Радецкий А.М., Горшкова В.П. Решение контрольных и самостоятельных работ по химии за 9 класс
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51
x
152
52
2
52
=
⋅
o
152 52
2 52
⋅
⋅
]
ω
(Cr
2
O
3
) = 100% – 10% = 90%
m
l_hj
K
r
2
O
3
) =
ij
23
m
76
(Cr O ) 0,9
=
ω
= 84,4
]
Hl\_l
m(Cr
2
O
3
) = 84,4
]
<ZjbZgl
1.
+6 -2
3
hdbkebl_ev
Mo O
+
0
2
\hkklZgh\bl_ev
3H
→
Mo
0
+3H
2
+1
O
–2
2.
160
]
80
]
CuO
+ C
→
o]
64
]
Cu
+ CO
160
80
64
=
x
o
64 160
80
⋅
]
m
ij
(Cu) = m
l_hj
(Cu)
⋅ η
(Cu) = 0,85
⋅
128 = 108,8
]
Hl\_l
m(Cu) = 108,8
]
JZ[hlZObfbq_kdb_k\hckl\Zf_lZeeh\
<ZjbZgl
Z
Na
0
+ S
0
→
Na
2
+1
S
–2
Na
e
Na
S
e
S
0
1
0
2
1
2
−
→
+
→
−
+
−
−
2
1
2Na + S
→
Na
2
S
[
Fe + H
2
SO
4
→
FeSO
4
+ H
2
↑
\
2Al + 3Br
2
→
2AlBr
3
2.
[
Zn + Hg(NO
3
)
2
→
Hg
↓
+ Zn(NO
3
)
2
Zn + Hg
2+
→
Zn
2+
+ Hg
\
Mg + Pb(NO
3
)
2
→
Mg(NO
3
)
2
+ Pb
↓
Mg + Pb
2+
→
Pb + Mg
2+
<ZjbZgl
Z
Cu
0
+ O
2
0
→
Cu
2+
O
–2
Cu
e
Cu
O
e
O
0
2
2
2
2
4
2
−
→
+
→
−
+
−
−
2
1
52
2Cu + O
2
→ 2CuO
[
2Fe + 3Cl
2
t
→
R
2FeCl
3
\
Ba + 2H
2
O → Ba(OH)
2
+ H
2
2.
Z
Cu + 2AgNO
3
→ Cu(NO
3
)
2
+ 2Ag↓
Cu + 2Ag
+
→ Cu
2+
+ 2Ag↓
[
Fe + Hg(NO
3
)
2
→ Fe(NO
3
)
2
+ Hg↓
Fe + Hg
2
+
→ Fe
2
+
+ Hg↓
<ZjbZgl
Z
Al
0
+ I
2
0
→ Al
+
3
I
3
–1
Al
e
Al
I
e
I
0
3
2
0
1
3
2
2
−
→
+
→
−
+
−
−
2
3
2Al + 3I
2
→ 2AlI
3
[
Cu + S → CuS
\
2Na + 2H
2
O → 2NaOH + H
2
2.
Z
Fe + CuCl
2
→ Cu↓ + FeCl
2
Fe + Cu
2+
→ Cu↓ + Fe
2+
\
2Al + 3Pb(NO
3
)
2
→ 3Pb↓ + 2Al(NO
3
)
3
2Al + 3Pb
2+
→ 3Pb↓ + 2Al
+3
<ZjbZgl
Z
Mg
0
+ H
+1
Cl
–1
→ Mg
+2
Cl
2
–1
+ H
2
0
Mg
e
Mg
H
e
H
0
2
1
2
0
2
2
2
−
→
+
→
−
+
+
−
1
1
Mg + 2HCl → MgCl
2
+ H
2
[
2Zn + O
2
→ 2ZnO
\
Ca + Cl
2
→ CaCl
2
2.
Z
Mg + Hg(NO
3
)
2
→ Hg↓ + Mg(NO
3
)
2
Mg + Hg
2+
→ Hg↓ + MgNO
3
)
2
\
Zn + FeSO
4
→ Fe + ZnSO
4
Zn + Fe
2+
→ Fe + Zn
2+
JZ[hlZWe_dljhebaDhjjhabyf_lZeeh\
<ZjbZgl
Z
A
( )
+
2Cl
–
– 2e
–
→ Cl
2
0
K
( )
−
K
+
+ 1e
–
→ K
0
2KCl
j\
we a
−
→
2K + Cl
2
53
[
A
( )
+
2Br
–
– 2e
–
→ Br
2
0
K
( )
−
Zn
2+
+ 2e
–
→ Zn
0
2H
2
O + 2e
–
→ H
2
+ 2OH
–
ZnBr
2
+ H
2
O
we a
−
→
Zn + Br
2
↓ + Zn(OH)
2
+ H
2
↑
2.
Dhjjhaby²kZfhijhba\hevgucijhp_kkjZajmr_gbyf_lZeeZih^
\ha^_ckl\b_fhdjm`Zxs_ckj_^u
AZf_^e_gbxdhjjhabbkihkh[k l\ m_l
Zs_ehqghklvkj_^u
[ihgb`_gb_l_fi_jZlmju
\ihdjulb_aZsblgufbkehyfb
JZajmrblkydjurdZld`_e_ah[he__Zdlb\gucf_lZee
<ZjbZgl
Z
A
( )
+
2I
–
– 2e
–
→ I
2
0
K
( )
−
2H
2
O + 2e
–
→ H
2
+ 2OH
–
2NaI + 2H
2
O
we a
−
→
I
2
↓ + H
2
↑ + 2NaOH
[
A
( )
+
2Cl
–
– 2e
–
→ Cl
2
0
K
( )
−
Cu
2+
+ 2e
–
→ Cu
0
CuCl
2
jj
we a
−
→
Cu + Cl
2
2.
JZaebqZxlZObfbq_kdmxdhjjhabx²\ hkgh\_e_`bl
obfbq_kdh_\aZbfh^_ckl\b_[We_dljhobfbq_kdmxdhjjhabx²
gZjy^mk obfbq_kdbfijhp_kkhfgZ[ex^Zxlkywe_dljbq_kdb_
ijhp_kku
Ld\hagbdZ_l]Zev\Zgbq_kdbcwe_f_gl
Fe-Sn
b ijhl_dZ_lijhp_kk
dhjjhabb
<ZjbZgl
Z
A
( )
+
2Br
–
– 2e
–
→ Br
2
0
K
( )
−
2H
2
O + 2e
–
→ H
2
+ 2OH
–
2KBr + 2H
2
O
we a
−
→
Br
2
↓ + H
2
↑ + 2KOH
[
A
( )
+
2Cl
–
– 2e
–
→ Cl
2
0
K
( )
−
Ca
2+
+ 2e
–
→ Ca
0
CaCl
2
j\
we a
−
→
Ca + Cl
2
2.
Mkbe_gbxdhjjhabbkihkh[kl\ mxl
ZIh\ur_gb_dbkehlghklbkj_^u
54
[Ih\ur_gb_l_fi_jZlmju
\DhglZdlk f_lZeehf
:exfbgbcb pbgdjZajmrZlky[uklj__ldwlh[he__Zdlb\gu_
f_lZeeu
<ZjbZgl
Z
A
( )
+
2Cl
–
– 2e
–
→ Cl
2
0
1
K
( )
−
Na
+
+ 1e
–
→ Na 2
2NaCl
w
ea
−
→
2Na + Cl
2
[
A
( )
+
2Br
–
– 2e
–
→ Br
2
0
K
( )
−
Fe
2+
+ 2e
–
→ Fe
0
FeBr
2
j\
we a
−
→
Fe + Br
2
2. 1)
GZg_k_gb_aZsblguoihdjulbc
bkihevah\Zgb_Zahlghcdbkehlu
PbgdjZajmrZ_lky[uk lj__ihlhfmqlhhg[he__Zdlb\guc
f_lZeeq_f`_e_ ah
JZ[hlZS_ehqgu_b s_ehqgha_f_evgu_f_lZeeu
<ZjbZgl
1. Ca
Z
→
CaO
[
→
CaCO
3
\
→
Ca(NO
3
)
2
Z
2Ca + O
2
→ 2CaO
[
CaO + CO
2
→ CaCO
3
\
CaCO
3
+ 2HNO
3
→ Ca(NO
3
)
2
+ H
2
O + CO
2
↑
2.
S_ehqgu_f_lZeeuijhy\eyxl\hkklZgh\bl_evgu_k\hckl\Z
2Li
0
+ Cl
2
0
→ 2Li
+1
Cl
–1
Li
0
– 1e
–
→ Li
+1
\hkklZgh\bl_ev
Ca
0
+ S
0
→ Ca
+2
S
–2
Ca
0
– 2e
–
→ Ca
+2
\hkklZgh\bl_ev
3.
20
]
KOH
+
25
]
3
HNO
→ KNO
3
+ H
2
O
n(KOH) =
m KOH
M KOH
(
)
(
)
=
20
56
= 0,357
fhev
n(HNO
3
) =
m HNO
M HNO
(
)
(
)
3
3
25
63
=
= 0,397
fhev
Ld
HNO
3
gZoh^blky\ba[uld_lhj Zkq _l\_^_fih
KOH.
20
]
56
]
KOH
+ HNO
3
→
o]
3
101
]
KNO
+ H
2
O
55
20
56
101
=
x
x =
20 101
56
⋅
= 36,1
]
Hl\_l
m(KNO
3
) = 36,1
]
<ZjbZgl
Z
2Na + S
→
Na
2
S
[
Ca + Cl
2
→
CaCl
2
\
2Li + 2H
2
O
→
2LiOH + H
2
↑
2. Ca(OH)
2
+ CO
2
→
CaCO
3
↓
+ H
2
O
2KOH + CO
2
→
K
2
CO
3
+ H
2
O
Ba(OH)
2
+ CO
2
→
BaCO
3
↓
+ H
2
O
;Zjblh\Zy\h^Zlh`_[m^_lfmlg_lvgZ\ha^mo_
3. CaCO
3
t
→
R
CaO + CO
2
ω
(CaCO
3
) = 100 – 8% = 92%
m(CaCO
3
) = 1000
⋅
0,92 = 920
]
920
]
3
100
]
CaCO
t
→
R
o]
56
]
CaO
+ CO
2
920
100
56
=
x
x =
920 56
100
⋅
= 515,2
]
Hl\_l
m(CaO) = 515,2
]
<ZjbZgl
1. Na
Z
→
NaOH
[
→
NaHCO
3
\
→
Na
2
SO
4
Z
2Na + 2H
2
O
→
2NaOH + H
2
[
NaOH + CO
2
→
NaHCO
3
\
2NaHCO
3
+ H
2
SO
4
→
Na
2
SO
4
+ 2H
2
O + 2CO
2
↑
2.
@_kldhklv\h ^u²kh\hdmighklvk\hckl\h[mkeh\e_gguo
gZebqb_f\\h^_bhgh\
Mg
2+
b
Ca
2+
;
kmffZjgh_kh^_j`Zgb_wlbo
bhgh\gZau\Z_lkyh[s_c`_kldhklvxJZaebqZxldZj[hgZlgmxb
g_dZj[hgZlgmx`_kldhklv
Ca(HCO
3
)
2
t
→
R
CaCO
3
↓ + H
2
O + CO
2
Ca(HCO
3
)
2
+ Ca(OH)
2
→ 2CaCO
3
↓ + 2H
2
O
3.
11,5
]
2Na
+
78
]
2
2H O
→ 2NaOH + H
2
↑
n(Na) =
1D
0
1D
P
=
= 0,5
fhev
56
n(H
2
O) =
2
+
0
2
+
P
=
= 4,33
fhev
LdG
2
H gZoh^blky\ba[uld_jZkq _l\_^_fih
Na.
11,5
]
223
]
2Na
⋅
+ 2H
2
O
→
o]
240
]
2NaOH
⋅
+ H
2
↑
[
⋅
=
x =
⋅
⋅
⋅
= 20
]
Hl\_l
m(NaOH) = 20
]
<ZjbZgl
1. NaOH + HCl
→
NaCl + H
2
O
NaOH + CO
2
→
NaHCO
3
2NaOH + H
2
SO
4
→
Na
2
SO
4
+ 2H
2
O
2.
IjbgZ]j_\ZgbbfjZfhjjZaeZ]Z_lky
:
CaCO
3
→
CaO + CO
2
CaO + H
2
O
→
Ca(OH)
2
3.
o]
223
]
2Na
⋅
+ 2H
2
O
→
2NaOH +
3
e
2
22,4
e
H
↑
[
=
⋅
x =
⋅
⋅
=
]
η
G
2
) =
beb
Hl\_l
η
G
2
) = 89,3%.
JZ[hlZ:exfbgbcb_]hkh_^bg_gby
<ZjbZgl
1.
:O&O
2
→
2AlCl
3
2Al + 3H
2
SO
4
→
Al
2
(SO
4
)
3
+ 3H
2
2Al + Cr
2
O
3
→
Al
2
O
3
+ 2Cr
2Al + 3CuSO
4
→
Al
2
(SO
4
)
3
+ 3Cu
2. Al(OH)
3
Z
→
AlCl
3
Al(OH)
3
Al
2
O
3
Z
Al(OH)
3
+ 3HCl
→
AlCl
3
+ 3H
2
O
[
AlCl
3
+ 3NaOH
→
Al(OH)
3
+ 3NaCl
57
\
2Al(OH)
3
Al
2
O
3
+ 3H
2
O
3.
:exfbgbc²k_j_[jbklh[_eucf_lZeee_]dbchq_gvfy]dbc
<ZjbZgl
1.
23
h
dbkebl_ev
Fe O
+
\hkklZgh\bl_ev
2Al
→Al
2
O
3
+2Fe
Al
2
(SO
4
)
3
+ 3BaCl
2
→ 2AlCl
3
+ 3BaSO
4
↓
Al(OH)
3
+ NaOH → Na[Al(OH)
4
]
\hkklZgh\bl_ev
4Al
+
2
hdbkebl_ev
3O
→2Al
2
O
3
2. Al(NO
3
)
3
→ Al(OH)
3
Al(NO
3
)
3
+ 3NaOH → Al(OH)
3
↓ + 3NaNO
3
Al
2
O
3
→ Al
2
(SO
4
)
3
Al
2
O
3
+ 3H
2
SO
4
→ Al
2
(SO
4
)
3
+ 3H
2
O
Al → AlCl
3
2Al + 3Cl
2
→ 2AlCl
3
3.
:exfbgbcijbf_gy_lky\ kZfhe_lhkljh_gbb^eyba]hlh\e_gby
ijh\h^h\\ f_lZeemj]bbijbwlhfbkihevamxlky_]he_]dh klv
\ukhdZywe_dljhijh\h^ghklvb \hkklZ gh\b l_evgZykihkh[ghklv
<ZjbZgl
1. 2Al + 3Cl
2
→ 2AlCl
3
2Al + 6HCl → 2AlCl
3
+ 3H
2
Al + FeCl
3
→ AlCl
3
+ Fe
Al
2
(SO
4
)
3
+ 3BaCl
2
→ 2AlCl
3
+ 3BaSO
4
↓
2. Al
Z
→
Al
2
O
3
[
→
Al
2
(SO
4
)
3
\
→
Al(OH)
3
Z
4Al + 3O
2
→ 2Al
2
O
3
[
Al
2
O
3
+ 3H
2
SO
4
→ Al
2
(SO
4
)
3
+ 3H
2
O
\
Al
2
(SO
4
)
3
+ 6NaOH → 2Al(OH)
3
↓ + 3Na
2
SO
4
3.
:exfbgbch[jZam_lijhqgmxhdkb^gmxie_gdmgZk\h_c
ih\_joghklbihwlhfmhgg_hdbkey_lky^Ze__
<ZjbZgl
1. 2Al + 3Br
2
→ 2AlBr
3
2Al + 3S → Al
2
S
3
2Al + 6H
2
O → 2Al(OH)
3
+ 3H
2
↑
2Al + 2NaOH + 6H
2
O → 2Na[Al(OH)
4
] + 3H
2
↑
2. 2Al + 6HCl → 2AlCl
3
+ 3H
2
↑
AlCl
3
+ 3KOH → Al(OH)
3
↓
D&O
3HCl + Al(OH)
3
→ AlCl
3
+ 3H
2
O
3.
GZih\_joghklbjZkdZe_ ggh]hZexfbgbyf]gh\_gghh[jZam_lk y
hdkb^gZyie_gdZdhlhjZyaZljm^gy_l^_nhjfZpbx
t°
58
JZ[hlZ Blh]h\Zyihl_f_
V
<ZjbZgl
1. Ca(OH)
2
+ CO
2
→ CaCO
3
+ H
2
O
Ca(OH)
2
+ 2HCl → CaCl
2
+ 2H
2
O
Ca(OH)
2
+ ZnCl
2
→ Zn(OH)
2
↓ + CaCl
2
2.
A
( )
+
2Cl
–
– 2e
–
→ Cl
2
0
K
( )
−
Fe
2+
+ 2e
–
→ Fe
0
2H
2
O + 2e
–
→ H
2
+ 2OH
–
FeCl
2
+ H
2
O
we a
−
→
Cl
2
+ Fe + H
2
+ Fe(OH)
2
3.
1) 4Al + 3O
2
→ 2Al
2
O
3
2) Al
2
O
3
+ 6HNO
3
→ 2Al(NO
3
)
3
+ 3H
2
O
3) Al(NO
3
)
3
+ 3NaOH → Al(OH)
3
+ 3NaNO
3
4) 2Al + 6H
2
O → 2Al(OH)
3
+ 3H
2
↑
5) Al(OH)
3
+ 3HCl → AlCl
3
+ 3H
2
O
4.
:exfbgbc²kZfucjZkijhkljZg_ggu cf_lZeeb hlghkbl_evgh
obfbq_kdbbg_jl_gaZkq_lh[jZam_fhchdkb^ghcie_gdb
:exfhl_jfby²kihkh[\hkklZgh\ e_ gbyf_lZeeh\babohdkb^h\
ijbf_gyxl\ f_lZeemj]bb
5.
140
]
56
]
CaO
+ H
2
O →
o]
2
74
]
Ca(OH)
[
=
x =
⋅
=
]
η(Ca(OH)
2
) =
= 0,984
beb
Hl\_l
η(Ca(OH)
2
) = 98,4%.
<ZjbZgl
1. Al
2
O
3
y\ey_lkyZf nh l_jgufhdkb^hf
Al
2
O
3
+ 2NaOH → 2NaAlO
2
+ H
2
O
Al
2
O
3
+ 6HCl → 2AlCl
3
+ 3H
2
O
2.
A
( )
+
2Cl
–
– 2e
–
→ Cl
2
0
A
l
2
O
3
A
l
(
N
O
3
)
3
A
l
(
O
H
)
3
2)
3)
A
l
(
O
H
)
3
A
l
C
l
3
5)
Al
1)
4)
59
K
( )
−
Na
+
+ 1e
–
→ Na
0
2NaCl
j
\
we a
−
→
2Na + Cl
2
3. Ca
1)
→
CaO
2)
→
Ca(OH)
2
3)
→
CaCO
4)
→
Ca(HCO
3
)
2
5
)
→
CaCl
2
1) 2Ca + O
2
→ 2CaO
2) CaO + H
2
O → Ca(OH)
2
3) Ca(OH)
2
+ CO
2
→ CaCO
3
↓ + H
2
O
4) CaCO
3
+ H
2
O + CO
2
→ Ca(HCO
3
)
2
5) Ca(HCO
3
)
2
+ 2HCl → CaCl
2
+ 2H
2
O + 2CO
2
↑
4.
DZevpbc²hq_gvfy]dbcb j_Zdpbhgghkihkh[gucf_lZee
5. Ca(OH)
2
+ Na
2
CO
3
→ CaCO
3
↓ + 2NaOH
ω(Na
2
CO
3
) = 100% – 5% = 95%
m(Na
2
CO
3
) = 80 ⋅ 0,95 = 76
]
Ca(OH)
2
+
76
]
23
106
]
Na CO
→
o]
3
100
]
CaCO
↓ + 2NaOH
[
=
x =
⋅
= 71,7
]
Hl\_l
m(CaCO
3
) = 71,7
]
<ZjbZgl
1. Al(OH)
3
+ NaOH → Na[Al(OH)
4
]
Al(OH)
3
+ 3HCl → AlCl
3
+ 3H
2
O
2Al(OH)
3
t
→
R
Al
2
O
3
+ 3H
2
O
2.
A
( )
+
2I
–
– 2e
–
→ I
2
0
K
( )
−
K
+
+ 1e
–
→ K
2KI
j\
we a
−
→
2K + I
2
3. CaCl
2
1)
→
CaCO
3
2)
→
Ca(HCO
3
)
2
3)
→
CaCO
3
4)
→
Ca(NO
3
)
2
5)
→
CaCO
3
1) CaCl
2
+ Na
2
CO
3
→ CaCO
3
+ 2NaCl
2) CaCO
3
+ H
2
O + CO
2
→ Ca(HCO
3
)
2
3) Ca(HCO
3
)
2
t
→
R
CaCO
3
↓ + CO
2
+ H
2
O
4) CaCO
3
+ 2HNO
3
→ Ca(NO
3
)
2
+ H
2
O + CO
2
5) Ca(NO
3
)
2
+ Na
2
CO
3
→ CaCO
3
+ 2NaNO
3
4.
@_kldhklvh[mkeh\e_gZkh^_j`Zgb_f\\h^_bhgh\
Mg
2+
b &D
2+
.
Ca(HCO
3
)
2
t
→
R
CaCO
3
+ H
2
O + CO
2
↑
Ijhbkoh^blh[jZah\Zgb_gZdbibb aZkhj_gb_dhleZ
60
5. CaCO
3
→ CaO + CO
2
↑
ω(CaCO
3
) = 100% – 20% = 80%
m(CaCO
3
) = 500 ⋅ 0,8 = 400
]
40
0
]
3
10
0
]
CaCO
→
o
]
56
]
CaO
+ CO
2
↑
[
=
x =
⋅
= 224
]
Hl\_l]
<ZjbZgl
1.
KZHy\ey_lkyhkgh\gufhdkb^hf
Z
CaO + CO
2
→ CaCO
3
[
CaO + H
2
O → Ca(OH)
2
\
CaO + 2HCl → CaCl
2
+ H
2
O
2.
A
( )
+
2Br
–
– 2e
–
→ Br
2
0
K
( )
−
Cu
2+
+ 2e
–
→ Cu
0
CuBr
2
jj
we a
−
→
Cu + Br
2
3.
A
l
2
O
3
A
l
2
(
S
O
4
)
3
2
A
l
A
l
C
l
3
A
l
(
O
H
)
3
5
3
1
4
1) 4Al + 3O
2
→ 2Al
2
O
3
2) Al
2
O
3
+ 3H
2
SO
4
→ Al
2
(SO
4
)
3
+ 3H
2
O
3) Al
2
O
3
+ 6HCl → 2AlCl
3
+ 3H
2
O
4) 2Al + 6HCl → 2AlCl
3
+ 3H
2
O
5) AlCl
3
+ 3NaOH → Al(OH)
3
+ 3NaCl
4.
< ijbjh^_Zexfbgbc\klj_qZ_lky\ \b^_
Al
2
O
3
\ khklZ\_]ebgu
[hdkblh\g_n_ebgZbqbklh]hdhjmg^Z
5. Ca(OH)
2
+
3,15
]
3
263
]
2HNO
⋅
→
o]
32
164
]
Ca(NO )
+ 2H
2
O
[
=
⋅
x =
⋅
⋅
= 4,1