Радецкий А.М., Горшкова В.П. Решение контрольных и самостоятельных работ по химии за 9 класс
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21
^+
2
SO
4
+Ca(OH)
2
→
CaSO
4
+2H
2
O
H
+
+OH
-
→
H
2
O
K_jgZydbkehlZbkihevam_lky^eyijhba\h^kl\Zm^h[j_ gbck__
ihfhsvxihemqZxlnhknhjgmxdbkehlmb\ijhfure_gghklb
hj]Zgbq_kdh]hkbgl_aZ
Ijbeb\\h[_ijh[bjdbjZkl\hj%D&O
2
\h^ghcbagbogZ[ex^Z_f
\uiZ^_gb_[_eh]hhkZ^dZ
BaCl
2
+Na
2
SO
4
→
BaSO
4
↓
+2NaCl
A^_kv[ue1D
2
SO
4
JZ[hlZKdhjhkl vobfbq_kdboj_Zdpbc
Obfbq_kdh_jZ\gh\_kb_
<ZjbZgl
1. H
2
+Cl
2
→
2HCl
J_Zdpbyb^_lf]gh\_gghkh\aju\hf
H
2
+Br
2
→
2HBr
J_Zdpbyb^_lf_^e_ggh
H
2
+I
2
→
2HI
J_Zdpbyb^_llhevdhijbgZ]j_\Zgbb
2. CO
]
+H
2
O
]
→
←
CO
]
+H
]
+Q
Z\e_\hldj_Zdpbywdahl_jfbqgZ
[\ijZ\hldm\_ebqb\Z_lkydhgp_gljZpbybkoh^guo\_s_kl\
<ZjbZgl
1. H
2
+Cl
2
→
2HCl
?keb dhgp_gljZpby wlbo \_s_kl\ fZeZ lh j_Zdpby b^_l f_^e_ggh
_keb dhgp_gljZpby ^hklZlhqgh \_ebdZ lh j_Zdpby ijhbkoh^bl
f]gh\_gghkh\aju\hf
2. 2H
]
+O
]
→
←
2H
2
O
]
+Q
Z\ijZ\hldj_Zdpbywdahl_jfbqgZ
[\ijZ\hldj_Zdpbyijhl_dZ_lk mf_gvr_gb_fh[t_fZkbkl_fu
<ZjbZgl
1. Zn+2HCl
→
ZnCl
2
+H
2
↑
Ijbh[uqguomkeh\byoj_Zdpbyijhl_dZ_lf_^e_ gghZijb
gZ]j_\ZgbbgZqbgZ_lky[mjgh_\u^_e_gb_\h^hjh^Z
2. N
]
+3H
]
→
←
2NH
]
+Q
Z\ijZ\hldj_Zdpbywdahl_jfbqgZ
[\e_\hldj_Zdpbyijhl_dZ_lk mf_gvr_gb_fh[t_fZkbkl_fu
22
<ZjbZgl
1. 3H
2
+N
2
→
2NH
3
;_adZlZebaZlhjZ kdhjhklv wlhcj_Zdpbb qj_a\uqZcgh fZeZ gh ijb
^h[Z\e_ gbbdZlZebaZlhjZ__ kdhjhklvagZqbl_evghm\_ebqb\Z_lky
2. 2HBr
]
→
←
H
]
+Br
]
-Q
Z\ijZ\hldj_Zdpbywg^hl_jfbqgZ
[\ijZ\hldm\_ebqb\Z_lkydhgp_gljZpbybkoh^gh]h\_s_kl\Z
JZ[hlZBlh]h\Zyihl_f_ ,,
<ZjbZgl
1. H
2
SO
4
+2KOH
→
K
2
SO
4
+2H
2
O
3H
2
SO
4
+2Al
→
Al
2
(SO
4
)
3
+3H
2
H
2
SO
4
+FeO
→
FeSO
4
+H
2
O
H
2
SO
4
+Ba(NO
3
)
2
→
BaSO
4
↓
+2HNO
3
H
2
SO
4
+Zn(OH)
2
→
ZnSO
4
+2H
2
O
2. S
1
→
SO
2
2
→
SO
3
3
→
H
2
SO
4
4
→
SO
2
1) S+O
2
→
SO
2
2) 2SO
2
+O
2
→
2SO
3
3) SO
3
+H
2
O
→
H
2
SO
4
4) H
2
SO
4
+Na
2
SO
3
→
Na
2
SO
4
+H
2
O+SO
2
↑
3. 2SO
]
+O
]
→
←
2SO
]
+Q
Z\e_\hldj_Zdpbywdahl_jfbqgZ
[\e_\hldj_Zdpbyijhl_dZ_lk mf_gvr_gb_fh[t_fZkbkl_fu
M\_ebq_gb_kdhjhklbj_Zdpbbh[`b]ZibjblZ^hklb]Z_lkyiml_f
gZ]j_\Zgbyihklhyggh]hlhdZdbkehjh^Zm\_ebq_gbyiehsZ^b
khijbdhkgh\_gbyj_Z]_glh\b dZlZebaZlhjZ
<ZjbZgl
1. 3H
2
SO
4
+2Fe(OH)
3
→
Fe
2
(SO
4
)
3
+6H
2
O
H
2
SO
4
+BaCl
2
→
BaSO
4
↓
+2HCl
H
2
SO
4
+2LiOH
→
Li
2
SO
4
+2H
2
O
H
2
SO
4
+MgO
→
MgSO
4
+H
2
O
H
2
SO
4
+Zn
→
ZnSO
4
+H
2
↑
1 3 4
2. H
2
S
1
←
S
3
→
FeS
4
→
SO
2
↓
2
Na
2
S
1) H
2
+S
→
H
2
S
23
2) S+2Na
→
Na
2
S
3) S+Fe
→
FeS
4) 4FeS+7O
2
→
2Fe
2
O
3
+4SO
2
Zijbm\_ebq_gbbl_fi_jZlmjukdhjhklvj_Zdpbbm\_ebqb\Z_lky
[ijbm\_ebq_gbbdhgp_gljZpbbbkoh^guo\_s_kl\kdhjhklv
j_Zdpbbm\_ebqb\Z_lkyIjbf_jukfl_fZ ,,jZ[hlZ\ZjbZgl
\hijhk
4. 4FeS
2
+11O
2
t
→
R
2Fe
2
O
3
+8SO
2
Wlhlijhp_kkhkgh\ZggZijbf_g_gbbijbgpbiZl_iehh[f_gZijb
wlhf^hklb]Z_lkyfZdk bfZevguc\uoh^ijh^mdlZ
<ZjbZgl
1. H
2
SO
4
+2NaOH
→
Na
2
SO
4
+2H
2
O
H
2
SO
4
+CuO
→
CuSO
4
+H
2
O
H
2
SO
4
+Mg
→
MgSO
4
+H
2
H
2
SO
4
+BaCl
2
→
BaSO
4
+2HCl
3H
2
SO
4
+2Al(OH)
3
→
Al
2
(SO
4
)
3
+6H
2
O
2. S
1
→
ZnS
2
→
SO
2
3
→
SO
3
4
→
H
2
SO
4
1. S+Zn
→
ZnS
2) 2ZnS+3O
2
→
2ZnO+2SO
2
3) 2SO
2
+O
2
→
2SO
3
4) SO
3
+H
2
O
→
H
2
SO
4
ZIjbijZ\bevghfih^[hj_dZlZebaZlhjZkdhjhklvj_Zdpbb
agZqbl_evgh\h aj ZklZ_l
[JZagu_\_s_kl\Zj_Z]bj mxlk h^gbfbl_f`_\_s_kl\hfk jZaghc
kdhjhklvx
Ijbf_jukfL_fZ,,jZ[hlZ\ZjbZgl\hijhk
Ijbf_kbkh^_j`Zsb_ky\ k_j g bklhf]Za_fh]mly\eylvky
bg]b[blhjZfbj _Zd pbb62
2
+O
2
→
←
2SO
3
ijbwlhfagZqbl_evgh
kgb`Z_lky__ k dhjhklv
<ZjbZgl
1. H
2
SO
4
+Ca(OH)
2
→
CaSO
4
+2H
2
O
H
2
SO
4
+ZnO
→
ZnSO
4
+H
2
O
H
2
SO
4
+Fe
→
FeSO
4
+H
2
↑
H
2
SO
4
+Ba(NO
3
)
2
→
BaSO
4
↓
+2HNO
3
H
2
SO
4
+Cu(OH)
2
→
CuSO
4
+2H
2
O
2. SO
]
+H
2
O
`
→
←
H
2
SO
`
+Q
Z\e_\hldj_Zdpbywdahl_jfbqgZ
[\ijZ\hldm\_ebqb\Z_lkydhgp_gljZpbybkoh^gh]h\_s_kl\Z
24
3. H
2
S
1
←
S
3
→
SO
2
4
→
SO
3
↓
2
Al
2
S
3
1) S+H
2
→
H
2
S
2) 3S+2Al
→
Al
2
S
3
3) S+O
2
→
SO
2
4) 2SO
2
+O
2
→
2SO
3
Ld62
3
j_Z]bjm_lkgZqZeZk\h^ygufiZjhflhagZqbl_evgZy
qZklv+
2
SO
4
ihemqZ_lky\ \b^_lmfZgZdhg^_gkbjh\Zlvdhlhjuc
g_m^h[gh
25
L_fZ,,,Ih^]jmiiZZahlZ
JZ[hlZ:ahl:ffbZdKhebZffhgby
<ZjbZgl
Z
0
0
2
hdbkebl_ev \hkklZgh\bl_ev
N+ Li
→
Li
+
N
-3
hdbkebl_ev\hkklZgh\bl_ev
[1+
4
Cl+NaOH
→
NaCl+NH
3
↑
+H
2
O
\1+
4
Cl+AgNO
3
→
AgCl
↓
+NH
4
NO
3
Cl
-
+Ag
+
→
AgCl
↓
2. H
2
+Cl
2
→
2HCl
Pt
3H
2
+N
2
→
2NH
3
NH
3
+HCl
→
NH
4
Cl
N
2
:N
00
N: N
≡
N
:ahl±[_kp\_lguc]Za[_a\dmkZ[_aaZiZoZ
t
ie
=-210
0
K
t
dbi
=-196
0
K
<ZjbZgl
Z
00
22
hdbkebl_ev \hklZgh\bl_ev
N+3H
→
2N
-3
H
+
[1+
3
+H
2
SO
4
→
(NH
4
)
2
SO
4
\1+
4
)
2
SO
4
+Ba(NO
3
)
2
→
BaSO
4
↓
+2NH
4
NO
3
SO
−
+Ba
2+
→
BaSO
4
↓
>Zggu_jZkl\hjufh`ghjZkihagZlvijbihfhsbjZkl\hjh\%D&O
2
b 1D
2
SiO
3
.
1) BaCl
2
+Na
2
SO
4
→
BaSO
4
↓
+2NaCl
BaCl
2
+H
2
SO
4
→
BaSO
4
↓
+2HCl
NH
4
Cl+BaCl
2
→
/
2) H
2
SO
4
+Na
2
SiO
3
→
Na
2
SO
4
+H
2
SiO
3
↓
Na
2
SO
4
+Na
2
SiO
3
→
/
3. NH
3
+H
+
→
NH
4
+
N 2s
2
2p
3
↑↓
↓
↓
↓
ς
ς
H
+
1s
0
H1s
1
↑
H1s
1
↑
H1s
1
↑
H
↓
H
→
N
←
H c
\yabdh\Ze_glgu_iheyjgu_
26
NH
3
+H
+
→
NH
3
H
+
k\yavih^hghjghZdp_ilhjghfmf_oZgbafm
III I
N
-3
H
+
NH
3
IV I
(N
-3
H
+
)
+
(NH
4
)
+
<ZjbZgl
Z1
+3Mg
0
→
Mg
+
N
−
hdbkebl_ev\hkklZgh\bl_ev
[1+
4
)
2
CO
3
+2HNO
3
→
2NH
4
NO
3
+CO
2
+H
2
O
CO
−
+2H
+
→
CO
2
+H
2
O
2. SO
3
+H
2
O
→
H
2
SO
4
N
2
+3H
2
Pt
→
2NH
3
2NH
3
+H
2
SO
4
→
(NH
4
)2SO
4
3. N 2s
2
2p
3
↑↓
↓
↓
↓
ς
ς
ς
N 2s
2
2p
3
↑↓
↓
↓
↓
K\yabdh\Ze_glgZyg_iheyjgZy
N
<ZjbZgl
Z
t
00
22
\hkklZgh\bl_ev hdbkebl_ev
N+O
→
R
2N
+2
O
-2
[1+
3
+HCl
→
NH
4
Cl
\1+
4
)
2
SO
4
+Ba(OH)
2
→
BaSO
4
↓
+2NH
3
+2H
2
O
2NH
+
+SO
−
+Ba
2+
+2OH
-
→
BaSO
4
+2NH
3
+2H
2
O
>Zggu_\_s_kl\Zfh`ghjZkihagZlvijbihfhsbjZkl\hjh\%D&O
2
b
NaOH
1) BaCl
2
+(NH
4
)
2
SO
4
→
2NH
4
Cl+BaSO
4
↓
NH
4
Cl+BaCl
2
→
/
NaCl+BaCl
2
→
/
2) NH
4
Cl+NaOH
→
NaCl+NH
3
↑
+H
2
O
NaCl+Na
HG
→
/
3. NH
3
H:
N
:H H
→
N
←
H
H
↑
H
:ffbZd²]Zakj_adbfaZiZohf
t
ie
=-78
0
K
t
dbi
=-33
0
K
27
JZ[hlZ:ahlgZydbkehlZb__kheb
<ZjbZgl
Z&D2+12
3
→
Ca(NO
3
)
2
+H
2
O
CaO+2H
+
→
Ca
2+
+H
2
O
[1D2++12
3
→
NaNO
3
+H
2
O
OH
-
+H
+
→
H
2
O
\.
2
CO
3
+2HNO
3
→
2KNO
3
+H
2
O+CO
2
CO
−
+2H
+
→
H
2
O+CO
2
2. Cu
0
+H
+1
N
+5
O
−
dhgp
→
Cu
+2
(N
+5
O
−
)
2
+N
+4
O
−
+H
+
O
-2
Cu
0
-2e
-
→
Cu
2+
1
N
+5
+1e
-
→
N
+4
2
Cu+4HNO
3
→
Cu(NO
3
)
2
+2NO
2
+2H
2
O
3. N
2
Z
→
NO
[
→
NO
2
\
→
HNO
3
Z1
2
+O
2
→
2NO
[122
2
→
2NO
2
\12
2
+2H
2
O+O
2
→
4HNO
3
<ZjbZgl
Z
Ca(OH)
2
+2HNO
3
→
Ca(NO
3
)
2
+2H
2
O
OH
-
+H
+
→
H
2
O
[&X2+12
3
→
Cu(NO
3
)
2
+H
2
O
CuO+2H
+
→
Cu
2+
+H
2
O
\3E
+2
(N
+5
O
−
)
2
→
Pb
+2
O
-2
+N
+4
O
−
+O
N
+5
+1e
-
→
N
+4
4
2O
-2
-4e
-
→
O
1
2Pb(NO
3
)
2
→
2PbO+4NO
2
+O
2
>Zggu_\_s_kl\Zfh`ghjZkihagZlvijbihfhsbjZkl\hjh\
gbljZlZ[Zjby
Ba(NO
3
)
2
b
NaOH
Ba(NO
3
)
2
+Na
2
SO
4
→
2NaNO
3
+BaSO
4
↓
NH
4
NO
3
+Ba(NO
3
)
2
→
/
Ba(NO
3
)
2
+NaNO
3
→
/
NH
4
NO
3
+NaOH
→
NaNO
3
+NH
3
↑
+H
2
O
NaNO
3
+NaOH
→
/
<ZjbZgl
Z+12
3
+Na
2
O
→
2NaNO
3
+H
2
O
2H
+
+Na
2
O
→
2Na
+
+H
2
O
[.2++12
3
→
KNO
3
+H
2
O
28
OH
-
+H
+
→
H
2
O
\1D
2
CO
3
+2HNO
3
→
2NaNO
3
+H
2
O+CO
2
CO
−
+2H
+
→
H
2
O+CO
2
2. Hg
0
+H
+
N
+
5
O
−
dhgp
→
Hg
+
2
(N
+
5
O
−
)
2
+N
+
4
O
−
+H
+
O
-
2
Hg
0
-2e
-
→
Hg
+
2
1
N
+5
+1e
-
→
N
+4
2
Hg+4HNO
3
→
Hg(NO
3
)
2
+2NO
2
+2H
2
O
3. KNO
3
Z
→
HNO
3
[
→
Cu(NO
3
)
2
\
→
NO
2
Z.12
3
+H
2
SO
4
→
K
2
SO
4
+2HNO
3
[+12
3
+CuO
→
Cu(NO
3
)
2
+H
2
O
t
0
\&X12
3
)
2
→
2CuO+4NO
2
+O
2
<ZjbZgl
Z
Ba(OH)
2
+2HNO
3
→
Ba(NO
3
)
2
+2H
2
O
OH
-
+H
+
→
H
2
O
[&D&2
3
+2HNO
3
→
Ca(NO
3
)
2
+H
2
O+CO
2
CaCO
3
+2H
+
→
Ca
2+
+H
2
O+CO
2
\+12
3
+MgO
→
Mg(NO
3
)
2
+H
2
O
2H
+
+MgO
→
Mg
2+
+H
2
O
t
0
2. Ag
+1
N
+5
O
−
→
Ag
0
+N
+4
O
−
+O
Ag
+1
+1e
-
→
Ag
0
N
+5
+1e
-
→
N
+4
4 2
2O
-2
-4e
-
→
O
2 1
t
0
2AgNO
3
→
2Ag+2NO
2
+O
2
>Zggu_\_s_kl\Zfh`ghjZkihagZlvijbihfhsbjZkl\hjh\
Ba(NO
3
)
2
b $J12
3
1) Ba(NO
3
)
2
+H
2
SO
4
→
BaSO
4
↓
+2HNO
3
Ba(NO
3
)
2
+HCl
→
/
Ba(NO
3
)
2
+HNO
3
→
/
2) AgNO
3
+HCl
→
AgCl
↓
+HNO
3
AgNO
3
+HNO
3
→
/
JZ[hlZNhknhjb_]hkh_^bg_gby
<ZjbZgl
Z32
2
→
2P
2
O
5
29
[3+
3
+HCl
→
PH
4
Cl
\+
3
PO
4
+3KOH
→
K
3
PO
4
+3H
2
O
H
+
+OH
-
→
H
2
O
KbevghgZ]j_\ZydjZkgucnhknhj\ hlkmlkl\b_\ha^moZihemqZ_f
[_eucnhknhjWlhagZqblqlhwlhZeehljhigu_fh^bnbdZpbb
h^gh]hwe_f_glZ
NhknZlgZljbyhij_^_ebfijbihfhsbeZdfmkZkj_^Z_]h
jZkl\hjZs_ehqgZyeZdfmkkbg__l: dbkehluhij_^_ebfijb
ihfhsbjZkl\hjZ$J12
3
.
AgNO
3
+HCl
→
AgCl
↓
+HNO
3
[_euc
2AgNO
3
+H
3
PO
4
→
/
<ZjbZgl
Z3&O
2
→
2PCl
5
[30J
→
Mg
3
P
2
\1D
3
PO
4
+3AgNO
3
→
3NaNO
3
+Ag
3
PO
4
↓
PO
−
+3Ag
+
→
Ag
3
PO
4
↓
Nhknhjhq_gvj_Zdpbhgghkihkh[gh_\_s_kl\hZ fhe_dmeZZahlZ
mklhcqb\Zb obfbq_kdbfZehZdlb \gZ
3. P
Z
→
Ca
3
P
2
2
GH
[
→
PH
3
\
→
P
2
O
5
Z3&D
→
Ca
3
P
2
[&D
3
P
2
+6H
2
O
→
3Ca(OH)
2
+2PH
3
\3+
3
+4O
2
→
P
2
O
5
+3H
2
O
<ZjbZgl
Z3&D
→
Ca
3
P
2
t
0
[&D
3
(PO
4
)
2
+3H
2
SO
4
→
3CaSO
4
+2H
3
PO
4
\1D2++
3
PO
4
→
Na
3
PO
4
+3H
2
O
OH
-
+H
+
→
H
2
O
Ijbk]hjZgbb[_eh]hb djZkgh]hnhknhjZ\ h^bgZdh\uomkeh\byo
h[jZamxlkyh^bgZdh\u_\_s_kl\Z
<h^mfh`ghhij_^_eblvijbihfhsbeZdfmkZ\h^Zbf__l
g_cljZevgmxkj _^ mbeZdfmk\g_chklZg_lkynbhe_lh\uf\
dbkehlZohgihdjZkg__ l+
NO
3
\ hlebqb_hl+
3
PO
4
,
jZkl\hjy_l
f_^v
Cu+HNO
3
→
Cu(NO
3
)
2
↓
+2NO
2
+2H
2
O
`_eluc
Cu+H
3
PO
4
→
/
30
<ZjbZgl
Z36
→
P
2
S
3
[3
2
O
5
+3H
2
O
→
2H
3
PO
4
\+
3
PO
4
+3Na
2
CO
3
→
2Na
3
PO
4
+3CO
2
+3H
2
O
2H
+
+CO
−
→
CO
2
+H
2
O
Fhe_dmeZnhknhjZ\ iZjZokhklhblbao Zlhfh\3
4
Z fhe_dmeZ
ZahlZbao 1
2
)
3. P
Z
→
P
2
O
5
[
→
H
3
PO
4
\
→
K
3
PO
4
Z32
2
→
2P
2
O
5
[3
2
O
5
+3H
2
O
→
2H
3
PO
4
\.2++
3
PO
4
→
K
3
PO
4
+3H
2
O
JZ[hlZJZkq_luihhij_^_e_gbxfZkkh\hcbebh[t_fghc^heb
\uoh^Zijh^mdlZj_Zdpbbhll_hj_lbq_kdb\hafh`gh]h
b h[jZlgu_aZ^Zqb
1. NaOH+
252
]
3
63
]
HNO
→
o]
3
85
]
NaNO
+H
2
O
252
o
63 85
=
o
⋅
]
m
ij
(NaNO
3
)=
§
(NaNO
3
)
⋅
m
l_hj
(NaNO
3
)=0,9
⋅
]
Hl\_l
m
ij
(NaNO
3
]
2.
20
]
4
2 53,5
]
2NH Cl
⋅
+Ca(OH)
2
→
CaCl
2
+
oe
3
2 22,4
e
2NH
⋅
+2H
2
O
20
o
2 53,5 2 22,4
=
⋅⋅
o
⋅
⋅
⋅
e
V
ij
(NH
3
)=
§
(NH
3
)
⋅
V
l_hj
(NH
3
)=8,37
⋅
e
Hl\_l91+
3
e
3.
280
]
56
]
CaO
+H
2
O
→
o]
2
74
]
Ca(OH)
280
o
56 74
=
o
⋅
]
§
(Ca(OH)
2
))=
ij
l_hj
m(Ca(OH))
358
m (Ca(OH) ) 370
=
beb
Hl\_l
§
(Ca(OH)
2
))=96,76%
4.
20
]
4
2NH Cl
+
20
]
CaO
→
2NH
3
+CaCl
2
+H
2
O