Радецкий А.М., Горшкова В.П. Решение контрольных и самостоятельных работ по химии за 9 класс
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31
n(NH
4
Cl)=
4
m(CaO) 20
M(NH Cl) 53,5
=
fhev
n(CaO)=
m(CaO) 20
M(CaO) 56
=
fhev
Ld&D2gZoh^blky\ ba[uld_jZkq_l\_^_fih1+
4
Cl.
20
]
4
2 53,5
]
2NH Cl
⋅
+CaO
→
o]
3
217
]
2NH
⋅
+CaCl+H
2
O
20
o
2 53,5 2 17
=
⋅⋅
o
⋅
⋅
⋅
]
m
ij
(NH
3
)=
§
(NH
3
)
⋅
m
l_hj
(NH
3
)=0,98
⋅
]
Hl\_l
m
ij
(NH
3
]
5.
6,4
]
42
64
]
NH NO
t
→
R
oe
2
22,4
e
N
+2H
2
O
6,4
o
64 22,4
=
o
⋅
e
V
ij
(N
2
)=
§
(N
2
)
⋅
V
l_hj
=0,89
⋅
e
Hl\_l
V
ij
(N
2
e
6.
56
]
2
22,4
e
N
+3H
2
→
oe
3
2 22,4
e
2NH
⋅
56
o
22,4 2 22,4
=
⋅
o
⋅
⋅
e
V
ij
(NH
3
)=
§
(NH
3
)
⋅
V
l_hj
(NH
3
)=0,5
⋅
e
m(NH
3
)=
P
9
9
⋅
m(NH
3
)=
⋅
]
Hl\_l
V
ij
(NH
3
e
m(NH
3
]
7.
107
]
4
53,5
]
NH Cl
o
t
→
oe
3
22,4
e
NH
+HCl
o
=
o
⋅
e
§
(NH
3
)=
ij
l_hj
V
V
=
beb
Hl\_l
§
(NH
3
)=84,8%
32
8.
od]
31
d]
P
→
200
d]
34
98
d]
HPO
107
o
53,5 22,4
=
o
⋅
d]
m
i
j
(P)=
l
_hj
m
(P)
η
=
]
Hl\_lP3 ]
9.
33
]
42 4
132
]
(NH ) SO
+Ca(OH)
2
→
CaSO
4
+
oe
3
2 22,4
e
2NH
⋅
+2H
2
O
33
o
132 2 22,4
=
⋅
o
⋅
⋅
e
V
ij
=
§
(NH
3
)
⋅
V
l_hj
(NH
3
)=11,2
⋅
e
Hl\_l91+
3
e
10.
o]
3
dhgp
64
]
Cu+4HNO
→
Cu(NO
3
)
2
+
4
e
2
2 22,4
e
2NO
⋅
+2H
2
O
o
64 44,8
=
o
⋅
]
m
ij
(Cu)=
l_hj
m
(P)
η
=
]
Hl\_l
m(
&X ]
11.
30
]
4
2 53,5
]
2NH Cl
⋅
+Ca(OH)
2
→
CaCl
2
+
oe
3
2 22,4
e
2NH
⋅
↑
+2H
2
O
33
o
132 2 22,4
=
⋅
o
⋅
⋅
⋅
e
V
ij
(NH
3
)=V
l_hj
⋅
§
(NH
3
)=12,561
⋅
e
Hl\_l91+
3
e
12.
20,2
]
3
2101
]
2KNO
⋅
+H
2
SO
4
→
K
2
SO
4
+
o]
3
263
]
2HNO
⋅
20,2
o
2 101 2 63
=
⋅⋅
o
⋅
⋅
⋅
]
m
ij
(HNO
3
)=m
l_hj
⋅
§
(HNO
3
)=0,98
⋅
]
Hl\_l
m(HNO
3
]
33
13.
40
]
43
80
]
NH NO
o
t
→
o
e
2
22,4
e
NO
+2H
2
O
40
o
80 22,4
=
o
⋅
e
V
ij
(N
2
O)=V
l_hj
⋅
§
(N
2
O)=11,2
⋅
e
Hl\_l91
2
2 e
14.
80
]
342
310
]
Ca (PO )
+3H
2
SO
4
→
3CaSO
4
+
o]
34
298
]
2H PO
⋅
80
o
310 2 98
=
⋅
o
⋅
⋅
]
m
ij
(H
3
PO
4
)=
§
(H
3
PO
4
)
⋅
m
l_hj
=0,96
⋅
]
Hl\_l
m(H
3
PO
4
]
JZ[hlZBlh]h\Zyihl_f_,,,
<ZjbZgl
1.
IjbagZdbkjZ\g_gby :ahl Nhknhj[_euc
Z:]j_]Zlgh_
khklhygb_gm
]Za L\_j^h_\_s_kl\h
[Obfbq_kdb_
k\hckl\Z
1) N
2
+3H
2
→
2NH
3
2) N
2
+O
2
→
2NO
3) N
2
+3Ca
→
Ca
3
N
2
4) 2N
2
+3Si
→
Si
3
N
4
1)2P+8H
2
O
→
→
2H
3
PO
4
+5H
2
2) 2P+5O
2
→
2P
2
O
5
3) 2P+3Ca
→
Ca
3
P
2
4) 2P+5Cl
2
→
2PCl
5
\J_ZdpbhggZy
kihkh[ghklbb
^_ckl\b_gZ
hj]Zgbafq_eh\_dZ
Obfbq_kdbfZehZdlb\_g
mqZkl\m_l\ ijhp_kk_
^uoZgby
Obfbq_kdb\ukhdhZdlb\_g
y^h\bl
2. P
1)
→
P
2
O
5
2)
→
HPO
3
3)
→
H
3
PO
4
4)
→
Na
3
PO
4
5)
→
Ag
3
PO
4
1) 4P+5O
2
→
2P
2
O
5
2) P
2
O
5
+H
2
O
→
2HPO
3
3) HPO
3
+H
2
O
→
H
3
PO
4
4) H
3
PO
4
+3NaOH
→
Na
3
PO
4
+3H
2
O
5) Na
3
PO
4
+3AgNO
3
→
Ag
3
PO
4
↓
+3NaNO
3
34
3.
37
]
2
74
]
Ca(OH)
+(NH
4
)
2
SO
4
→
o]
3
217
]
2NH
⋅
+CaSO
4
+2H
2
O
37
o
74 2 17
=
⋅
o
⋅
⋅
]
§
(NH
3
)=
i
j
l_hj
m
m
=
beb
Hl\_l
§
(NH
3
)=88,24%
<ZjbZgl
1.
IjbagZdbkjZ\g_gby :ahl Nhknhj[_euc
ZKljh_gb_Zlhfh\
+7
N 1S
2
2S
2
2P
3
2 5
+15
P 1s
2
2s
2
2p
6
3s
2
3p
3
2 8 5
[<Z`g_crb_kl_i_gb
hdbke_gby
GZb[he__
jZkijhkljZg_gu
kh_^bg_gbykh
kl_i_gyfbhdbke_gby±
3, 0, +4, +5
GZb[he__jZkijhkljZg_gu
kh_^bg_gbykhkl_i_gyfb
hdbke_gby±
\KhklZ\b k\hckl\Z
hdkb^h\
H[jZam_lhdkb^u1
2
O,
NO, N
2
O
3
, NO
2
, N
2
O
5
.
GZb[he__mklhcqb\u
N
2
2±[_kp\_lguc]Zak
ijbylgufaZiZohfb
keZ^dh\Zluf\dmkhf
12±[_kp\_lguc]Za
[_a\dmkZb aZiZoZ12
2
±[mjuc]Za
H[jZam_lhdkb^u3
2
O
3
b
P
2
O
5
.
P
2
O
3
±[_eh_
djbklZeebq_kdh_\_s_kl\h
P
2
O
5
±[_eucihjhrhd
2. N
2
1)
→
NH
3
2)
→
NO
3)
→
NO
2
4)
→
HNO
3
5)
→
Ba(NO
3
)
2
1) N
2
+3H
2
Pt
→
2NH
3
2) 4NH
3
+5O
2
Pt
→
4NO+6H
2
O
3) 2NO+O
2
→
2NO
2
4) 4NO
2
+2H
2
O+O
2
→
4HNO
3
5) 2HNO
3
+Ba(OH)
2
→
Ba(NO
3
)
2
+2H
2
O
] o]
3. 2NaNO
3
+H
2
SO
4
→
Na
2
SO
4
+2HNO
3
2
⋅
]
2
⋅
]
⋅
=
⋅
o
o
⋅
⋅
⋅
]
m
ij
(HNO
3
)=
§
(HNO
3
)
⋅
m
l_hj
=12,6
⋅
]
Hl\_l
m(HNO
3
]
+7
+15
35
<ZjbZgl
1.
IjbagZdb
kjZ\g_gby
NH
3
PH
3
ZKljh_gb_
fhe_dmeu
H
→
N
←
H
↑
H
H-P-H
|
H
[Nbabq_kdb_
k\hckl\Z
=Za[_ap\_lZkj_adbfaZiZohf
e_]q_\ha^moZohjhrh
jZkl\hjbf \\h^_
;_kp\_lguc]Zak q_kghqguf
aZiZohfly`_e__\ha^moZ
fZehjZkl\hjbf\\h^_hq_gv
y^h\bl
\Obfbq_kdb_
k\hckl\Z
1) NH
3
+H
2
O
→
NH
4
OH
2) 4NH
3
+3O
2
→
N
2
+6H
2
O
3) NH
3
+HCl
→
NH
4
Cl
1) 2PH
3
+4O
2
→
P
2
O
5
+3H
2
O
2) PH
3
+HCl
→
PH
4
Cl
2. P
1)
→
CaP
2
2)
→
PH
3
3)
→
P
2
O
5
4)
→
HPO
3
5)
→
H
3
PO
4
1) 2P+3Ca
→
Ca
3
P
2
2) Ca
3
P
2
+6H
2
O
→
3Ca(OH)
2
+2PH
3
3) 2PH
3
+4O
2
→
P
2
O
5
+3H
2
O
4) P
2
O
5
+H
2
O
→
2HPO
3
5) HPO
3
+H
2
O
→
H
3
PO
4
3.
30
]
4
2 53,5
]
2NH Cl
⋅
+Ca(OH)
2
→
CaCl
2
+
oe
3
2 22,4
e
2NH
⋅
+2H
2
O
30
o
2 53,5 2 22,4
=
⋅⋅
o
⋅
⋅
⋅
]
V
ij
(NH
3
)=
§
(NH
3
)
⋅
V
l_hj
=0,92
⋅
e
Hl\_l91+
3
e
<ZjbZgl
1.
IjbagZdb
kjZ\g_gby
HNO
3
H
3
PO
4
ZNbabq_kdb_
k\hckl\Z
;_kp\_lgZy^ufysZykygZ
\ha^mo_`b^dhklv
t
ie
=-42
0
K
t
dbi
=83
0
K
;_eh_djbklZeebq_kdh_\_s_kl\h
t
ie
=42
0
K
[Obfbq_kdb_
k\hckl\Z
1) 8HNO
3
+3Cu
→
3Cu(NO
3
)
2
+
+2NO+4H
2
O
H
3
PO
4
+Cu
→
/
2) 10HNO
3
+4Zn
→
4Zn(NO
3
)
2
+
+NH
4
NO
3
+3H
2
O
3) HNO
3
+NaOH
→
NaNO
3
+H
2
O
1) 2H
3
PO
4
+3Zn
→
Zn
3
(PO
4
)
2
+3H
2
↑
2) H
2
PO
4
+3NaOH
→
Na
3
PO
4
+3H
2
O
2. Pb(NO
3
)
2
1)
→
NO
2
2)
→
HNO
3
3)
→
NH
4
NO
3
4)
→
NH
3
5)
→
(NH
4
)
2
SO
4
1) 2Pb(NO
3
)
2
t
→
R
2PbO+4NO
2
+O
2
36
2) 4NO
2
+2H
2
O+O
2
→
4HNO
3
3) HNO
3
+NH
3
→
NH
4
NO
3
4) NH
4
NO
3
+NaOH
→
NH
3
+H
2
O+NaNO
3
5) 2NH
3
+H
2
SO
4
→
(NH
4
)
2
SO
4
3.
62
]
431
]
4P
⋅
+5O
2
→
o]
25
2142
]
2P O
⋅
62
o
4 31 2 142
=
⋅⋅
o
⋅
⋅
⋅
]
§
(P
2
O
5
)=
ij
l_hj
m
m
=
beb
Hl\_l
§
(P
2
O
5
)=91,55%
37
L_fZ ,9Ih^]jmiiZm]e_jh^Z
JZ[hlZM]e_jh^Hdkb^um]e_jh^Z
<ZjbZgl
1.
K
1s
2
2s
2
2p
2
↑↓
↑↓
↑
↑
K
*
1s
2
2s
1
2p
3
↑↓
↑
↑
↑
↑
I_j_oh^ \hafh`_g ld bf__lky k\h[h^gZy hj[blZev ijb wlhf
jZagbpZ\wg_j]bbf_`^m
2s-
b
p-
hj[blZeyfbg_\bebdZ
Z
CuO+C
→
Cu+CO
[&&2
2
→
2CO
t
0
\&D&2
3
→
CaO+CO
2
↑
3. 2CO+O
2
→
2CO
2
¹
(CO)=100%-5%=95%
V(CO)=V
]ZaZ
⋅
¹
(CO)=20
⋅
e
e oe
2CO + O
2
→
2CO
2
2
⋅
e e
o
=
⋅
o
⋅
e
Hl\_l92
2
e
<ZjbZgl
M]e_jh^\ ijbjh^_\klj_qZ_lky\\b^_
ZZefZaZ±hq_gvl\_j^h_ijhqgh_\_s_kl\hk \ukhdhc
t
ie
djbklZeebq_kdZyj_r_ldZ±ZlhfgZy
[]jZnblZ±l\_j^h_\_s_kl\h`bjgh_gZhsmivk \ukhdhc
t
ie
we_dljhb l_iehijh\h^_gdjbklZeebq_kdZyj_r_ldZZlhfgZy
\dZj[bgZ±l\_j^h_\_s_kl\hl\_j`_]jZn blZghfy]q_ZefZaZ
q_jgucf_edhdjbklZeebq_kdbcihjhrhdh[eZ^Z_l
ihemijh\h^gbdh\ufbk\hckl\ZfbDjbklZeebq_kdZyj _r_ldZ
ZlhfgZy
Z&2
2
+Ca(OH)
2
→
CaCO
3
+H
2
O
[&+
2
t
→
R
CH
4
\&2
2
→
CO
2
3. CaCO
3
+2HNO
3
→
K
a(NO
3
)
2
+H
2
O+CO
2
¹
(CaCO
3
)=100%-8%=92%
m(CaCO
3
)=0,92
⋅
]
38
55,2
]
3
10
0
]
CaCO
+2HNO
3
→
Ca(NO
3
)
2
+H
2
O+
o
e
2
22,4
e
CO
55,2
o
100 22,4
=
o
⋅
e
Hl\_l9&2
2
e
<ZjbZgl
M]e_jh^\ kh_^bg_gbyoijhy\e y_lkl_i_gbhdbke_gby±
+2, +4
C
-4
H
+
, C
−
H
+
, C
+2
O
-2
, C
+4
O
−
Z
CO+FeO
→
Fe+CO
2
[&$O
→
Al
4
C
3
\&2
2
+CaO
→
CaCO
3
3. C+O
2
→
CO
2
¹
(C)=100%-6%=94%
m(C)=0,94
⋅
]
] oe
C+O
2
→
CO
2
] e
o
=
o
⋅
e
Hl\_l9&2
2
e
<ZjbZgl
:^khj[pby²ijhp_kkih]ehs_gbyih\_joghklvx\_s_kl\Z
^jm]h]h\_s_kl\Z
>_khj[pby²ijhp_kk\u^_e_gby\_s_kl\Zk_]hih\_joghklb
:^khj[pbhggZ ykihkh[ghklv^j_\_kgh]hm]eyijbf_gy_lky\
f_^bpbg_ijbhljZ\e_gbyo^eyih]ehs_gbyy^h\
Z&22
2
→
2CO
2
[&6
→
CS
2
\&D&2
3
+2HCl
→
CaCl
2
+H
2
O+CO
2
3. CaCO
3
t
→
R
CaO+CO
2
¹
(CaCO
3
)=100%-10%=90%
m(CaCO
3
)=0,9
⋅
]
] oe
450
]
3
100
]
CaCO
→
CaO+
oe
2
22,4
e
CO
450
o
100 22,4
=
o
⋅
e
39
Hl\_l9&2
2
e
JZ[hlZM]hevgZydbkehlZb__kheb
<ZjbZgl
1. CaCO
3
1)
→
CO
2
2)
→
CaCO
2
3)
→
Ca(HCO
3
)
2
4)
→
CaCl
2
1) CaCO
3
t
→
R
CaO+CO
2
2) CO
2
+Ca(OH)
2
→
CaCO
3
+H
2
O
3) CaCO
3
+H
2
O+CO
2
→
Ca(HCO
3
)
2
4) Ca(HCO
3
)
2
+2HCl
→
CaCl
2
+2H
2
O+2CO
2
Kh^_j`bfh_ijh[bjhdfh`ghjZkihagZlvijbihfhsbjZkl\hjZ
AgNO
3
.
AgNO
3
+KCl
→
[_euc
AgCl
↓
+KNO
3
2AgNO
3
+K
2
CO
3
→
23
k\_leh`_eluc
Ag CO
↓
+2KNO
3
3AgNO
3
+K
3
PO
4
→
34
`_eluc
Ag PO
↓
+3KNO
3
3. CO
2
+H
2
O
→
H
2
CO
3
H[jZamxsZykydbkehlZhdjZrb\Z_leZdfmk\djZkgucp\_lijb
gZ]j_\ZgbbdbkehlZjZaeZ]Z_lkyihwlhfmeZdfmk\gh\vklZgh\blky
nbhe_lh\uf
H
2
CO
3
→
H
2
O+CO
2
↑
<ZjbZgl
1. CO
1)
→
CO
2
2)
→
KHCO
3
3)
→
K
2
CO
3
4)
→
CO
2
1) 2CO+O
2
→
2CO
2
2) CO
2
+KOH
→
KHCO
3
3) KHCO
3
+KOH
→
K
2
CO
3
+H
2
O
4) K
2
CO
3
+2HCl
→
2KCl+H
2
O+CO
2
IjbijhimkdZgbbm]e_dbkeh]h]ZaZq_j_aba\_kldh\mx\h^m
kgZqZeZgZ[ex^Z_lky__ ihfmlg_gb_Z ihlhfjZkl\hj_gb_hkZ^dZ
Ca(OH)
2
+CO
2
→
CaCO
3
↓
+H
2
O
CaCO
3
+CO
2
+H
2
O
→
Ca(HCO
3
)
2
<h^gu_jZkl\hjudZj[hgZlh \dZebyb gZljbybf_xls_ehqgmx
kj_^mldwlbkhebh[jZah\Zgukbevgufhkgh\Zgb_fbkeZ[hc
dbkehlhcb\\h^_]b^jhebamxlky
<ZjbZgl
1. Ca(HCO
3
)
2
1)
→
CaCO
3
2)
→
CO
2
3)
→
Na
2
CO
3
4)
→
H
2
CO
3
1) Ca(HCO
3
)
2
t
→
R
CaCO
3
+H
2
O+CO
2
40
2) CaCO
3
+2HCl
→
CaCl
2
+H
2
O+CO
2
3) CO
2
+2NaOH
→
Na
2
CO
3
+H
2
O
4) Na
2
CO
3
+2HCl
→
2NaCl+H
2
CO
3
2. CaCO
3
t
→
R
CaO+CO
2
CO
2
+KOH
→
KHCO
3
2KHCO
3
t
→
R
K
2
CO
3
+H
2
O+CO
2
Khebm]hevghcdbkehlufh`ghhlebqblvijbihfhsbj_Zdpbbk
kbevghcdbkehlhcijbwlhf[m^_l\u^_eylvkym]e_dbkeuc]Za
<ZjbZgl
1. C
1)
→
CO
2
2)
→
H
2
CO
3
3)
→
Na
2
CO
3
4)
→
NaNO
3
1) C+O
2
→
CO
2
2) CO
2
+H
2
O
→
H
2
CO
3
3) H
2
CO
3
+2NaOH
→
Na
2
CO
3
+2H
2
O
4) Na
2
CO
3
+2HNO
3
→
2NaNO
3
+H
2
O+CO
2
↑
Z
Ba(OH)
2
+Na
2
CO
3
→
BaCO
3
↓
+2NaOH
[+
2
SO
4
+Na
2
CO
3
→
Na
2
SO
4
+H
2
O+CO
2
↑
Ihemqblvkh^mlZdbfh[jZahfg_evayld&D&2
3
±g_jZkl\hjbfZy
\\h^_khev
JZ[hlZDj_fgbcb_]hkh_^bg_gby
<ZjbZgl
JZ^bmkZlhfZdj_fgby[hevr_jZ^bmkZZlhfZm]e_jh^Z_]h
we_dljhgukeZ[__ijbly]b\Zxlkyd y^jml_hge_]q_bohl^Z_l
ihlhfm_]hg_f_lZeebq_kdb_k\hckl\Z\ujZ`_gukeZ[__
2. SiO
2
Z
→
Si
[
→
SiO
2
\
→
Na
2
SiO
3
Z6L2
2
+C
t
→
R
Si+CO
2
[6L2
2
→
SiO
2
\6L2
2
+2NaOH
→
Na
2
SiO
3
+H
2
O
3. Mg
2
Si+4HCl
→
2MgCl
2
+SiH
4
¹
(Mg
2
Si)=100%-5%=95%
m(Mg
2
Si)=0,95
⋅
]
152
]
2
76
]
Mg Si
+4HCl
→
2MgCl
2
+
o]
4
32
]
SiH
152
o
76 32
=
o
⋅
]
Hl\_l
m(SiH
4
]