^

Nghiem
hop
li:
a = 1; R = 39:
kali
(K)
Do do cong
thufc
can tim la KOH.
b) Goi nguyen to can tim la M:
Cong
thuc
oxit:
M2OX,
M20y
Cong
thilfc
hidroxit:
M(OH)^,
M(OH)y
16x
16y
X
100% <r> M = 39a
Theo de bai, ta c6:
o
2M
+ 16x
•
2M + 16y

x
M + 8y 20
=
20 : 27
Mat
khac:
M
+ 8x y 27
27Mx
+ 216xy = 20My + 160xy
27Mx
+ 56xy = 20My
17x
17y
(1)
o
<=>
M
+ 17x
•
M + 17y
M
+ 17y :
=
107 : 135
107
Tif
(1) va (2), ta c6:
<=>
<=>

Bang
bien
luan:
y
M + 17X 135
135xM
+ 2295xy = 107My + 1819xy
135xM
+ 476xy = 107My
5(20My
- 56xy) + 476xy = 107My
lOOMy
- 280xy + 476xy = 107My
7My
= 196xy <=> M = 28x
(2)
X
1
2
3 "
M
28
56
84
Vay
M = 56
(Fe).
Vay
cong
thijfc

oxit:
FeO; Fe203
cong
thiic
hidroxit:
Fe(0H)2;
Fe(0H)3
25
Cdu
VI.
a)
HCl
+
AgNOa
(mol)
0,25
HCl
+
NaOH
(mol)
0,15 <- 0,15
35,876
—> AgCU
+
HNO3
0,25
^ NaCl
+ H2O
(1)
(2)

Trifffng
hap 2: Cx < Cy
0,015
0,025
=
0,02
15
25
3 5
=
40
<=>
:
V
V
V
<=>
o
<=>
Ta
c6: n,
143,5
TCf phan
ijfng
(1) va (2), ta c6:
0,25 + 0,15 0,4
=
0,25
(mol) wk
n^^oH = 0'^ x 0,3 = 0,15

(mol)
v
+v
=
0,2M
(Vi
V + V = 2)
b)
Phan ufng
(mol)
(mol)
2HC1
+ Fe
0,025
V
2HC1
+ Fe
0,015
FeCl2
+ Hat
0,025
2V
->
FeCl2
+ H2t
0,015
V
2V' •
Theo
cau (a): -

n^^^.,^
=
n^^^^^
=
Cx-V
=
0,25
(mol)
-•
nHci/,2)
=
"NaOH
= CY-V = 0,15 (mol)
Trong
V
lit dung djch HCl thi
c6
0,25
mol
HCl.
0,1
ht
dung dich HCl thi
c6 mol
HCl.
Tiiang
tLf:
Trong
V
lit dung dich HCl

c6
0,15 mql HCl.
0,1
lit dung dich HCl
c6 mol
HCl.
V
Theo
de: n
H„
0,448
22,4
=
0,02
(mol)
TriTdrng
hdp 1:
Neu
Cx > Cy
0,025 0,015^-^,32 c ^
<=>
2V
2V
5V'
- 3V
=
8
V.V
V
V

=
8; V + V = 2 (V, V < 2)
2V'
2V V'
3V
- 5V' = 8W', V + V = 2
3V
- 5(2 - V) = 8V(2 - V)
3V
- 10 + 5V 16V 8V^
V,
=1,7
V2 = -0,72
(loai)
Vi
= 1,7 => V = 0,3
=^
C^ = 0,147 va Cy = 0,5.
<^
8V - 8V - 10 = 0 =^
(3)
Cau
vn.
a)
Cac
phan ufng:
NaOH
(mol)
(4)
5(2

- V) - 3V = 8V(2 - V) « 10 - 5V - 3V = 16V - SV^
«
8V^ - 24V + 10 - 0
\5
V2 = 2,5 > 2 (loai)
V
= 0,5 => V = 1,5 ^ Cx = 0,5M va Cy = 0,1M
(mol)
(mol)
(mol)
0,0025CA
<
NaOH
+
0,005CB
<-
0,005CB
2NaOH
+
H2SO4
—
HCl
>
NaCl
+
PI2O
0,0025CA
—> NaNOs
+ H2O
0,005CB

0,0025CA
HNO3
-
>
Na2S04
+
2H2O
0,01Cc
<- 0,005Cc -> 0,005Cc
H2SO4
+
BaCla
>
BaS04i
+
2H2O
0,004 <- 0,004
The
tich
cua
dung dich (B):
100 ml = 0,1
lit
Goi CA
la
nong
do mol cua
dung dich HCl.
So
mol cua

HCl trong
25 ml
dung dich
B:
0^.0,01x0,025
0,1
=
0,0025CA
Goi CB
la
nong
do mol cua
dung dich
HNO3
=>
So mol
ciia
HNO3
trong
25 ml
dung dich
B:
CB.0,02 xb,
025
0,1
=
0,005CB
Goi
Cc la
nong

do mol cua
dung dich
H2SO4
=>
So mol
ciia
H2SO4
trong
25 ml
dung dich
B:
Cg.0,02
X
0,025
0,1
=
0,005Cc
26
I
rtl nifll np
THI
Hnr
ciwu
nini
unA
unp
n
"^LSIMI
THI HOC SINH GIOI HOA HOC 9
(1)

(2)
(3)
(4)
27
Khoi
iLforng
dung
dich
NaOH
= 8 x 1,25 = 10 (gam)
So mol cua
NaOH:
^ = 0,02 (mol)
100 X 40
0 Q*^?
So mol cua
BaS04:
—— = 0,004 (mol)
233
So' mol cua H2SO4 tham gia phan
ufng
(4):
X
0,02x0,04
0,1
=
0,004
=^
0,008Cc
=

0,004
^Cc = 0,5 (mol/1)
So mol cua
NaOH
trung
h6a
trong
phan
ufng
1, 2, 3 1^:
0,0025CA
+ 0,005CB = 0,02 -
O.OlCc
= 0,015
« 25CA +
50CB
= 150 5CA +
IOCB
= 30
«
CA+2CB = 6 (I
Va: mniuo'i =
niNaCi
+ mNaNo, + ™Na,so,
<=> 1,365 = 0,0025CA X 58,5 + 0,005CB X 85 +
0,005Cc
x 142
1,365 = 0,14625CA + 0,425CB +
0,355
(1!

The (I) vao (H) ta c6:
0,14625(6
- 2CB) + 0,425CB = 1,01
o
0,8775
- 0,2925CB + 0,425CB = 1,01
o 0,1325CB =
0,1325
=^ CB = 1 (mol/1)
Suy ra: CA = 6 - 2CB = 4 (mol)
b) Cac phan
ufng
trung
hoa:
NaOH
+ HCl >
NaCl
+ H2O (1
(mol)
a < a
2NaOH
+ H2SO4 >
Na2S04
+ 2H2O (2
(mol)
2c <- c
NaOH
+ HNO3 > NaNOs + H2O (3
(mol)
b <- b

Ba(0H)2 + 2HC1 > BaCla + 2H2O (4
Goi
a, b, c Ian Ixxat Ik so mol ciia HCl, HNO3, H2SO4 tham gia phan
L^ng
(1), (2), (3).
a', b', c' Ian
liiat
la so mol cua HCl, HNO3, H2SO4 tham gia phan
ang (4), (5), (6).
, CAX
0,01x0,05 ,
iiHci
= a + a' = = 0,02 mol
,
, , CB X 0,02x0,05 • ,
nHN03
= b + b' = — = 0,01 mol
,
CcxO,02xO,05
,
nH2S04 = c + c =
-^-^1^
= 0,005 mol
a + b + 2c = 0,8V (vdi V la the
tich
ciia
dung
dich
C).
^

+ K +c' = 0,2V
2 2
2]n,^^
= a + a' + b + b' + c + c' =
0,035
(0,8V - 2c - b) + a' + b + b' + c' + c = 0,035
^
0,8V - c + (0,4V - 2c') + c' = 0,035
=^ 1,2V - (c + c') = 0,035
1,2V - 0,005 = 0,035
=:> V - 0,03333 (lit) = 33,33 ml.
DE
SO 6
OE
THl
HOC
SINK GIOI
HOA HOC
9,
CAP
TP.
HO CHJ
MINH
NAM HOC
2000
-
2001
Cdu I.
Viet
phan

ling de
hieu
dien
chuoi
bien
hoa sau:
CaO > Ca(0H)2 ^
CaCOs
CaCOs
CaClo
Ca(N03)2
(mol)
(mol)
Ba(0H)2
+ 2HNO3 > Ba(N03)2 + 2H2O
b'
—
<-
2
b'
Ba(0H)2 + H2SO4 >
BaS04
+ 2H2O
(mol)
c' «- c'
Ldl
GlAl
Oi
THl HOC SINH Gidl HOA HOC
9

Cdu
11.
a)
Tdch hon hap gdm BaCOs, BaS04, KCl, MgCh
bdng
phuang
phdp
hoa hoc.
b) Cho cac hoa
chat:
Na, MgCh, FeCls, FeCk,
AICI3.
Chi dung
them
H2O
hay
nhdn
biet
chung.
Cdu in. Cho
phuang
trinh
phan
ling c6
dang
sau: BaCh + ? = NaCl + ?
Hay
viet
4
phuang

trinh
phan
ling xdy ra. Biet rdng cac
phan
i2ng deu
xdy ra
hodn
todn.
.au rV.d
25°C
nguai ta da. hda tan 450 gam kali nitrat vao trong 500 gain
niiac cat (dung dich A). Biet rdng dp tan ciia nitrat kali la 32 gam „
20°C.
Hay xdc dinh khoi lugng kali nitrat tdch ra khoi dung dich khj
Idm
lanh dung dich A den
20°C.
~lau y. Cho 3 gam hdn hap hai kim loqi vun nguycn
chat
Id
nhom
v,i
magie
tdc dung hct vai
H2SO4
lodng thi thu duqc 3,36 lit mot
chat
khi a
dieu kien ticu cliudn.
Xdc dinh thdnh phdn phdn tram ve khoi lugng ciia nlwm vd

magic
trong hdn hap.
^du VL Khi cho a gam Fe vdo trong 400 ml dung dich HCl, aau khi phun
ling
ket thuc dem c6 can dung dich thu dicgc 6,2 gam
chat
rdn X.
Ncu cho hon hap gom a gam Fe vd b gam Mg vdo trong 400 ml dun/-
dich
HCl thi sau
kJii
phdn ling ket thiic thu diigc 896 ml khi H2 (dicii
kien chuan) vd c6 can dung dich thi dugc 6,68 gam clidt rdn Y.
Tilth
a, b, nong do phdn td gam ciia dung dich HCl vd thdnh phdn kiwi
lugng cdc
chat
trong X, Y.
(Gid
sii Mg khong phdn ling vai nitdc vd khi phdn I'ing vol axit thi Mij'
pJidn
ling trade, het Mg mai den Fe. Cho
biet
cdc phdn ling dcu xdy ru
hodn todn).
Biet: H=1;N= 14; O ^16; Mg = 24, Al = 27; CI = 35,5; K = 39; Ca = 40; Fe = 56
LCilGIAI
du I. Chuoi bien hoa:
CaCOs
^-^^—> CaO + CO.T

CaCOa
+ 2HC1 > CaCh +
COgt
+ H2O
CaO + 2HC1 >
CaCl2
+ H2O
CaO + H2O >
Ca(0H)2
Ca(0H)2
+ CO2 ->
CaCOsi
+ IlaO
CaCl2
+
2AgN03
> Ca(N03)2 + 2AgCU
Ca(0H)2
+
2HNO3
^ Ca(N03)2 +
2H2O
Ca(N03)2 +
Na2C03
^
CaC03>l
+ 2NaN03
Cau II-
a) Scf do tach:
BaCOy

BaSO.
+HC1
IdC
BaCl. -^Mi_^BaCO. i
loc
BaSO^
i
BaCO,,
BaSO^
+H;,o
KCl
MgCl^
KCl
MgCl^
+K011
vL/a du
KCl
'Mg(OH)^ i
+ HC1
->MgCl,
b) Trich mSi chat 1 it lam mau thuf.
+) Cho ni/orc Ian
liJOt
vao cac mau
thiJf
tren. Mau nao c6 khi bay ra la
natri.
Na + H2O
-> NaOH + -Hat
2

+) Cho
dung dich NaOH
Ian
liigft
vao cac mau
thtf tren
thi:
- MSU
CO
ket tua mau
trSng
la
MgCl2
MgCl2
+
2NaOH
> Mg(0H)2i + 2
NaCl
- Mau nao c6 ket
tiia
mau
trang xanh,
de lau hoa nau do la
FeCl2.
FeCla
+
NaOH
> Fe(0H)2i +
2NaCl
trdng

xanh
2Fe(OH)2
+ + H2O
-> 2Fe(OH)3i
ndu do
Cdu III,
- Mau cho ket tua mau nau do la
FeCls.
FeCla
+
3NaOH
>
Fe(0H)3i
+
3NaCl
- Mau cho ket tua keo
trang
la AICI3.
AICI3 +
3NaOH
-> A1(0H)3^ +
3NaCl
Neu NaOH
dii
thi
ket
tiaa
tan dan:
A1(0H)3
+

NaOH
>
NaAlOa
+
2H2O
BaCl2
+
Na2C03
>
BaC03i
+
2NaCl
BaCl2
+ Na2S04 —>
BaS04i
+
2NaCl
3BaCl2
+ 2Na3P04 >
Ba3(P04)2i
+
6NaCl
BaCla
+ Na2S03 >
BaSOgi
+
2NaCl
(hoac
BaCl2
+

NaaSiOa
>
BaSiOgi
+
2NaCl
LOi
r.iAi,
Cdu
IV. Bap so:
Kho'i Itfang
kali
nitrat
t^ch ra
khoi dung dich:
290 (gam)
KNO3.
Cdu
V. Goi a va b Ian
liigt
la so' mol cua Mg Al.
Phan
ijfng:
Mg +
H2SO4
> MgS04 + Hgt (1)
(mol)
a ->• a
2A1
+ 3H2SO4 >
Al2(S04)3

+ 3H2t (2)
3b
(mol)
b ->
Theo
de bai, ta c6 he
phuong
trinh:
2
24a + 27b = 3
a . = 0,15
Giai
he ta
dtfOc:
a = 0,05 va b = —
3
Vay:
%mMg = ^' x 100 = 40% va
%mAi
= 60%.
3
Cdu
VL
+) Tinh a va thanh phan cua (X).
Thi
nghiem
1: Khi cho a gam Fe + 400 ml
dung dich HCl,
c6 can
dung

dich
thi thu
difgc
6,2 gam chat r^n. Neu Fe
phan
ijfng
het
tufc
chat rSn
Chi
la FeCl2 ^ n^^ =
np^^i,
= ^ = 0M^9, (mol) (*)
Thi
nghiem
2: Cho a gam Fe va b gam Mg vao 400 ml
dung dich HCl,
c6
can dung dich
thi thu
diigc
6,68 gam chat r^n va
0,896
lit H2
(dktc).
Phan
Lfng:
Mg + 2HC1 >
MgCla
+ H2t (1;

(mol)
X X
Fe + 2HC1 > FeCla + Hat (2)
(mol)
y y
(Vdi
X,
y Ian
liTOt
1^
so'
mol
cua Mg va Fe
tham
gia
phan iJtng tren).
TCr
(1), (2) ^ nj, = X + y = « x + y = 0,04 (**)
"2 22,4
So sanh (*) va (**) ta
thay: Khoi lugng
kim
loai
d thi
nghiem
2
nhieu
han
d thi
nghiem

1
nhifng
so mol H2 thu
diioc
lai it hon. Do do
trong
thi
nghiem
(1) thi Fe
di/
va HCl het so mol H2 cf thi
nghiem
1 la:
0,04
(mol).
Fe + 2HC1 > FeCla + Hgt (3)
(mol)
0,04 0,08 0,04 ^0,04 i
Tif (3)
ra^^cA.,
= 0,04 x 127 = 5,08 (gam)
Khoi
lugng
Fe
phan uTng:
0,04 x 56 = 2,24 (gam)
va khoi lircfng
Fe
diT:
6,2 - 5,08 = 1,12 (gam)

Vay:
a = 2,24 + 1,12 = 3,36
(gam).
+) Tinh b va thanh phan cua (Y)
Neu
khoi lircfng
Mg
dung
du thi:
mcLatrdn = 0,04 X 95 + 3,36 = 7,16 (gam)
Theo
de bai: nichat rin = 6,68 < 7,16 => Mg
khong du
nen Fe da
phan
ijrng.
Phan
Lfng:
Mg + 2HC1 >
MgCh
+ H2t (4)
(mol)
X -> x X
Fe + 2HC1 > FeCla + Hat
(mol)
y -> y y
(5)
Tif
(4), (5) ta
CO

he
phtfcfng
trinh:
nj^
= X + y = 0, 04
n^ran
=
n^MgCl,
+
^FeCl,
+ ^Fed, = 6,68
CO
X
+ y = 0,04
95x
+
127y
+
(s, 36 - 56y) = 6,68
|x
+ y = 0,04 fx = 0,02
[95x
+ 71y = 3,32 ly = 0,02
Vay:
niMg
= 0,02 x 24 = 0,48 (gam)
"^Fecij
= X 127 = 2,54 (gam)
mpedLT
= 3,36 - 0,02 x 56 = 2,24 (gam) ^ b = 0,48

(gam).
DES0
7
Ki
THI CHON
HOC
SINK GIQIHOA
HOC 9 TP. QUY
NHdN, TJNH MW MW
NAM
HOC
2000
-
2001
Cdu
1.
^) Chi dung nuac hay nhdn biet 3 bgt kim loai: Ba, Al vd Ag.
^) Til cdc chat sau:
Na^O,
HCl, H2O, Al c6 the dieu che
dugc
n/ulng
clid't
fnai ndo md kJiong dung them phiCang tien ndo khdc. Viet phan ling
•
minh hoa.
2. Viet cdc phuang trinh phan ling tlieo sa do sau:
A
-> B
+ Y

D
+z,t°
A
^'•^'t
C Id chat ket tua mdu do ndu vd A, B, C, D, X, Y, Z Id ki hieu
I'Cng
cong thitc 1 chat.
Cdu
3. Bern nij gam hon hap
ZnCOs,
Zn dun nong ngodi khong ' i dc
plidn ling xdy ra hodn todn, thu dugc nig gam chat rdn.
Biet nil = mo.
Tinh
% khoi lugng
ZnCOs
trong hSn hap ddu.
Cuu
4. Bem dung dich chiia 0,1 mol sdt clorua tdc dung vai dung dich
NaOH
du thu
dugc
9,05 gam ket tua.
Xdc
dinh cong
thiCc
sdt clorua vd tinh hieu sudt phdn ling.
Cdu
5. Bem 46,4 gain Fe^Oy tdc dung vdi Ho dun nong thu
dugc

chat rdn B
gom Fe vd Fe^Oy du. Bem chat rdn B tdc dung het vai dung dich HNO.
lodng du thu dUgc dung dich C c6 chUa 145,2 gam mud'i Fe(N03)3 va
a mol NO thodt ra. Tat cd phdn Ung xdy ra hodn todn.
a)
Xdc dinh cong thUc Fe^Oy.
h)
Biet a = 0,52, tinh khoi lUgng
tiCng
chat trong B.
Ldi
GIAI
Cdu
1
a)
Cho 3 kim
loai
vao 3 coc
niicfc.
-
Tan CO bot khi bay len la Ba:
Ba +
2H2O
>
Ba(0H)2
+ Hat '
-
Kliong
tan la Al va Ag.
-

Cho 2 kim
loai
Al va Ag Ian
liicft
vao hai coc chufa
dung
dich
Ba(0H)2:
Kim
loai
nao tan c6 bot khi bay len la
Al:
2A1
+
Ba(0H)2
+
2H2O >
Ba(A102)2
+ Sllat
-
Khong
tan la Ag.
3.
ZnCOo
-> ZnO + CO2
.
2Zn + O2
2ZnO
I
(1)

(2)
Goi
X,
y Ian
lircft
la so mol
ZnCOg
va Zn
trong
h6n hop
dau.
Vi mj =
m2.
Khoi
liiong
CO2
thoat
ra 0 (1) =
khoi
lugng
oxi
tham
gia a (2)
=> 44x = 16y ^ - = ~
y
11
Vay
%
khoi
iLforng

ZnCOg
= 41,15%.
Cdu
4. Neu la FeCl2: FeCl2 +
2NaOH
>
Fe(0H)2i
+ 2NaCl
Khi
phan
ufng
xay ra 100%
thi
khoi
lugng
ket
tiia
Fe(0H)2
= 9g < 9,05g
=i> V6 11, vay do la FeCls.
FeCla +
3NaOH
>
Fe(0H)3i
+ SNaCl
=>
Hieu
sua't (H) = = 84,58%
Cdu
5.

a)
Khoi
liiang Fe c6
trong
145,2 gam FeCNOsJa:
145,2 X 56
b)
NasO
+
H2O
>
2NaOH
2A1
+
2NaOH
+
2H2O
> 2NaA102 +
3H2t
NaAlOa
+ HCl +
H2O
>
Al(0H)3i
+
NaCl
NaOH
+ HCl >
NaCl
+ H2O

Al
+ 6HC1 >
2AICI3
+
3H2T
242
=
33,6 gam
Cdu
2. Phan
ijfng:
2Fe +
3CI2
FeCl,
+
3NaOH
-> 2FeCl3
>
Fe(0H)3i
+
3NaCl
2Fe(OH)3
FeaOa +
3H2O
A:
Fe;
X:
CI2;
Fe203
+

3H2
B:
FeCla;
Y:
NaOH;
>
2Fe +
3H2O
C:
Fe(0H)3;
D:
Fe203;
Z:
H2
Khoi
iLforng
oxi
trong
46,4 gam Fe^Oy = 46,4 - 33,6 = 12,8 gam
,
X 33,6 X 16 3 , „ ^
y ^12:^ = 1
-^""^^^^^^^^^^^
b)
Fe +
4HNO3
> Fe(N03)3 + NOt +
2H2O
3Fe304
+

28HNO3
> 9Fe(N03)3 + NOT +
UlUO
Goi X,
y Ian
liigt
la so mol
ciia
Fe va
Fe304
trong
hon hop.
So mol
NO:
x + ^ = 0,52
3
Bao
toan
Fe: 56x + 168y = 33,6
=>
X
= 0,51 mol ^
mpe
=
28,56
(gam)
V',-
A,
B, D, X, Y, Z
CO

the khac,
nhUng
C
phai
la
Fe(0H)3.
y
= 0,03 mol ^
m^^
^ = 6,96
(gam).
Wl Rli,
„S
DE SO 8
flE
THI
HOC SINH GlDl HOA HOC 9, CAP TP. HQ CHI IVIINH NAM HOC 2001 -
ZOOZ
Cdu
1: Bo
tiic
ud can
bdng
cdc
phitang trinh phdn ling
sau:
Cdu
CaCh
+ ?
Ba(HC03)2

+ ?
CaS03
+ ?
HCl
FeClo
FeCh
Ca3(P04)2^
+ ?
^
BaCOaJ-
+ ?
->
SO. f + ? + ?
+
?
+
?
+
?
->
NallCOs
+ ?
->
FeCh
FeCU
Cdu
2:
Vict
cdc
pliUang

trinh phdn
I'tng
de
bleu dien chuoi bicn lioa
sau:
FcS2
^ SO2 -> SO3 ^
H2SO4
-> SO2
Na2S03
^
BaSOa
Cdu
3:
a)
Cho 6
dung dich
gom:
NaCl, BaCU,
CUSO4,
NaOH, MgCh, AgNO,.
KJioiig
dung
them
hoa
chat
ndo
khdc,
hay
nhan

biet
chung.
b)
Cho 3
dung dich: BaCU, BafNOi)., Ba(HC03)2.
Chi
diCgc
si'i
dung
them
mot hoa
chat,
hay
nlidn
bict
cluing.
c)
Cho hdn hap khi gom CO2, SOo.
Bdng
pliuang
phdp
hoa hoc, hay
tdch
rieng
CO2.
Cdu
4: A vd B Id 2
loai clidt
chi
cliiia

cdc
nguyen
to X, Y.
Thdn.Ii
pJidn
phdn tram
cua
nguyen
to X
trong
A vd B Idn
liigt
Id 30,4% vd
25,97/.
Neu
cong
thi'ic phdn
til cua A la XY2, thi
cong
thiic phdn
tii cua B
Id
gi?
Cdu
5: De gia
tang nong
do
cila
50 gam
dung dich CuSO^

57c len gap luu
Idn,
CO boil
hoc
sinh
da
thUc
Jiien
bdng
bdn
cdch khdc
nJiau:
Hoc sinh
A: dun
nong dung dich
de Idm bay hai
phdn ni'ca lugng nitac.
Hoc sinh
B:
them
2,78 gam
CuSO^ khan
vdo
dung dich.
Hoc sinh
C:
them
4,63 gam
tinh
the

CUSO4.5H2O
vdo
dung dich.
Hoc sinh
D:
them
50 gam
dung dich CuSO^ 157c
vdo
dung dich.
Hoi
hoc
sinJi
ndo da lam
diing, gidi thich.
Cdu
6: Hop
chat
A bi
phdn hiiy
a
nhiet
do coo
thco
phaang trinh plidn
dug:
L6\I
1:
Bo
tuc

va
can bkng:
3CaCl2 +
2Na3P04
>
Ca3(P04),i
+
6NaCI
Ba(HC03)2
+
Ba(0H)2
>
2BaC03i
+ 2H2O
CaSOa
+
2HC1
>
SOat
+ CaCl2 + H2O
HCl
+
NaaCOa
>
NaHC03
+
NaCl
2FeCl2
+ CI2 >
2FeCl3

2FeCl3
+
Fe
>
SFeClg
Cdu
2:
Viet
phUcfng
trinh
4FeS2
+
IIO2
2Fe203
+ 8SO2
2SO2 + O2 ^
450*^
C
:i
2SO3
SO3 + H2O
a)-
2A
B
+ 2D + 4E
San
phdm
tgo
thdnh
deu a the khi,

khoi lugng
mol
trung blnh
cua hon
hgp
khi sau
phdn ting
Id
22,86 (gimol).
Tinh
khoi lugng
mol cua A.
Cho
so
lieu:
H = 1; O =
16;
S = 32;
Cu
=
6"-^.
Hoc sinh
c6 the
sii dung bdng
do
tan vd bdng
he
thong tudn hodn
cdc
nguyen

to
hoa
hoc.
>
H2SO4
2H2SO4
+
Cu
>
CUSO4
+ S02t +
2H2O
SO2 +
2NaOH
>
Na2S03
+ H2O
Na2S03
+ BaCl2 >
BaSOsi
+ 2
NaCl
Nhan
biet
CUSO4:
mau
xanh
Cho
mSu
CUSO4

tac dung vdi cac mau c5n lai
c6 1
mau cho ket tua
xanh
lam
la
NaOH,
mot mau cho ket
tiia
trSng
la
BaCl2.
Lay
mSu NaOH cho tac dung
vdi
cac mau con lai: mau cho ket tua
trSng
la
MgCla,
mSu cho ket tua den
1^
AgNOa, mau khong tao ket
tua
la
NaCl.
CUSO4
+
2NaOH
CUSO4
+ BaCl2 -

-> Cu(0H)2i
+
Na2S04
MgCl2
+
2Na0H
—
AgNOg
+
NaOH
-
2AgOH
> Agp + up
•>
BaS04i
+
CuCla
-> Mg(0H)2i
+
2NaCl
^ AgOH
+
NaNO,
b)
- Dun
nong
mau cho ket tua la
Ba(HC03)2
Ba(HC03)2
^ BaCOgi + CO2T + H2O

- Hai mau con lai cho tac dung
vdri
dung dich AgNOs. Mau cho i
BaCl2 mSu con lai la Ba(N03)2.
BaCl2 + 2AgN03 > Ba(N03)2 + 2AgCU
c)
- Cho qua dung dich thuoc tim SO2 bi hap thu, con CO2 bay ra (cho
qua dung dich Br2 chi c6 SO2 bi hap thu).
- Co the dot
chay
SO2 SO3 hap thu vao Ba(0H)2 sau do cho HCl
vao de giai phong CO2.
Cdu4:
-
Cong
thufc A: XY2
-
Cong
thufc B: XaYb
- Trong A: %X = 30,4% %Y = 100% - 30,4% = 69,6%
- Trong B:
TCrd),
(2)
X _ 30,4 _ X 60,8
2Y 69,6
Y
69,6
%X = 25,9% %Y
=
74,1%

aX _ 25,9 _
X
25,9b
bY 74,1
Y
74,1a
60,8
25,9b
a _ 2
69,6 74,1a
b
5
Cdu
J
Cong
thufc cua B la: X2Y5.
5
2,5
.50 = 2,5 gam
(n,„so_
= ^ =
0,015625
mol)
^H,o
= 50 - 2,5 = 47,5 gam
Nong do can dat
di/cfc
la 10%.
Hoc sinh A lam bay hoi phan
nijfa

luong niidc
m^^Q
biioai
= =
23,75
= m^^o cbn
mdungd.ch
sau = 50 -
23,75
=
26,25
gam
[CuS04]% =
I
T^M^rJ
X 100 = 9,52% ^ 10%
• Hoc sinh B: m
126,25y
=
2,5 + 2,78 = 5,28 gam
sai
CUSO4
mdungdich
= 50 + 2,78 =
52,78
gam
' 5,28 '
[CuS04]%
=
52,78

J
100 = 10% => dung
2.
11
Hoc sinh C:
mp^so^
them =
160 X 4,63
250
=
2,9632
gam
I^CuSO.
dung
dich sau
—
2,5 +
2,9632
=
5,4632
gam
niduag
djch
= 50 + 4,63 = 54,63g
diing
[CuS04]%
= -5^^^ X 100 = 10%
Hoc sinh D:
54, 63
-

ii_
™CuS04 them vAo -
.50 = 7,5 gam
Cdu
6: 2A
Dat
2a (mol)
™cuS04 dung
dich
sau = 2,5 + 7,5 = 10 gam
nidung
dich
= 50 + 50 = 100 gam
[CuS04]% = 10%) ^ dung
>
B + 2D + 4E
a 2a 4a
Neu
lay 2a mol A nhiet phan se tao thanh 7a mol khi.
Dinh
luat
bao to^n khoi liicfng cho:
MA
X
2a =
22,86
x 7a =>
MA
= 80.
OE SO 9

BE THI HOC SINH GIOI HQA HOC 9, TP HAI
PHONG
(BANG
A) NAM HOC Z001 - Z002
Cdu
I
1. Cho bang phdn loai cdc chat:
1
2
3 4
5 6
7 8
HI
NO
CO
O2
Fe
Cu(0H)2
CH4
KOH
H2SO,
Na20
NO
SO2 N2
KOH
C6H22O6
Ba(0H)2
H2S
CO2
CH4 Br2

NaOH
CCI4
NaOH
Hay cho
blet
vi tri (1), (2), (3), (4)., (5), (6), (7), (8) la cdc ti( gi.
•2.
Neu hien tUgng, viet phUpng
trinh
phdn ling cho cdc thi nghieni sau:
a)
Nhung dinh sdt da cqo sach gi vao dung dich CuSOj.
Sue
khi SO2 vao dung dich
Ca(HC0a)2.
^)
Dan khi etilen qua dung dich nUac brom.
^' Cho day chuyen hoa sau:
Fe
->A -^B ->C-^Fe ->E->-F-^D
dinh
A, B, C, D, E, F. Viet cdc phiiang
trinh
phdn ring.
I un/t unr a
39
Cau
II.
1. Dung dich Boocdo dung chong nam cho cay dugc pha
theo

tl le:
1kg CUSO4.5H2O + 10kg voi song (CaO) + 100 lit nuac
Hay tinh thdnh phdn %
theo
khoi lUgng cdc chat c6 trong dung die/,
Boocdo. Viet cdc phiiang
trinh
phdn ling.
2. Tie glucoza va cdc chat v6 ca can thiet, viet cdc phuang
trinh
phcn,
ling
dieu che: etylaxetat.
Cdu
III.
1. Ba khi A, B, C c6 phdn ti2 khoi bdng nhau vd bdng 28 dvC. A, B cu
the hi dot chdy trong khong khi, sdn phdm sinh ra deu c6 khi COo, li
CO
the
kill}
dugc CuO a nhiet do coo, C la thdnh phdn quan
trgiii:
trong phdn bon hoa hoc. Xdc dinh cong thilc phdn ti2 cua A, B, C, vii')
cdc phuang
trinh
phdn iCng.
2. (M), (N), (P), (Q), (R), (X) Id nhUng hap chat hUu ca dugc
biet
den
trong chuang

trinh
hoa hoc pho thong cap trung hoc ca sd. (P) vd (Ni
CO
cung cong
thvCc
phdn tit.
- Ve khoi lugng phdn tit (klpt):
klpt (N) = -klpt (M); klpt (X) = 3klpt (R) - Gklpt (Q)
2
2,3 gam (N) hay 1,5 gam (Q) c6 the tinh bdng the tich cua 1,6 gam 0^
cung dieu kien.
- Ve tinh chat: (M), (N), (R) cd phdn itng vai Na, tic 0,1 mol (M) c6 the
cJio
the tich H2 Ian nhdt la 3,36 lit (dktc), chi c6 (R) phdn itng dugc
vai
dung dich NaOH. Tit (X) cd the dieu che ra (N), (R), (Q) c6 phdn
ling
vai CI2 (chieu sang). Xdc dinh cong thitc cdu tgo cua (M), (N), (P'.
(Q),
(R) vd cong thUc phdn tii cua (X).
Gidi
thich.
Cdu
IV. Hoa tan mudi nitrat cHa mot kim logi hoa tri 2 vao nudc dugc 200
lu'i
dung
dich (A). Cho vao dung dich (A) 200 ml dung dich K3PO4, phan iint-
xay ra vita du, thu dugc ket tua (B) vd dung dich (C).
Klioi
lugng ket tua (I'

vd khoi lugng mudi nitrat trong dung dich (A) khdc nhau 3,64 gam.
1. Tim nong do mol/lit cua dung dich (A) vd (C), gid thiet the tich dwni
dich
khong thay doi do pha trgn vd the tich ket tua khong ddng ke.
2. Cho dung dich NaOH (lay du) vao 100 ml dung dich (A) thu dugc kH
tua
(D), Igc lay ket tua (D) roi dem nung den khoi lugng khong do'
can dugc 2,4 gam chat rdn. Xdc
dinli
kim logi trong mudi nitrat.
40
1^1
niAi np
TMi
unr ciuu mni un/t unr Q
V
Doi
c/'idy
hodn todn 1,344 lit (dktc) hon hgp 3 hidrocacbon the khi:
C
H'>n
+
2! CinH2m', CkH2k -
2-
Sau phdn Ung, dan lion hgp sdn piidm Idn
lugt
Q'^'^
H2SO4 fdgc), dung dich NaOH (du) thdy khoi lugng H2SO4
(dd^^)
^'^"^ ^'^^ khoi lugng dung dich NaOH tang 7,04 gam.

1
Tinh
thdnh phdn %
theo
the tich hSn hgp 3 hidrocacbon,
biet
the tich
hidrocacbon CkH2k-2 trong hon hgp gap 3 Idn the tich CJl2„+2-
2 Xdc dinh cong thiic phdn ti2 3 hidrocacbon,
biet
rhng cd 2 hidrocacbon
CO
so nguyen tit cacbon bdng nhau vd bdng so nguyen tit cacbon
cua hidrocacon con Igi.
Ldl
GIAI
Cdu
I.
1.
Ten goi cac vi
tri:
(1):
axit;
(2): oxit; (3): oxit khong tao muoi; (4); chat khi;
(5): dcfn chat; (6):
baza;
(7): chat hau ccf; (8):
bazcf
kiem.
2. Viet phan

ijfng:
a) Fe + CUSO4 >
FeS04
+ Cui
- Dung dich mau xanh bi nhat dan.
- Co ket
tiia
cua. dong xuat hien.
b)
SO2 +
Ca(HC03)2
> CaSOgl- + 2C02t + H2O
(cd ket tua va c6. khi)
hay
2SO2
+
Ca(HC03)2
>
Ca(HS03)2
+
2CO2T
(c6 khi bay ra) '
c)
CH2 = CH2 + Br2 >
CH2Br-CH2Br
(mat mau ndu do dung dich nifdc brom)
3. Thiic hien day chuyen hoa:
Pe yi_> FeClg : > Fe(0H)3
Fe203
Fe

FeCl2
Fe(0H)2
>
FeS04
>
FeCl2.
Phan
ling:
1) 2Fe +
3CI2
2FeCl3
2) FeClg +
3NaOH
> Fe(OH)34 +
3NaCl
nau
do
THI HQC
SINH
Gini
HfiA
Hnn Q " 41
3)
2Fe(OH)3
->
FeaOa
+
SHgO
4)
FeaOg

+ SCO
5)
Fe +
2HC1
-
> 2Fe + 3CO2T
FeCl2 + H2T
6)
FeCl2 +
2NaOH
>
Fe(0H)2i
+
2NaCl
tvang
xauh
7)
Fe(0H)2
+
H2SO4
>
FeS04
+
2H2O
Cac phan ufng dieu che:
,y
„ len men
riiau
• L6ni2*-'6
>

2C2H5OH
+ 2C02t
len men
giam
.
C2H5OH
+ O2
^
^ ^
dam
dac
•
CH3COOH
+
C2H5OH
^ \ ==
m
8)
FeS04
+
BaCla
10
>
FeCl2
+
BaS04i
10*
Cdu
1.
Theo

de:
ncao
= — x
1000
= -—
(mol);
n„ =
1000
10^
56
10^
18
56
250
250
(mol)
(mol)
Phan ijfng: CaO
+ H2O
>
Ca(0H)2
Ca(0H)2
+
CUSO4
>
Cu(0H)2i
+
CaS04
(1)
Khoi iLTcfng dung dich

Boocdo
thu
dirge
la: 100
+ 1 +
10
=
111
(kg)
_ 10^
•^4
250
(mol)
m
C"(o")2
250
X
98 =
0,392.10^ (gam)
%m
Cu(OH),
0,392.10'^
111 X1000
100
=
0,3539fc.
.
TCrd)
n
CaSO,,

m
CaSO,
10'
250
10^
" 250
(mol)
136
=
544 (gam)
%m
544
CaS04
111x1000
X 100
=
0,49%
Tir(l)
n
- 121 _
1^
Ca(OH)^
d.r - 56 250
m
CalOH)^
dii
10*
%m
10^^
56

250
12918
(mol)
X
74 *
12918 (gam)
X
100 =
11,64%
Ca(OH).^dLJ
111x1000
%m„ ^ =
100%
-
(0,353%
+
0,49%
+
11,64%)
=
87,517%.
>
CH3COOH
+ H2O
^
CH3COOC2H5
+ H2O
(etyl
axetat)
III.

-
A,
B,
C
deu
CO M =
28 dvC.
-
B CO
the kliLf
dagc
CuO
d
nhiet do cao
va
khi chay tao
CO2
=> (B): CO.
-
C la
thanh phan quan trong trong phan bon => (C):
N2.
-
(A)
khi ch^y tao
CO2
(MA
=
28) => (A):
C2H4:

(etylen).
Phan ufng:
C2H4
+ 3O2
2C0
+ O2 -
200,1
+
2H2O
->
2CO9T
2.
Ta c6:
n
CO
+
CuO
1,6
->
Cu +
CO2T.
CO,,
32
= 0,05 (mol)
Do do: phan ttf khoi cua
(N):
2,3
0,05
1,5
= 46 (gam)

phan
ttjf
khoi cua
(Q):
=
30 (gam)
0, 05
Mat khac,
(N)
tac dung
vdi Na
nen
(N)
la:
C2H5OH.
(Q)
tac dung vdi
CI2
chieu sang nen (Q) la:
C2H6.
•
Theo
de: phan ttf khoi cua
(N) = -
khoi liJgng phan tii cua
M
2
phan
tijf
khoi ciia

(M)
la: 2.46
= 92
(gam)
3,36
0,1 mol
M + Na
3,36 lit
H2
22,4
= 0,15 (mol)
> nM
: = 2 : 3 => (M) la
ri/gu
3
Ian rugu (ri/gu
3
chiirc)
=^ (M): CH„ -CH-CH,
II
I-
OH
OH OH
P
CO
cong
thufc
C2H6O
nhifng khong phan ufng vdi
Na va

NaOH.
=^
(P):
CH3-O-CH3
(ete).
R
viia
phan ufng vdi
Na va
viTa phan ufng vdi
NaOH.
=>
Cong
thufc (R):
CH3COOH
(MR
= 60g)
Va phan tuf khoi cua
(X)
la: Mx
=
3MR
= 3 X 60 =
180 (gam)
va
(X)
dieu che
ra
C2H5OH
va

CH3COOH
=>
(X):
C6H12OC
(glucozcf).
Cdu rv.
1. Phan
ling:
3M(N03)2 +
2K3PO4
>
M3(P04)2i
+ 6KNO3 (li
(Vdi
M la kim
loai
hoa tri II)
TU
phan
ufng
(1): Su khac nhau ve
kho'i
lijgng
la do thay 6N()
(M
= 372) bang
2PO4
(M = 190).
^ ri , > = —= 0,02 (mol)
M3(PO.,)2

372-190
(1) ^M(N03)^ = ^''M,(^O,).^ = 3 X 0,02 = 0,06 (mol)
Vay:
- = = 0,3M
0,02 X 6
'KNOS
0,2 + 0,2
2.
Phan
iJfng:
M(N03)2
+ 2NaOH > M(0H)2l + 2NaN03 (2)
(mol)
0,03 0,03
M(0H)2
—> MO + H2O , (3)
(mol)
0,03 0,03
Tii
(2), (3) muo = 0,03 x (M + 16) = 2,4
=> M = 64; Dong (Cu).
Cdu V.
Dat:
A:
C„H2n+2
; B: CniH2,n ; C: CkH2k;-2-
Ta c6: n
hon
hep 3
khi

= " ^'^^
1.
Tinh
phan
tram
the"
tich
moi hidrocacbon.
Cac phan
ufng:
C„H2n.2 + O2 '—^
nC02
+ (n +
DHgO
(D
\ J
(mol)
a an a(n + 1)
C„,H2:n + — O2 mCOs +
mllaO
(21
2
CkH2k-2 + [^^J O2 ^
kC02
+ (k - DHsO (3,*
(mol)
3a . 3ak 3a(k - 1)
Ta co:
Binh
diing

H2SO4 tang
chinh
la
khoi
lugng
cua I-I2O.
m
H,0
= 2,52 gam => n
H.,0
2,52
18
= 0,14 (mol)
Binh
dung
NaOH
tang
chinh
la
khoi
lifgng
ciia CO2.
m.
= 7,04 gam => n
CO,
7,04
44
= 0,16 (mol)
Theo de: nc = 3nA, thi so mol H2O giam va nho han so' mol CO2 mot
lirgng

bang 2 Ian so mol cua A.
0,16-0,14
HA
=
= 0,01 (mol)
=:> nc = 0,03 (mol)
=> nB = 0,06 - 0,04 = 0,02 (mol)
Vi
la chat khi nen %V = %n
Vay:
%VA
= X 100 =
16,67%;
%VB
=
0,06
0,02
X
100 =
33,33%
0, 06
%Vc = 50%.
. Tim CTPT cua 3 hidrocacbon:
TCr
(1), (2), (3) :=> ^ n^o^ = 0,01n + 0,02m +
0,03k
= 0,16
n
+ 2m + 3k = 16
Do

la chat khi nen n, m, k < 4 va m, k > 2.
Vi
m, n, k la nguyen va chSn nen nghiem hgp li la: n = 2; m = 4; k = 2
^ 3 hidrocacbon can tim la: (A): C2H6; (B): C4H8 va (C): C2H2
(*)
DE SO 10
flE
THI
CHQN
HOC
SINH
GIQ!
HOA
HOC
9,
TJNH
BJNH
OjNH
NAM
HOC
2001
-
2002
1- Chi
dugc
dung CO2 vd HoO, hay trinh bay
each
phdn
biet
4 Ig

chila
^
chdt
rdn: K2CO3, BaCOj, KNO3, BaSO^. Viet
phdn
iCng de niinh hga.
2. Sue a (mol) COo vdo dung dich
chi'ia
1 mol CafOWs. -
Tinh
so mol CaCOs tgo thdnh ting vai gid tri a = 0; a = 1; a ^ 2.
Ve
duang
bleu
dien
so mol CaCOs tgo thdnh
theo
so mol CO2 da cho.
®" 3.
iV/ue^
phdn
m 1 gam hSn hgp Mg, MgCOs
ngodi
khong
khi den khi
phdn
ling xong ta thu
dugc
7712
ga77i

mot
chat
ra'n. Biet 7711 = 77x2. Tinh %
lugng Mg
t7-o7ig
hdn hap ddu.
:
THI
I
Cdu 4. Hoa tan hon hap
Na20,
NaHCOs, BaCl2,
NH4CI
c6 cung so mol vaa
nuac
du, dun nong nlie thu dugc dung
dich
A vd ket tiia BaCOs- Hoi
dung
dich
A
chiia
gi? Viet phdn
iCng
ininh
hga.
Cdu 5. Trgn 11,2 gam bgt Fe vd 4 gam bgt S trong
chen
sii dem
?iuiig

khong
CO
khong
khi de
phdn
iCng
xdy ra tao FeS vai
hieu
sudt 80%. Lay'
chat
rdn tim
dugc
trong
chen
sii cho tdc dung viia du vai V lit dung dich
HCl
IM,
thodt
ra a (mol) hon hap khi vd m (gam)
chat
rdn
khong
tan.
a)
Viet tat cd
phdn
ling xdy ra.
b)
Ttnh gid tri V, a, m.
Cdu 6. Dem hSn hgp gom 0,1 mol Mg vd 0,2 mol Al tdc dung vdi mot

lugng H2SO4 d4c, nong viia du thu
dugc
hon hgp mudi,
0,075
mol S va
0,175 mol SO2.
a)
Tinh
khdi
lugng hdn hgp mudi tgo thdnh.
h)
Tinh sd mol
H2SO4
phdn
iJCng
viia du.
LCil
GIAI
Cdu 1
-
Lay moi lo mot it cho vao 4 co'c.
-
Che
niidfc
vao 4 co'c, phan
diioc
2
nhom:
nhom
(I) tan:

dung
dich
K2CO3, KNO3;
nhom
(II)
khong
tan; BaCOg, BaS04.
-
Sue khi CO2 vao 2 coc
nhom
(II):
Neu
tan la coc chtTa BaCOs, phan
tfng
xay ra tao ra
dung
dich
BadlCOa)^:
CO2
+ H2O + BaCOg > Ba(HC03)2
Coc
khong
tan chufa
BaS04.
-
Lay it
dung
dich
Ba(HC03)2 nho vao 2 coc cua
nhom

(I).
Neu
coc nao tao ra ket tua trSng do la coc chufa K2CO3, phan
ufng
tao
raBaCOg.
K2CO3
+ Ba(HC0g)2 > BaCOg + 2KHCO3
Coc
khong
ket
tiia
chufa
KNOg.
Cdu 2.
a)
Tinh
so mol CaCOg: Khi a = 0
khong
c6 phan
ufng,
so mol CaCOg = 0.
Khi
a = 1 phan
ufng
xay ra:
CO2
+ Ca(0H)2 > CaCOg + H2O (D
=> So mol CaCOg = 1.
Khi

a = 2 phan
ufng
xay ra:
2CO2
+ Ca(0H)2 > Ca(HC03)2 (2'
=> So mol CaCOg = 0.
Triicfng
hop 2 c6 the
viet
phan
ufng
hoa tan het CaCOg.
If)
Chon 3
diem:
n
^'
"^CaCO.j
= 0
C02 " •'
'•'CaCO;,
= 1
C02 ~ ^
^CaCOg
= 0
C&u3.
Phan
ufng
xay ra:
Mg+

^02
MgCOg
—
MgO
MgO
+ C02t
>oi
a, b Ian
iLTcft
la so mol cua Mg va
MgCOg
trong
hSn hgp dau.
Vi
mi = m2 nen
khoi
lirgng
O2 (1) =
khoi
liigng
CO2 (2) =^ 16a = 44b
(1)
(2)
a
b
11
m
•Mg
m.
^MgCOg

Can 4.
Cac phan
ufng
xay ra:
NaaO
+ H2O
11.24 _ 264
4.84 ~ 336
Klio'i
liigng
Mg = 44%
NaliCOg
+
NaOH
BaCl2 + NaaCOa -
NH4CI
+
NaOH
-
2NaOH
> NagCOg + H2O
(1)
(2)
(3)
(4)
Vi
so mol cua 4 chat: NagO, BaClg,
NaHCOg,
NH4CI bSng
nhau,

nen
theo (1), (2), (3), (4)
dung
dich
A chi chila
NaCl.
Cdu 5.
a) Phan
ufng:
Fe + S '°
-> BaCOgi + 2NaCl
NaCl
+ H2O +
NHgt
FeS
(1)
Vi
CO
hieu
suat 80% nen chat ran gom: FeS, Fe
di/,
S diT
FeS + 2HC1 >
FeCl2
+ HaSt
Fe + 2HC1 >
FeCl2
+ H2t
(2)
(3)

^)
Theo (1), neu
hieu
suat
bkng
100% thi S het, Fe dif.
Khi
hieu
suat 80% thi: S dii = 0,8 gam = a; Sd mol S phan
ufng
= Sd
oiol
Fe phan
ufng
= So mol FeS
sinh
ra = 0,1 mol.
Vay
sdmol
Fe = 0,1 mol.
(2) va (3) ^ Sd mol HgS = Sd mol FeS = 0,1 mol.
Sd mol H2 = Sd mol Fe = 0,1 mol =:> a = 0,2 mol, sd mol HCI
dung
la
0.4 mol V = 0,4 lit.
Cdu
6.
a) Tinh khS'i Itfgng hon
hcfp
muol:

CiJ
0,1 mol Mg tao ra 0,1 mol MgS04 (bao toan khoi lirgng)
Cur
0,2 mol Al tao ra 0,1 mol
Al2(S04)3
(bao toan khoi lifgng)
Vay khoi
li/crng
hon hgp muoi
=
(120.0,1
+
342.0,1)
=
46,2 gam
b) Tinh
so
mol H2SO4 dac n6ng da dung vCra du.
Phan
ufng
xay ra:
3Mg
+
4H2SO4
>
3MgS04
+ Si +
4H2O
(1)
Mg

+
2H2SO4
>
MgS04
+
S02t
+
2H2O
(2)
2A1
+
4H2SO4
>
Al2(S04)3
+ Si +
4H2O
(3)
2A1
+
6H2SO4
>
Al2(S04)3
+
3S02t
+
6H2O
(4)
Theo
(1) va
(3) ^

^H^SO,
dung
=
^.ng
=
(0,075
x
4) mol
= 0,3
(mol)
Theo
(2)
va
(4)
=>
n^^^gg^
^^^^
= 2n^^^ =
(0,175
x 2)
mol
=
0,35 (mol)
Vay so' mol H2SO4
da
dung viTa du 1^ 0,65 (mol).
DES0
11
DE
THI HOC

SINH
Gidl
HOA HOC 9
(VONG
1),
TJNH
KHANH
HOA NAM HOC
2001
-2002
Cdu
I.
1.
Hay
thuc
hien
cdc
phep
tinh
sau (yeu cdu ghi day du dan vi
troii;;
phep
tinh):
a)
Tinh
so mol do
trong 7,19 gam
do;
b)
Tinh

so mol O vd O2
trong
8 gam oxi;
c)
Tinh
khoi lugng cua 0,05
mol kem;
d)
Tinh
khoi lugng cua 0,75
mol
nudc;
d)
Tinh
so
nguyen
tii C
trong 0,02 gam
C;
g)
Tinh
so
phdn
tvC
CO2 trong 1,1 gam
khi
CO2;
h)
Tinh
so g cua 1

nguyen tii Na;
i)
Tinh
so g cua
1 phdn
tii
SO2.
2.
Mot hdn hgp X gom FeO vd
Fe203
cd
khoi lugng
Id 30,4
gam. Nun.^
hon
hgp nay
trong
mot
binh
kin cd
chica
22,4 lit CO
(dktc).
Kho'
lugng
hon hgp khi
thu dugc sau
khi
nung
la

36 gam.
a)
Hay xdc
dinh thdnh phdn
hon hgp khi.
Biet rdng
hdn hgp X
b\
khv[
hodn toan thdnh
Fe.
b)
Tinh
khoi lugng
Fe
thu dugc vd khoi lugng cua mSi
oxit
sdt trong X.
Co
6 to mat
nhan diCng
cdc
dung dich khong mdu Id:
iVa^SOj
(1);
MasCOs
(2);
BaCls
(3);
BafNOaJs

(4);
AgNOa
(5);
MgCls
(6).
Bdng phuang
plidp
hoa hoc vd
khong dung
them
cdc hoa
chat
khdc
hay
trinh bay
each
nhan
biet
cdc
dung dich tren,
biet
rdng chung
deu c6
nong
do du
Ian
de
cdc
ket tua it tan
cUng

c6 the tao
thdnh. (Khong
can
viet
phuang trinh
phdn ling).
Cdu
3-
^°<^
dinh nong
do cua cdc
muoi
NaHCOa
vd
Na2C03 trong
mot
dung dich
hdn hgp cua
chung (dung dich
A),
nguai
ta lam cdc thi
nghieni nhu sau:
Thi
nghiem
1: Lay 25 ml
dung dich
A cho tdc
dung
vai 100 nd

dung
dich
HCl IM (du) dun
nong
hdn hgp, sau dd
trung
hoa
lugng axit
du
bdng lugng vita
du la 14 ml
dung dich NaOH 2M.
Thi
nghiem
2:
Lai lay
25 ml
dung dich
A, cho
tdc dung vai lilang
du
dung
dich
BaCl2.
Loc bo ket
tua mdi
too
thdnh, thu
lay
nUdc Igc

vd
nuac nla gap
Igi
roi cho tdc dung vai lugng vica du la
26
ml dung dich HCl
IM.
1. Viet cdc phuang trinh phdn ling
xay ra
vd gidi thich vdn
tdt.
2.
Tinh
nong
do mol cua moi
muoi
trong dung dich
A.
Cdu
4.
Mot
dung dich axit axetic CH3COOH
cd C% =
10%.
Lay 300 gam
dung dich axit
nay cho tdc
dung vai
300 ml
dung dich NaOH

2M tao ra
dung dich
A.
Dung dich
A
cd tinh axit hay
baza?
Tinh
nong
dd % cdc
chat
tan
trong dung dich
A,
biet
rdng dung dich
NaOH2M
cd
d =
1,2 giml.
Ghi
chu: Hoc sink dugc
phep
svt dung bdng tudn hodn, bdng tinh tan, gido vien
coi tin khong gidi thich gi them.
LCll GIAI
Cdu
1
7
1'

1.
a)
nci
=
—-— (mol)
35,5
8
8
b)
no = —
(mol); n^^
= —
(mol)
c)
mzn
= 0,05
X
65,38
=
3,27 gam
d)
mj^^Q
=
0,75 mol
x 18
g/mol
=
13,5 gam
iSlSi^li^
THI HOC SINH

G|6'HOA
HQC 9
d)
nc = =
1,66.10^^
(mol)
g)
n
So nguyen tuf
C =
6,02.10^^
nguyen ti:f/mol.l,66.10^ mol
=
9,993.10^°
nguyen tuf.
1,1
CO2
44
=
0,025
(mol)
So phan
td CO2 =
0,025
x
6,02.10''
=
1,505.10^^
phan tit.
h)

MNa
= 23
g/mol
niNa
23
=
3,82.10"^^
g/nguyen ttf
6,02.10'
23
i)
MgQ =
64,06
g/mol =>
=
64,06
-,23
=
10,64.10'"'
g/nguyen tuf
6,02.10'
2, a) Ta c6: nco (ban ddu) = 1 mol => mco (ban
dsuj
= 28 gam.
Do tang
khdi
liiOng:
36 - 28 = 8 gam = mo =>
n^^
= 0,5 mol.

Vay CO
0,5
mol
CO
ket hop vdi
0,5
mol
O
cho
ra 0,5
mol CO2.
Th^nh
phan hon hap khi
la: 0,5 mol CO va 0,5 mol CO2.
b)
Ta
c6: n^^
(c6
trong FeO
va
Fe203)
=
n^^ lay
ra (d
tren)
= 0,5
mol.
Goi
a =
npeo

va b =
np^^p^.
Vay trong
X c6 (a + 3b)
mol
O.
Do do:
a + 3b = 0,5 (1)
Va: 72a
+
160b
=
30,4
(2)
Giai
(1) va (2)
dirge:
a = 0,2
mol FeO
va b = 0,1
mol
FeaOg.
Vay:
mpeo
= 0,2 x 72 =
14,4 gam
"^Fe.,03
= 0,1 X 160 = 16 gam.
Theo
dinh luat

bao
toan nguyen
to
thi:
npe (thu
dLToc)
=
npe (trong FeO)
+
npe (trong
Fe203)
=
a + 2b = 0,4
mol.
Vay: nipe (thu
duoc)
= 0,4 x 56 =
22,4 gam.
Cdu
2
1. Lay mot dung dich bat ki
cho vao 5
dung dich
con
lai,
ta c6
bang sau:
Na2S04
Na2C03
BaCl2

Ba(N03)2
AgNOa
MgCl2
Na2S04
-
-
i
i
i
-
Na2C03
- -
i
i i
BaClz
i
i
-
-
i
-
Ba(N03)2
i
i
- - - -
AgNOa
i
i
-
-

i
MgCl2
-
i
-
-
i
-
50
LOI GIAI at THI HOC SINH GIQI HOA HOC 9
Tii
bang
tren
ta
thay:
Dung
dich
nao cho vao tao
ra
4
Ian ket tua
la
dung
dich
Na2C03
va
• AgNOa
(cap
dung
dich

1).
Dung
dich
nao cho vao tao
ra
3
Ian ket tua
la
dung
dich
Na2S04
va
BaCla
(cap
dung
dich
2).
Dung
dich
nao cho
vao
tao ra 2
Ian
ket
tiia
la
dung
dich
MgCls
va

Ba(N03)2
(cap
dung
dich
3).
+) Lay mot trong hai chat
a
c&p dung
dich
3
Ian lirgt
cho vao 2
dung
dich
6 cap 2, neu c6 tao
ra ket
tiia:
thi chat
cho vao la
Ba(N03)2,
con
lai
la
MgCl2.
Chat
tao
ra ket tua
a cap 2 la
Na2S04,
con

lai
la
BaCl2.
+) Lay Ba(N03)2
da
tim
diTOc
a
cap
3
cho vao hai dung
dich
d cap 1,
neu CO ket tua thi:
Chat
tao ra
ket
tiia
vcfi Ba(N03)2
la
Na2C03 con
lai
la
AgNOa.
Cdu
3
Bat
so
mol Na2C03
va

NaHCOa trong
25
ml dung
dich
A
Ian liTot
la x, y.
Doi
v(5i thi nghiem
1, ta c6:
Na2C0a
+
2HC1
>
2NaCl
+
COgt
+
H2O
(1)
(mol)
X 2x
I
NaHCOa
+
HCl
>
NaCl
+
CO.T

+
H2O
(2)
1^
(mol) y y
H||
HCl (dii)
+
NaOH
>
NaCl
+
H2O
(3)
So mol HCl trong
100
ml dung
dich
la:
0,1 x 1 = 0,1
mol
So mol HCl dir sau phan ufng
(1) va (2)
la: 0,014
x 2 =
0,028
mol.
k So
mol HCl da tac dung
vdri

dung
dich
A
la:
2x
+ y =
0,1
-
0,028
=
0,072
(4)
Boi
vdfi thi nghiem
2:
BaCl2
+
NaaCOa
>
BaCOai
+
2NaCl
(5)
Sau
khi loc
bo
ket tua, lay niidfc loc,
ni/dc
riia
chufa NaHCOa cho tac dung

vefi
dung
dich
HCl.
^ •
||^|^'
NaHCOa
+
HCl
>
NaCl
+ CO2 +
H2O
(6)
(mol)
y y
^iai
ra
ta
c6:
y =
0,026
x 1,0 =
0,026
mol
x
=
0,023
mol
nong

do
mol cua NaHCOa la:
0,023
:
0,025
=
0,92M.
nong
do
mol
cua
NaHCOg la:
0,026
:
0,025
=
1,04M.
iJ^iililnf

Cdu
4
300 gam
CH3COOH
10%
chOra
30 gam
CH3COOH
^
So mol
CH3COOH

=
0,5mol.
300
ml
dung dich NaOH
2M
chufa nNaOH
= 0,3 x 2 = 0,6
mol.
Phan
ling:
CH3COOH +
NaOH
>
CHaCOONa
+ H2O
(mol)
0,5 0,5 0,5
Ti
le
phan
lifng
la 1 : 1.
Vay
sau
phan ufng
con
NaOH
= 0,6 - 0,5 = 0,1 (mol)
dung dich

A c6
tinh
bazcf.
Tinh
khoi lufOng dung dich
A:
m
(dung dich
A) = m
(dung dich
CH3COOH) + m
(dung dich NaOlI
= 300 +
Vd,.>gdkh
X d = 300 + 300 X 1,2 = 660 gam
0,5
X 82 X 100%
i
a)
V
\
C%
(CHgCOONa)
=
660
=
6,21%
«(NaOH)
=2iiLf^,
0,6%.

660
DE SO 12
DE
THI
HOC
SINH
GIQI
HOA
HOC 9, CAP
TP.
HO CHJ
MINH
NAM HOC 2002
-
2003
Cdu
I.
a)
Cho
biet
dp tan mot
loai muoi bii'n
doi
theo
nhiet
dp
theo
bang
sau
neu

lam
Iqnh dung dich
c6
nong
dp %
ciia muoi
do Id 22% tie
50°Cl
thi
phqm
vi de bdt ddu
xudt
Men ket
tinh
cua
muoi
do
trong khodni^
nhiet
do
ndo?
Nhiet
dp
10°
C
20°C
30°C
40°C
Dp
tan

(trong lOOg nicac)
n,3g
15,2g
18,9g
24,8g
36,7g
b)
Co 6
dung dich dugc ddnh
so
ngdu nhien
tic 1 den 6, moi
dung c/?'
chiia
mpt
chdt
gSm:
BaCh,
H2SO4, NaOH,
HCl,
MgCl., NazCOa.
La>-
luat thuc hien
cdc thi
nghiem
ud
dugc
ket qua sau:
Thi
nghiem

1:
Dung dich
(2) cho ket tua vai cdc
dung dich
(3) vd
(4'i'
Thi
nghiem
2:
Dung dich
(6) cho ket tua vai cdc
dung dich
(1) vd
^4'
Thi
nghiem
3:
Dung dich
(4) cho khi bay len khi tdc
dung
vai
dung
dich
(3) vd (5).
Hay
xdc
dinh
s6 cua cdc
dung dich.
IL

Viet
5
phuang
tri/ih
phan icng dieu
che
CaCOs.
Cho
4 kim
loai
sau: Be, Na, Al, Ca. Hoi kim
loai
ndo khi tqo hap kim
vdi
Mg thi khi dot
chdy
hap kim do,
oxlt
tqo nen c6
khoi lugng
gap
doi khoi lugng
hgp kim ban ddu.
Cho
biet:
Kim
loai
Be
Na Al
Ca

Hda
tri.
H
I
HI
H
Khoi
lugng nguyen
ti'i
9
23
27
40
i
c)
Dot
chdy
hodn todn
5,6 lit
(dktc)
hdn hgp khi gom H2 vd H2S. sdn
phdm
tqo
thdnh
la
nuac
vd mpt
chdt
khi.
Lugng

khi nay tdc
dung viCa
du
vdi 100 ml
dung dich NaOH
2M de tqo
thdnh muoi trung
hda.
Tinh
khoi lugng inol trung binh ciia
hon hgp khi ban ddu.
Cdu
HI. Dot
chdy
1,76 gam
sunfua
cua mpt kim
loqi
MeS (kim
loqi
Me
trong
cdc hgp
chat
chi cd hai hda' tri 2 vd 3)
trong
oxi du. Hda tan
chdt
rdn
sau

phan ilng bdng
mpt
lugng
viia
du
dung dich
H2SO4
29,4%. Dung
dich
tqo nen c6
nong
dp
34,5%.
Lam
Iqnh dung dich
thi thu
dugc
2,9
gam
tinh
the
hidrat
Idng
xud'ng
vd
dung dich
con Iqi cd
nong
dp
muoi

Id 23%. Xdc
dinh
cong
thUc
cua
tinh
the
hidrat.
Cdu
rV. Cho m gam hon hgp
CaCOs
vd FeS tdc
dung
vUa du vdi V ml
dung
dich
HCl (D ^ 1,1 g/ml) thi thu
dugc
a lit hon hgp
klii
X
(dktc)
c6
khoi lugng
mol
trung binh
Id
40,67 (gam/mol)
vd
dung dich

Y cd
khoi
ugng
la b gam.
<.)
Tinh
a
theo
m, V, b.
b)
Ap
dung:
Cho m = 1,44 gam; V = 400 ml; b =
440,83
gam.
Tinh
a.
c)
Neu chi sii
dung
mpt dU
kien khoi lugng
mol
trung binh ciia
X la
^^0,67 (gam/mol),
hay
tinh
%
theo

khoi lugng
cua hdn hgp ban ddu.
Cdu
V. Mpt hon hgp A gom
Na2C03
vd
NaoSOs
cho tdc
dung
vdi
dung
(iich
HCl du thu
dugc
hon hgp khi X cd
klidi
lugng
mol
trung binh
Id
56 (gam/mol).
C/io
0,224
tit
(dktc)
khi X di qua 1 lit
dung dich Ba(0H)2-
•Sau
thi
nghiem phdi diing

50 ml
dung dich
HCl 0,2M de
trung
hda
^"(0H)2du
'
Tinh
% so mol mdi khi
trong
X vd %
khoi lugng
moi
chdt trong
hon
hgp
A.
Tinh
nong
dp
mol/lit
cua
dung dich Ba(0H)2
ban ddu.
''^'10
sd
lieu:
H= 1,0= 16, Na = 23. S =
.32,
CI = 35,5, Ca = 40, Fe = 56, Ba = 137

VTHI
HOC
SINH
Rini
HOA HOC 9 53