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![](https://rs.olm.vn/images/avt/0.png?1311)
Vậy D là \(H_2\)
Từ trái qua
\(KClO_3\rightarrow O_2\rightarrow ZnO\rightarrow H^2\rightarrow Fe\rightarrow Fe_2\left(SO_4\right)_3\)
1) \(2KClO_3\rightarrow2KCl+3O_2\)
2) \(O_2+2Zn\rightarrow2ZnO\)
3) \(ZnO+H_2\rightarrow Zn+H_2O\)
4) \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
5) \(3H_2+Fe_2O_3\rightarrow2Fe+3H_2O\)
6) \(2Fe+6H_2SO_{4\left(đ,t^o\right)}\rightarrow Fe_2\left(SO_4\right)_3+6H_2O+3SO_2\uparrow\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B là \(O_2\)
C là \(H_2O\)
D là \(KOH\)
E là \(Cu\left(OH\right)_2\)
Từ trái qua
1) \(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)
2) \(O_2+2H_2\underrightarrow{t^0}2H_2O\)
3) \(H_2O+K_2O\rightarrow2KOH\)
4) \(2KOH+Cu\left(NO_3\right)_2\rightarrow2KNO_3+Cu\left(OH\right)_2\)
5) \(Cu\left(OH\right)_2\underrightarrow{t^0}CuO+H_2O\)
\(B:O_2\\ C:CuO\\ D:CuCl_2\\ E:Cu\left(OH\right)_2\)
PTHH:
\(KMnO_4\rightarrow B:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
\(B\rightarrow C:O_2+2Cu\underrightarrow{t^o}2CuO\)
\(C\rightarrow D:CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(D\rightarrow E:CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)
\(E\rightarrow D:Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) SO2 + Na2O -> Na2SO3
Na2SO3 + H2SO4 -> Na2SO4 + SO2 + H2O
Na2SO4 + Ba(OH)2 -> BaSO4 \(\downarrow\)+ 2NaOH
2NaOH + CO2 -> Na2CO3 + H2O
b) CaO + CO2 -> CaCO3
CaCO3 \(\rightarrow^{t^o}\) CaO + CO2
CaO + H2O -> Ca ( OH)2
Ca ( OH)2 + CO2 -> CaCO3 + H2O
CaCO3 + H2SO4 -> CaSO4 + CO2 + H2O
c) 2Fe + 3Cl2 \(\rightarrow^{t^o}\) 2FeCl3
FeCl3 + 3NaOH -> Fe(OH)3 \(\downarrow\) + 3NaCl
2Fe(OH)3 \(\rightarrow^{t^o}\) Fe2O3 + 3H2O
Fe2O3 + 3H2SO4 -> Fe2(SO4)3 + 3H2O
Fe2(SO4)3 + 3BaCl2 -> 2FeCl3 + 3BaSO4\(\downarrow\)
d) Fe + 2HCl -> FeCl2 + H2
FeCl2 + 2AgNO3 -> 2AgCl\(\downarrow\) + Fe(NO3)2
Fe(NO3)2 + 2NaOH -> Fe(OH)2\(\downarrow\) + 2NaNO3
Fe(OH)2 + H2SO4 -> FeSO4 + 2H2O
![](https://rs.olm.vn/images/avt/0.png?1311)
bên chỗ hocmai, bn Nguyễn Thị Ngọc Bảo đã giúp bn rồi nên có lẽ bn k cần nx nhỉ
\(A\rightarrow B\rightarrow C\rightarrow D\rightarrow Cu\) Ta có thứ tự sau: \(CuO\rightarrow Cu\left(OH\right)_2\rightarrow CuCl_2\rightarrow Cu\left(NO_3\right)_2\rightarrow Cu\)PTHH: \(CuO+H_2O\rightarrow Cu\left(OH\right)_2\) \(Cu\left(OH\right)_2+2HCl\rightarrow CuCl_2+2H_2O\) \(CuCl_2+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2AgCl\downarrow\) \(Cu\left(NO_3\right)_2+Fe\rightarrow Fe\left(NO_3\right)_2+Cu\downarrow\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
\(2AlCl_3+3Ba\left(OH\right)_2\rightarrow2Al\left(OH\right)_3\downarrow+3BaCl_2\)
\(2Al\left(OH\right)_3\underrightarrow{t^o}Al_2O_3+3H_2O\)
\(2Al+3S\underrightarrow{t^o}Al_2S_3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 4FeS2+11O2\(\overset{t^0}{\rightarrow}\)2Fe2O3+8SO2
2SO2+O2\(\overset{V_2O_5,t^0}{\rightarrow}2SO_3\)
SO3+H2O\(\rightarrow\)H2SO4
Cu+ 2H2SO4đ\(\overset{t^0}{\rightarrow}\)CuSO4+SO2+H2O
b)
AlCl3+3NaOH\(\rightarrow\)Al(OH)3+3NaCl
2Al(OH)3\(\overset{t^0}{\rightarrow}Al_2O_3+3H_2O\)
Al2O3+3H2SO4\(\rightarrow\)Al2(SO4)3+3H2O
Al2(SO4)3+3BaCl2\(\rightarrow\)3BaSO4+2AlCl3
![](https://rs.olm.vn/images/avt/0.png?1311)
1, (1) 4Al+ 3O2--(to)---> 2Al2O3
(2) Al2O3+6HCl---> 2AlCl3+ 3H2O
(3) AlCl3+3NaOH---->Al(OH)3+ 3NaCl
(4) 2Al(OH)3---(to)---> Al2O3+ 3H2O
(5)2Al2O3---(to+ đpnc)--> 4Al+ 3O2
(6) 2Al+ 3Cl2---> 2AlCl3
2, (1)Fe+ CuSO4-->Cu+ FeSO4
(2) FeSO4+ 2NaOH--> Fe(OH)2+ Na2SO4
(3) Fe(OH)2 +2HCl---> FeCl2+ 2H2O
Bài cuối tự làm nhé!
1.
a;
2H2O -> 2H2 + O2
H2 + CuO -> Cu + H2O
Cu +4 HNO3 -> Cu(NO3)2 + 2NO2 + 2H2O
2Cu(NO3)2 -> 2CuO + 4NO2 + O2
CuO + CO -> Cu + CO2
Ca + 2H2O -> Ca(OH)2 + H2
Ca(OH)2 + CO2 -> CaCO3 + H2O
nCaCO3=0,05(mol)
Ta có:
nCaCO3=nCa=,05(mol)
mCa=40.0,05=2(g)