Радецкий А.М., Горшкова В.П. Решение контрольных и самостоятельных работ по химии за 9 класс
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11
Al(NO
3
)
3
+
16
]
340
]
3NaOH
⋅
→
o]
3
78
]
Al(OH)
↓
+3NaNO
3
16
o
340 78
=
⋅
o
=
⋅
⋅
=10,4
]Hl\_l
m(Al(OH)
3
)=10,4
]
3. Na
2
CO
3
+2HNO
3
→
2NaNO
3
+H
2
O+CO
2
m(HNO
3
)=
ω
(HNO
3
)
⋅
m
jjZ
=0,2
⋅
400=80
]
n(HNO
3
)=
3
3
m(HNO )
80
M(HNO ) 63
=
=1,27
fhev
n(Na
2
CO
3
)=
2
3
23
m(Na CO )
53
M(Na CO ) 106
=
=0,5
fhev
Ld
HNO
3
gZoh^blky\ba[uld_lhjZkq_l \_^_fih
Na
2
CO
3
53
23
106
Na CO
+2HNO
3
→
2NaNO
3
+H
2
O+
oe
2
22,4
e
CO
53
o
106 22,4
=
o
=
16 78
340
⋅
⋅
e
Hl\_l
V(CO
2
)=11,2
e
4.
94,5
]
20
]
3
MgO+2HNO
→
Mg(NO
3
)
2
+H
2
O
n(MgO)=
m(MgO) 20
M(MgO) 40
=
=0,5
fhev
n(HNO
3
)=
3
3
m(HNO )
94,5
M(HNO ) 63
=
=1,5
fhev
Ld
HNO
3
gZoh^blky\ba[uld_lhjZkq_l \_^_fih
MgO
20
]
40
]
MgO
+2HNO
3
→
o]
32
148
]
Mg(NO )
+H
2
O
53
o
106 22,4
=
o
=
20 148
40
⋅
=74
]
Hl\_l
m(Mg(NO
3
)
2
)=
]
5.
40
]]
4
CuSO + Fe
→
FeSO
4
+Cu
n(CuSO
4
)=
4
4
m(CuSO ) 40
M(CuSO ) 160
=
=0,25
fhev
n(Fe)=
m(Fe) 12
M(Fe) 56
=
=0,21
fhev
Ld
CuSO
4
gZoh^blky\ ba[uld_lhihke_hdhgqZgbyj_Zd pbbhg
hklZg_lky\ jZkl\hj_
12
Hl\_lhklZg_lky
6. BaCl
2
+Na
2
SO
4
→
2NaCl+BaSO
4
↓
m(BaCl
2
)=
ω
(BaCl
2
)
⋅
m
jjZ
=0,05
⋅
15=0,75
]
n(BaCl
2
)=
2
2
m(BaCl ) 0,75
M(BaCl ) 208
=
=0,0036
fhev
m(Na
2
SO
4
)=
ω
(Na
2
SO
4
)
⋅
m
jjZ
=0,08
⋅
10=0,8
]
n(Na
2
SO
4
)=
24
24
m(Na SO ) 0,8
M(Na SO ) 142
=
=0,0056
fhev
Ld
Na
2
SO
4
gZoh^blky\ba[uld_jZkq_l\_^_fih
BaCl
2
.
0,75
]
2
208
]
BaCl
+Na
2
SO
4
→
2NaCl+
o]
4
233
]
BaSO
↓
0,75
o
208 233
=
o
=
0,75 233
208
⋅
=0,84
]
Hl\_l
m(BaSO
4
)=0,84
]
7. Fe
2
(SO
4
)
3
+6NaOH
→
2Fe(OH)
3
+3Na
2
SO
4
m(Fe
2
(SO
4
)
3
)=
ω
(Fe
2
(SO
4
)
3
)
⋅
m
jjZ
=0,05
⋅
400=20
]
n(Fe
2
(SO
4
)
3
)=
243
243
m(Fe (SO ) )
20
M(Fe (SO ) ) 400
=
=0,05
fhev
m(NaOH)=
ω
(NaOH)
⋅
m
jjZ
=0,03
⋅
200=6
]
n(NaOH)=
m(NaOH) 6
M(NaOH) 40
=
=0,15
fhev
Ld
Fe
2
(SO
4
)
3
gZoh^blky\ba[uld_lhjZkq_l\_^_fih
NaOH.
Fe
2
(SO
4
)
3
+
6
]
640
]
6NaOH
⋅
→
o]
3
2107
]
2Fe(OH)
⋅
↓
+3Na
2
SO
4
6
o
6 40 2 107
=
⋅⋅
o
=
⋅
⋅
⋅
=5,35
]
Hl\_l
m(Fe(OH)
3
)=5,35
]
8.
16
]]
4
CuSO + Fe
→
FeSO
4
+Cu
n(CuSO
4
)=
4
4
m(CuSO ) 16
M(CuSO ) 160
=
=0,1
fhev
n(Fe)=
m(Fe) 12
M(Fe) 56
=
=0,21
fhev
Ld
Fe
gZoh^blky\ba[uld_lhjZkq_l\_^_fih
CuSO
4
13
16
]
4
160
]
CuSO
+Fe
→
FeSO
4
+
o]
64
]
Cu
16
o
160 64
=
o
=
⋅
=6,4
]
Hl\_l
m(Cu)=6,4
]
9. 2NaOH+CuSO
4
→
Cu(OH)
2
+Na
2
SO
4
m(NaOH)=0,2
⋅
200=40
]
n(NaOH)=
m(NaOH) 40
M(NaOH) 40
=
=1
fhev
n(CuSO
4
)=
fhev
Ld
CuSO
4
gZoh^blky\ ba[uld_lhjZkq_l\_^_fih
NaOH.
40
]
240
]
2NaOH
⋅
+CuSO
4
→
Na
2
SO
4
+
o]
2
98
]
Cu(OH)
16
o
160 64
=
o
=
⋅
⋅
=49
]
Hl\_l
m(Cu(OH)
2
)=49
]
10. 2HCl+Na
2
CO
3
→
2NaCl+H
2
O+CO
2
m(HCl)=
ω
(HCl)
⋅
m
jjZ
=0,3
⋅
150=45
]
n(HCl)=
m(HCl) 45
M(HCl) 36,5
=
=1,23
fhev
n(Na
2
CO
3
)=
fhev
Ld
HCl
gZoh^blky\ ba[uld_jZkq_l\_^_fih
Na
2
CO
3
2HCl+
0,5
fhev
23
Na CO
→
2NaCl+H
2
O+
0,5
fhev
2
CO
n(Na
2
CO
3
)=n(CO
2
)=
fhev
V(CO
2
)=n(CO
2
)
⋅
V
m
=0,5
⋅
e
Hl\_l
V(CO
2
)=
e
11. H
2
SO
4
+Ba(NO
3
)
→
BaSO
4
+2HNO
3
m(H
2
SO
4
)=
ω
(H
2
SO
4
)
⋅
m
jjZ
=0,1
⋅
40=4
]
n(H
2
SO
4
)=
4
4
m(HSO ) 4
M(HSO ) 98
=
=0,041
fhev
n(Ba(NO
3
)
2
)=
32
32
m(Ba(NO ) )
2,61
M(Ba(NO ) ) 261
=
=0,01
fhev
Ld
H
2
SO
4
gZoh^blky\ba[uld_lhjZkq_l\_^_fih
Ba(NO
3
)
2
.
14
H
2
SO
4
+
2,61
32
261
Ba(NO )
→
o]
4
233
]
BaSO
+2HNO
3
2,61
o
261 233
=
o
=
⋅
=2,33
]
Hl\_l
m(BaSO
4
)=2,33
]
12.
10
]]
3
MgO+2HNO
→
Mg(NO
3
)
2
+H
2
O
n(MgO)=
m(MgO) 10
M(MgO) 40
=
=0,25
fhev
n(HNO
3
)=
3
3
m(HNO )
40
M(HNO ) 63
=
=0,635
fhev
Ld
HNO
3
gZoh^blky\ba[uld_lhjZkq_l \_^_fih
MgO
10
]
40
]
MgO
+2HNO
3
→
o]
32
148
]
Mg(NO )
+H
2
O
53
o
106 22,4
=
o
=
⋅
=37
]
Hl\_l
m(Mg(NO
3
)
2
)=3
]
13. H
2
SO
4
+BaCl
2
→
BaSO
4
+2HCl
m(H
2
SO
4
)=
ω
(H
2
SO
4
)
⋅
m
jjZ
=0,07
⋅
200=14
]
n(H
2
SO
4
)=
4
4
m(HSO ) 14
M(HSO ) 98
=
=0,143
fhev
n(BaCl
2
)=2
fhev
Ld
BaCl
2
gZoh^blky\ba[uld_lhjZkq_l\_^_fih
H
2
SO
4
.
BaCl
2
+
14
]
24
98
]
HSO
→
o]
4
233
]
BaSO
+2HCl
2,61
o
261 233
=
o
=
⋅
=33,29
]
Hl\_l
m(BaSO
4
)=33,29
]
14.
16,2
]
30
]
3
ZnO+2HNO
→
Zn(NO
3
)
2
+H
2
O
n(ZnO)=
m(ZnO) 16,2
M(ZnO) 81
=
=0,2
fhev
n(HNO
3
)=
3
3
m(HNO )
30
M(HNO ) 63
=
=0,48
fhev
Ld
HNO
3
gZoh^blky\ba[uld_jZkq_l \_^_fih
ZnO
15
16,2
]
81
]
ZnO
+2HNO
3
→
o]
32
189
]
Zn(NO )
+H
2
O
53
o
106 22,4
=
o
=
⋅
=37,8
]
Hl\_l
m(Zn(NO
3
)
2
)=37,8
]
JZ[hlZ Blh]h\Zyihl_f_
I
<ZjbZgl
Z
KCl
→
K
+
+Cl
-
[
H
2
SO
4
→
2H
+
+SO
−
\
Ca(OH)
2
→
Ca
2+
+2OH
-
]
Cu(NO
3
)
2
→
Cu
2+
+2NO
−
^
Al
2
(SO
4
)
3
→
2Al
3+
+2SO
−
Z
NaCl+AgNO
3
→
AgCl
↓
+NaNO
3
Cl
-
+Ag
+
→
AgCl
↓
[
2HNO
3
+Ca(OH)
2
→
Ca(NO
3
)
2
+2H
2
O
H
+
+OH
-
→
H
2
O
\+&O.
2
CO
3
→
2KCl+H
2
O+CO
2
2H
+
+CO
−
→
H
2
O+CO
2
3. 2HCl+Zn
→
ZnCl
2
+H
2
↑
2H
+
+Zn
→
Zn
2+
+H
2
↑
2HCl+CaO
→
CaCl
2
+H
2
O
2H
+
+CaO
→
Ca
2+
+H
2
O
2HCl+Cu(OH)
2
→
CuCl
2
+2H
2
O
2H
+
+ Cu(OH)
2
→
Cu
2+
+2H
2
O
4.
32
]]
4
CuSO + Fe
→
FeSO
4
+Cu
n(CuSO
4
)=
4
4
m(CuSO ) 32
M(CuSO ) 160
=
=0,2
fhev
n(Fe)=
m(Fe) 21
M(Fe) 56
=
=0,375
fhev
Ld
Fe
gZoh^blky\ba[uld_lhjZkq_l\_^_fih
CuSO
4
32
]
4
160
]
CuSO
+Fe
→
FeSO
4
+
o]
64
]
Cu
16
32
o
160 64
=
o
=
⋅
=12,8
]
Hl\_l
m(Cu)=12,8
]
<ZjbZgl
Z
NaOH
→
Na
+
+OH
-
[
HNO
3
→
H
+
+NO
−
\
MgCl
2
→
Mg
2
+
+2Cl
-
]
K
2
CO
3
→
2K
+
+CO
−
^
Fe(NO
3
)
3
→
Fe
3+
+3NO
−
Z
2KOH+Cu(NO
3
)
2
→
Cu(OH)
2
↓
+2KNO
3
2OH
-
+Cu
2+
→
Cu(OH)
2
[
2HCl+Ba(OH)
2
→
BaCl
2
+2H
2
O
H
+
+OH
-
→
H
2
O
\
Na
2
SO
4
+Ba(NO
3
)
2
→
BaSO
4
+2NaNO
3
SO
−
+Ba
2+
→
BaSO
4
3. Ca(OH)
2
+CuSO
4
→
CaSO
4
+Cu(OH)
2
↓
2OH
-
+Cu
2+
→
Cu(OH)
2
↓
3Ca(OH)
2
+2K
3
PO
4
→
Ca
3
(PO
4
)
2
↓
+6KOH
3Ca
2+
+2PO
−
→
Ca
3
(PO
4
)
2
↓
Ca(OH)
2
+CO
2
→
CaCO
3
+H
2
O
4.
5
]]
3
MgO+2HNO
→
Mg(NO
3
)
2
+H
2
O
n(MgO)=
m(MgO) 5
M(MgO) 40
=
=0,125
fhev
n(HNO
3
)=
3
3
m(HNO )
40
M(HNO ) 63
=
=0,635
fhev
Ld
HNO
3
gZoh^blky\ba[uld_jZkq_l\_^_fih
MgO
5
]
40
]
MgO
+2HNO
3
→
o]
32
148
]
Mg(NO )
+H
2
O
5
o
40 148
=
o
=
⋅
=18,5
]
Hl\_l
m(Mg(NO
3
)
2
)=18,5
]
<ZjbZgl
Z
H
3
PO
4
→
3H
+
+PO
−
[
Ba(OH)
2
→
Ba
2+
+2OH
-
17
\
FeSO
4
→
Fe
2+
+SO
−
]
AlCl
3
→
Al
3
+
+3Cl
-
^
Cu(NO
3
)
2
→
Cu
2+
+2NO
−
Z
FeCl
3
+3NaOH
→
Fe(OH)
3
↓
+3NaCl
Fe
3
+
+3OH
-
→
Fe(OH)
3
↓
[
H
2
SO
4
+2LiOH
→
Li
2
SO
4
+2H
2
O
H
+
+OH
-
→
H
2
O
\
2HNO
3
+Na
2
CO
3
→
2NaNO
3
+H
2
O+CO
2
2H
+
+CO
−
→
H
2
O+CO
2
3. 3H
2
SO
4
+2Al
→
Al
2
(SO
4
)
3
+3H
2
↑
6H
+
+2Al
→
2Al
3+
+3H
2
H
2
SO
4
+2KOH
→
K
2
SO
4
+2H
2
O
H
+
+OH
-
→
H
2
O
H
2
SO
4
+Ba(NO
3
)
2
→
BaSO
4
↓
+2HNO
3
SO
−
+Ba
2+
→
BaSO
4
↓
4.
20
]]
NaOH+HCl
→
NaCl+H
2
O
n(NaOH)=
m(NaOH) 20
M(NaOH) 40
=
=0,5
fhev
n(HCl)=
m(HCl) 33
M(HCl) 36,5
=
=0,9
fhev
Ld
HCl
gZoh^blky\ ba[uld_jZkq_l\_^_fih
NaOH
20
]
40
]
NaOH
+HCl
→
o]
58,5
]
NaCl
+H
2
O
5
o
40 148
=
o
=
⋅
=29,25
]
Hl\_l
m(NaCl)=29,5
]
<ZjbZgl
Z
CaCl
2
→
Ca
2+
+2Cl
-
[
KOH
→
K
+
+OH
-
\
HCl
→
H
+
+Cl
-
]
Fe
2
(SO
4
)
3
→
2Fe
3+
+3SO
−
^
Na
3
PO
4
→
3Na
+
+PO
−
Z
CuSO
4
+BaCl
2
→
CuCl
2
+BaSO
4
↓
SO
−
+Ba
2+
→
BaSO
4
↓
18
[
2HBr+Ca(OH)
2
→
CaBr
2
+2H
2
O
H
+
+OH
-
→
H
2
O
\
3ZnSO
4
+2K
3
PO
4
→
Zn
3
(PO
4
)
2
↓
+3K
2
SO
4
3Zn
2+
+2PO
−
→
Zn
3
(PO
4
)
2
↓
3. CuSO
4
+2NaOH
→
Cu(OH)
2
↓
+Na
2
SO
4
Cu
2+
+2OH
-
→
Cu(OH)
2
↓
3CuSO
4
+2Al
→
Al
2
(SO
4
)
3
+3Cu
3Cu
2
+
+2Al
→
2Al
3
+
+3Cu
3CuSO
4
+2Na
3
PO
4
→
Cu
3
(PO
4
)
2
↓
+3Na
2
SO
4
3Cu
2+
+2PO
−
→
Cu
3
(PO
4
)
2
↓
] ]
4. Zn+H
2
SO
4
→
ZnSO
4
+H
2
n(Zn)=
m(Zn) 6,5
M(Zn) 65
=
=0,1
fhev
n(H
2
SO
4
)=
24
24
m(H SO ) 20
M(H SO ) 98
=
=0,204
fhev
Ld
H
2
SO
4
gZoh^blky\ba[uld_jZkq_l\_^_fih
Zn.
6,5
]
65
]
Zn
+H
2
SO
4
→
ZnSO
4
+
o]
2
2
]
H
5
o
40 148
=
o
=
⋅
=0,2
]
Hl\_l
m(H
2
)=0,2
]
19
L_fZ ,,I h^ ]jmii Zdbkehjh^ZHkgh\gu_
aZdhghf_jghklbl_q_gbyobfbq_kdboj_Zdpbc
JZ[hlZDbkehjh^K_jZK_jgZydbkehlZ
<ZjbZgl
Z62
2
→
SO
2
[6
0
+Li
0
→
Li
+
S
-
2
S
0
+2e
-
→
S
-2
1
Li
0
-1e
-
→
Li
+1
2
2Li+S
→
Li
2
+S
\+
2
SO
4
+Mg
→
MgSO
4
+H
2
↑
]+
2
SO
4
+CaO
→
CaSO
4
+H
2
O
^+
2
SO
4
+Ba(OH)
2
→
BaSO
4
↓
+2H
2
O
2H
+
+SO
−
+Ba
2+
+2OH
-
→
BaSO
4
↓
+2H
2
O
2. :
O
::
O
: O=O
:
O
:
O
:
O:
IjbagZdbkjZ\g_gby
O
2
O
3
1.
:]j_]Zlgh_
khklhygb_gm
2.
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6.
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2
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3
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SO
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→
Na
2
SO
4
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2
SiO
3
;
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→
Al
2
S
3
[6
0
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0
→
C
+4
S
−
2S
0
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-
→
2S
-2
1
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0
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-
→
C
+4
1
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→
CS
2
O
O
O
20
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2
SO
4
→
ZnSO
4
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2
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2
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4
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→
MgSO
4
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2
O
^+
2
SO
4
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2
→
CuSO
4
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2
O
2H
+
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2
→
Cu
2+
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2
O
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kmevnZlh\b kmevnb^h\Hkgh\gu_h[eZkl bijbf_g_gby^ey
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2
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2
SO
4
→
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4
↓
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<ZjbZgl
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2
→
H
2
S
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0
+Zn
0
→
Zn
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S
-2
S+2e
-
→
S
-2
1
Zn
0
-2e
-
→
Zn
+2
1
S+Zn
→
ZnS
\+
2
SO
4
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→
FeSO
4
+H
2
]+
2
SO
4
+CuO
→
CuSO
4
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2
O
^+
2
SO
4
+2NaOH
→
Na
2
SO
4
+2H
2
O
H
+
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-
→
H
2
O
2. O
2
±dbkehjh^2
3
±hahg
S
8
±fhghdebggZyk_jZ
S
8
±jhf[bq_kdZyk_jZ
6±ieZklbq_kdZyk_jZ
Wlhy\e_gb_h[jZah\Zgbyg_kdhevdboijhkluo\_s_kl\h^gbf
we_f_glhfghkblgZa\Zgb_Zeehljhiby
Wlb\_s_kl\Zfh`ghjZkihagZlvijbihfhsbjZkl\hjZ %D&O
2
.
K
2
SO
4
+BaCl
2
→
BaSO
4
↓
+2KCl
[_euc
HCl+BaCl
2
→
/
<ZjbZgl
Z61D
→
Na
2
S
[6
0
+F
→
S
+6
F
−
S
0
-6e
-
→
S
+6
1
F
2
+2e
-
→
2F
-1
3
S+3F
2
→
SF
6
\+
2
SO
4
+2Al
→
Al
2
(SO
4
)
3
+3H
2
9
]+
2
SO
4
+Fe
2
O
3
→
Fe
2
(SO
4
)
3
+3H
2
O