Радецкий А.М., Горшкова В.П. Решение контрольных и самостоятельных работ по химии за 9 класс
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61
m
ij
(HNO
3
) = η(HNO
3
) ⋅ m
l_hj
= 4,1 ⋅
]
Hl\_l
m(HNO
3
)
2
= 4,02
]
62
L_fZ
VI
@_e_ahF_lZeemj]by
JZ[hlZ@_e_ahb_]hkh_^bg_gby
<ZjbZgl
1. 4Fe + 3O
2
→ 2Fe
2
O
3
hdkb^`_e_aZ
III)
2Fe + 3Cl
2
t
→
R
2FeCl
3
oehjb^`_e_aZ
III)
Fe + S
t
→
R
FeS
kmevnb^`_e_aZ
(II)
2Fe + 3Br
2
→ 2FeBr
3
[jhfb^`_e_aZ
(III)
2.
+26
Fe 1s
2
2s
2
2p
6
3s
2
3p
6
3d
6
4s
2
< kh_^bg_gbyo`_e_ahijhy\ey_lkl_i_gbhdbke_gby
3. 3Fe
2+
+ 2K
3
[Fe(CN)
6
] →
362
l_fghkbgbc
Fe [Fe(CN) ]
↓ + 6K
+
4Fe
3+
+ 3K
4
[Fe(CN)
6
] →
463
[bjxah\uc
Fe [Fe(CN) ]
↓
+ 12K
+
4.
32
]
4
CuSO
+
28
]
Fe
→ FeSO
4
+ Cu
n(Fe) =
= 0,5
fhev
n(CuSO
4
) =
= 0,2
fhev
Ld
CuSO
4
gZoh^blky\ g_^hklZld_jZkq_l\_^_fihg_fm
32
]
4
160
]
CuSO
+ Fe → FeSO
4
+
o]
64
]
Cu
[
=
x =
⋅
= 12,8
]
Hl\_l
m(Cu) = 12,8
]
<ZjbZgl
1. Fe(NO
3
)
3
1)
→
Fe(OH)
3
2)
→
Fe
2
O
3
3)
→
Fe
4)
→
FeCl
2
1) Fe(NO
3
)
3
+ 3NaOH → Fe(OH)
3
↓ + 3NaNO
3
2) 2Fe(OH)
3
t
→
R
Fe
2
O
3
+ 3H
2
O
3) Fe
2
O
3
+ 3H
2
→ 2Fe + 3H
2
O
63
4) Fe + 2HCl → FeCl
2
+ H
2
↑
2.
Kh_^bg_gby`_e_aZkhkl_i_gvxhdbke_gbyijhy\eyxl
\hkklZgh\bl_evgu_k\hckl\Zijbwlhfhdbkeyykv^hkl_i_gb
hdbke_gbyldboZlhfijbwlhfi_j_oh^bl\[he__mklhcqb\mx
nhjfm< g_dhlhjuokemqZyowlbkh_^bg_gbyfh]mlijhy\eylv
hdbkebl_evgu_k\hckl\Zdh ]^Zbagbo\ul_kgy_lkyf_lZeebq_kdh_
`_e_ah
3. Fe + 2HCl → FeCl
2
+ H
2
↑
FeCl
2
+ 2NaOH → Fe(OH)
2
+ 2NaCl
4Fe(OH)
2
+ 2H
2
O + O
2
→ 4Fe(OH)
3
4.
112
]
56
]
Fe
+ H
2
SO
4
→ FeSO
4
+
oe
2
22,4
e
H
[
=
x =
⋅
= 44,8
e
V
ij
G
2
) = η
G
2
) ⋅ V
l_hj
= 0,98 ⋅
e
Hl\_l
V(H
2
) = 43,9
e
<ZjbZgl
Z
Fe + H
2
SO
4
→ FeSO
4
+ H
2
[
Fe + CuSO
4
→ FeSO
4
+ Cu
\
Fe + 2HBr → FeBr
2
+ H
2
]
Fe + Pb(NO
3
)
2
→ Fe(NO
3
)
2
+ Pb
2.
)H2b )H
2
O
3
—
hkgh\gu_hdkb^u)H2+
2
b )H2+
3
—
hkgh\gu_
]b^jhdkb^u
3.
Ij_^eh`_ggu_\_s_kl\Zfh`ghjZkihagZlvijbihfhsb
jZkl\hjh\1D
2
SO
4
b
NaOH.
FeCl
2
+ 2NaOH →
2
a_e_guc
Fe(OH)
↓ + 2NaCl
BaCl
2
+ Na
2
SO
4
→
4
[_euc
BaSO
↓ + 2NaCl
< h klZ\r_ckyijh[bjd_[m^_l
NaCl.
4. 2HCl + FeCO
3
→ FeCl
2
+ H
2
O + CO
2
m(FeCO
3
) = 0,8 ⋅ 50 = 40
]
2HCl +
40
]
3
116
]
FeCO
→ FeCl
2
+ H
2
O +
oe
2
22,4
e
CO
[
=
64
x =
⋅
= 7,72
e
Hl\_l
V(CO
2
)=7
e
<ZjbZgl
1. Fe
1)
→
FeSO
4
2)
→
Fe(OH)
2
3)
→
Fe(OH)
3
4)
→
FeCl
3
1) Fe + H
2
SO
4
→
FeSO
4
+ H
2
2) FeSO
4
+ 2NaOH
→
Fe(OH)
2
↓
+ Na
2
SO
4
3) 4Fe(OH)
2
+ 2H
2
O + O
2
→
4Fe(OH)
3
4) Fe(OH)
3
+ 3HCl
→
FeCl
3
+ 3H
2
O
2.
<h^gu_jZkl\hjukhe_c`_e_aZ
(III)
bf_xldbkemxkj_^mldkheb
h[jZah\ZgukeZ[ufhkgh\Zgb_fb kbevghcdbkehlhcb\\h^_
]b^jhebamxlky
Z
FeSO
4
+ BaCl
2
→
FeCl
2
+ BaSO
4
↓
[
FeSO
4
+ Ba(NO
3
)
2
→
BaSO
4
↓
+ Fe(NO
3
)
2
4.
7
]
2Fe
+
18
]
2
3Cl
→
2FeCl
3
n(Fe) =
)H0
)HP
=
= 0,125
fhev
n(Cl
2
) =
&O
0
&O
P
=
= 0,253
fhev
Ld
Cl
2
gZoh^blky\ ba[uld_jZkq_l\_^_fih
Fe.
7
]
256
]
2Fe
⋅
+ 3Cl
2
→
o]
3
2162,5
]
2FeCl
⋅
[
⋅
=
⋅
x =
⋅
⋅
⋅
= 20,31
]
Hl\_l
m(FeCl
3
) = 20,31
]
JZ[hlZKihkh[uijhfure_ggh]hihemq_gbyf_lZeeh \
<ZjbZgl
Z`_e_agZyjm^Z)H
2
O
3
)
[
\
2C + O
2
→
2CO 1)
ihemq_gb_\hkklZgh\bl_ey
Fe
2
O
3
+ 3CO
→
2Fe + 3CO
2
2)
\hkklZgh\e_gb_
3) CaCO
3
t
→
R
CaO + CO
2
3)
reZdhh[jZah\Zgb_
65
CaO + SiO
2
→ CaSiO
3
3CaO + P
2
O
5
→ Ca
3
(PO
4
)
2
MnO + SiO
2
→ MnSiO
3
2.
o
]
12
]
C
+ O
2
→
0,1
]
2
44
]
CO
[
=
x =
⋅
= 0,0273
]
ω
K
P
&
P
kl
=
beb
Hl\_l
ω
K
<ZjbZgl
Zqm]mg
[
\
2Fe + O
2
→ 2FeO
Si + 2FeO → SiO
2
+ 2Fe
Mn + FeO → MnO + Fe
2P + 5FeO → 5Fe + P
2
O
5
C + FeO → Fe + CO
2. 2Al
2
O
3
→ 4Al + 3O
2
ω(Al
2
O
3
) = 100% – 0,4% = 99,6%
m(Al
2
O
3
) =0,996 ⋅ 50
l l
49,8
]
23
2102
]
2Al O
⋅
→
o]
427
]
4Al
⋅
+ 3O
2
[
⋅
=
⋅
x =
⋅
⋅
⋅
= 26,36
l
Hl\_l
m(Al) = 26,36
l
<ZjbZgl
Z
Al
2
O
3
b
Na
3
[AlF
6
] (
[hdkblu
[
\
Al
2
O
3
+ Na
2
CO
3
t
→
R
2NaAlO
2
+ CO
2
2NaAlO
2
+ CO
2
+ 3H
2
O → 2Al(OH)
3
+ Na
2
CO
3
2Al(OH)
3
→ Al
2
O
3
+ 3H
2
O
66
2Al
2
O
3
we a
−
→
4Al + 3O
2
:exfbgbcijbf_gy_lky\ kZfhe_lhkljh_gbbf_lZeemj]bbb^ey
ba]hlh\e_gbyijh\h^h\
2.
o]
56
]
Fe
+ H
2
SO
4
→ FeSO
4
+
1,96
e
2
22,4
e
H
[
=
x =
⋅
= 4,9
]
ω(Fe) =
= 0,98
beb
Hl\_l
ω(Fe) = 98%.
<ZjbZgl
Zqm]mg
[
\
2Fe + O
2
→ 2FeO
FeO + C → Fe + CO
2FeO + Si → 2Fe + SiO
2
5FeO + 2P → 5Fe + P
2
O
5
FeO + Mn → MnO + Fe
2.
] K²
]
Fe
3
C) — x%
x =
⋅
= 54%
Hl\_l
ω(Fe
3
C) = 54%.
67
AZ^Zqbk ijhba\h^kl\_ggufb f_`ij_^f_lguf
kh^_j`Zgb_f
G_f_lZeeu
1. SiO
2
+ 2C
t
→
R
Si + 2CO
m(C) = 80 ⋅ 0,92
] d]
od]
2
60
d]
SiO
+
73
,6
d
]
212
d]
2C
⋅
t
→
R
Si + 2CO
[
⋅
=
x =
⋅
⋅
= 184
d]
Hl\_l
m(SiO
2
) = 184
d]
2. 2Ca
3
(PO
4
)
2
+
od]
120
d]
10C
+ 6SiO
2
t
→
R
6CaSiO
3
+
3
2
mf
224
f
10CO
+
248
d]
4
124
d]
P
x
120
248
124
=
o
120 248
124
⋅
= 240
d]K
m
224 124
=
m
224 248
124
⋅
f
3
KH
m
l_hj
K
ij
m(C)
240
(C) 0,9
=
ω
d]
Hl\_l
m(C) = 266,7
d]
V(CO) = 448
f
3
.
3.
od]
4
124
d]
P
+ 5O
2
+ 6H
2
O →
400
d]
34
498
d]
4H PO
⋅
x
124
400
4
98
=
⋅
x =
124 400
4 98
⋅
⋅
= 126,5
d]
m
l_hj
J
ij
34
m(P)
126,5
(H PO ) 0,8
=
η
d]
68
Hl\_l
m(P) = 158,2
d]
4.
120
d]
342
310
d]
Ca (PO )
+ 3H
2
SO
4
t
→
R
3CaSO
4
+
od]
34
298
d]
2H PO
⋅
m(Ca
3
(PO
4
)
2
) = 0,6 ⋅ 200 = 120
d]
120
310
2
98
=
⋅
x
o
120 2
98
310
⋅
⋅
d]
m
ij
(H
3
PO
4
) = m
l_hj
⋅ η(H
3
PO
4
) = 0,8 ⋅ 75,87 = 60,7
d]
Hl\_l
m(H
3
PO
4
) = 60,7
d]
5. SO
2
+ NO
2
+ H
2
O = H
2
SO
4
+ NO
m
jjZ
= ρ ⋅ V = 1,66 ⋅ 100 = 166
d]
m(H
2
SO
4
) = ω ⋅ m
jjZ
= 166 ⋅
d]
od]
2
64
d]
SO
+ NO
2
+
md]
2
18
d]
HO
=
122,84
d]
24
98
d]
HSO
+ NO
x
64
122
84
98
=
,
o
64 122
84
98
⋅
,
d]
(SO
2
)
m
18 98
=
m
18 122
84
98
⋅
,
= 22,6
d]G
2
H
Hl\_l
m(SO
2
) = 80,22
d]
m(H
2
O) = 22,6
d]
6. ∆ω
J ±
ω
G
2
H
2 13%
2 51%
,
,
beb
Hl\_l
ω
G
2
H
7. m
gZ^a
=5 m
ih^a
m
h[s
= m
gZ^a
+m
ih^a
=6 m
ih^a
m
gZ^a
G
2
H
ω
gZ^a
Â
m
ih^a
=0,88 m
gZ^a
=4,4 m
ih^a
m
ih^a
G
2
H
ω
ih^a
Â
m
ih^a
=0,8 m
ih^a
m
h[s
=4,4 m
ih^a
+0,8 m
ih^a
=5,2 m
ih^a
m
h[s
G
2
H
h[s ih^a
h[s ih^a
m(
GHP
0,867 86,7%
m6m
===
Hl\_l
69
F_lZeeu
1. Ca(OH)
2
+
od]
23
106
d]
Na CO
→
10
d]
240
d]
2NaOH
⋅
+ CaCO
3
↓
x
106
10
2
40
=
⋅
o
106 10
2 40
⋅
⋅
d]
ω(Na
2
CO
3
) = 100% – 2% = 98%
m
kh^u
=
m Na
CO
Na
CO
(
)
(
)
,
,
2
3
2
3
13
25
0
98
ω
=
d]
Hl\_l
m
kh^u
d]
2. HgS + O
2
→ Hg + SO
2
ω(HgS) = 100% – 10% = 90%
m(HgS) = 0,9 ⋅ 600 = 540
d]
540
d]
233
d]
HgS
+ O
2
→ Hg +
od]
2
201
d]
SO
540
233
201
=
x
o
540 201
233
⋅
d]
Hl\_l
m(Hg) = 465,8
d]
3.
m
dZjl
=
m K
K
( )
( )
,
ω
=
3
0
006
]
Hl\_l
m
dZjl
]
4.
ω
ijbf
=
39 3%
38%
39
3%
,
,
−
beb
Hl\_l
ω
ijbf
= 3,31%.
5.
458
]
34
3 229
]
3Mn O
⋅
+ 8Al →
o]
9 55
]
9Mn
⋅
+ 4Al
2
O
3
+
md>`
2500
d>`
458
3 229
9
55
⋅
=
⋅
x
o
458 9
55
3 229
⋅
⋅
⋅
]
Mn)
70
458
m
3 229 2500
=
⋅
m
458 2500
3 229
⋅
⋅
= 1666,7
d>`
Hl\_l
m(Mn) = 330
]
Q = 1666,7
d>`
Z
m(Au) =
750
1000
⋅ 860 = 645
]
[
m(Ag) =
875
1000
⋅ 800 = 700
]
Hl\_lZ
m(Au) = 645
][
m(Ag) = 700
]
7.
od]
23
106
d]
Na CO
→
170
d]
2
62
d]
kl_deh1D 2
m(Na
2
O) = 0,17 ⋅ 1000 = 170
d]
x
106
170
62
=
o
106 170
62
⋅
d]
Hl\_l
m(Na
2
CO
3
) = 291
d]
8. 2Na
2
O
2
+ 2CO
2
→ 2Na
2
CO
3
+ O
2
ω(Na
2
O
2
) = 100% – 2,5% = 97,5%
m(Na
2
O
2
) = m
i_j
⋅ ω(Na
2
O
2
) = 0,975 ⋅
]
312
]
22
278
]
2Na O
⋅
+
oe
2
2 22,4
e
2CO
⋅
→ 2Na
2
CO
3
+
me
2
234
e
O
312
2 78
2
22
4
⋅
=
⋅
x
,
o
312 2
22
4
2
78
⋅
⋅
⋅
,
e
312
m
2 78 22,4
=
⋅
m
312 22
4
2 78
⋅
⋅
,
e
Hl\_l
V(CO
2
) = 89,6
e
V(O
2
) = 44,8
e
9. m(Ca(OH)
2
)=
]Zr
ba\
m(
KZHG d]
⋅ω = ⋅ =
od]
3
100
d]
CaCO
→
51,6
d]
2
74
d]
Ca(OH)