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a.Fe + 2HCl-----> FeCl2 + H2
FeCl2 + 2NaOH----->Fe(OH)2 + 2NaCl
Fe(OH)2+ H2SO4---->FeSO4 + 2H2O
FeSO4 + Ba(NO3)2----->Fe(NO3)2 + BaSO4
b.4Al + 3O2---t*-->2Al2O3
Al2O3 + 6HCl------>2AlCl3 + 3H2O
AlCl3 + 3NaOH----->Al(OH)3 + 3NaCl
2Al(OH)3----t*---> Al2O3 + 3H2O
c.2Cu+ O2--t*->2CuO
CuO + 2HCl---->CuCl2 + H2O
CuCl2 + 2NaOH----->Cu(OH)2 + 2NaCl
Cu(OH)2---t*---> CuO + H2O
d.2Cu+ O2--t*->2CuO
CuO + 2HCl---->CuCl2 + H2O
CuCl2 + 2AgNO3-----> Cu(NO3)2 + 2AgCl
Cu(NO3)2 + 2NaOH-----> Cu(OH)2 + 2NaNO3
Cu(OH)2---t*--->CuO + H2O
CuO + H2---t*--> Cu + H2O
1.
2Cu + O2 -(to) → 2 CuO
CuO + H2SO4 → CuSO4 + H2O
CuSO4 + BaCl2 → CuCl2 + BaSO4↓
CuCl2 + 2AgNO3 → Cu(NO3)2 + 2AgCl↓
Cu(NO3)2 + NaOH → Cu(OH)2↓ + 2NaNO3
Cu(OH)2 → CuO + H2O
CuO + CO → Cu + CO2
2.
4Al + 3O2 → 2Al2O3
Al2O3 + 6HCl → 2AlCl3 + 3H2↑
AlCl3 + 3NaOH(vừa đủ) → Al(OH)3 + 3NaCl
2Al(OH)3 → Al2O3 + 3H2O
2Al2O3-(đpnc) → 4Al + 3O2↑
2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2↑
A: Cu(OH)2 \(\underrightarrow{t^o}\) CuO + H2O
CuO + 2HCl → CuCl2 + H2O
CuCl2 +2 NaOH → Cu(OH)2 + NaCl
Cu(OH)2 + H2SO4 → CuSO4 + 2H2O
B: Mg(OH)2 \(\underrightarrow{t^o}\) 2MgO + H2O
MgO + 2HCl → MgCl2 + H2O
MgCl2 + 2NaOH → Mg(OH)2 + NaCl
Mg(OH)2 + H2SO4 → MgSO4 + 2H2O
1)
\(2Fe\left(OH\right)_3\) -> \(Fe_2O_3\) + \(3H_2O\)
\(Fe_2O_3\) + \(3CO\) -> \(2Fe\) + \(3CO_2\)
\(2Fe+3Cl_2\) -> \(2FeCl_3\)
\(FeCl_3+3AgNO_3\) -> \(3AgCl\) +\(Fe\left(NO_3\right)_3\)
2)
\(4FeS+7O_2\) -> \(2Fe_2O_3+4SO_2\)
\(Fe_2O_3+3CO\) -> \(2Fe+3CO_2\)
\(2Fe+3Cl\) ->\(2FeCl_3\)
\(2FeCl_3+3Ba\left(OH\right)_2\)->\(3BaCl_2+2Fe\left(OH\right)_3\)
3)
\(2Fe+3Cl_2\) -> \(2FeCl_3\)
\(2FeCl_3+3Ba\left(OH\right)_2\) -> \(3BaCl_2+2Fe\left(OH\right)_3\)
\(2Fe\left(OH\right)_3\) -> \(Fe_2O_3+3H_2O\)
\(Fe_2O_3+3CO\) -> \(2Fe+3CO_2\)
4)
\(2Cu+O_2\) -> \(2CuO\)
\(CuO+2HCl\) -> \(CuCl_2+2H_2O\)
\(CuCl_2+2NaOH\) -> \(2NaCl_2+Cu\left(OH\right)_2\)
\(Cu\left(OH\right)_2+H_2SO_4\) -> \(2H_2O+CuSO_4\)
CaCO3+HCl-->CaCl2+CO2+H2O
CaCl2+Na2CO3-->CaCO3+NaCl
CaCO3-->CaO+CO2
CaO+H2O-->Ca(OH)2
Ca(OH)2+HNO3-->Ca(NO3)2+H2O
E)
Cu+O2-->CuO
CuO+H2-->Cu+H2O
CuO+..->Cu(Oh)2
Cu(Oh)2+HCl-->CuCl2+H2o
G)Na2SO3+CaCL2-->CaSO3+NaCl
S+O2-->SO2
SO2+H2O-->H2SO3
H2SO3+Ca--->CaSO3+H2
CaSO3+HCl-->CaCL2+SO2+H2O
S8+O2-->SO3
SO3+H2O-->H2SO4
H2SO4+Fe2O3-->Fe2(SO4)3+H2
(1) Cu(OH)2 + MgSO4 → CuSO4 + Mg(OH)2
(2) CuSO4 + BaCl2 → CuCl2 + BaSO4
(3) CuCl2 + 2 AgNO3 → Cu(NO3)2 + 2 AgCl
(4) 3Cu(NO3)2 + 2Al → 2Al(NO3)3 + 3Cu
(5) Al(NO3)3 + 3 NaOH → Al(OH)3 + 3 NaNO3
(6) Al(OH)3 → Al2O3 + H2O (nhiệt phân hủy)
Sai sót mong bỏ qua 
\(CuSO_4\underrightarrow{1}CuCl_2\underrightarrow{2}Cu\left(OH\right)_2\underrightarrow{3}CuO\underrightarrow{4}\\ \left(1\right)CuSO_4+BaCl_2\rightarrow CuCl_2+BaSO_4\\ \left(2\right)CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\\ \left(3\right)Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\\ \left(4\right)CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
\(\text{BaCl2 + CuSO4 → CuCl2 + BaSO4}\)
\(\text{2NaOH + CuCl2 → Cu(OH)2 + 2NaCl}\)
\(\text{Cu(OH)2 → CuO + H2O}\)
\(\text{2CuO → 2Cu + O2}\)
1/ Cao + H2O------> Ca (OH)2
2/ Ca (OH)2 + Fe(NO3)2 -----> Ca (NO3)2+Fe(OH)2
3/ Ca(NO3)2 + H2SO4---> CaSO4+2HNO3
4/ CuCl2 + 2NaOH -----> Cu(OH)2+2NaCl
5/ Cu(OH)2 + Na2SO4-----> CuSO4+2NaOH
6/ CuSO4 +Ba(NO3)2 ------> Cu (NO3)2+BaSO4
7/ CuO +2HCl----> CuCl2+H2O
1. CaO + H2O = Ca(OH)2
2. Ca(OH)2 + 2HNO3 = Ca(NO3)2 + 2H2O
3. Không xảy ra pứ này
4. CuCl2 +2NaOH = 2NaCl + Cu(OH)2
5. Cu(OH)2 + H2SO4 = CuSO4 + 2H2O
6. CuSO4 + Ba(NO3)2 = Cu(NO3)2 + BaSO4
7. CuO + 2HCl = CuCl2 +H2O
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Bài của Duong Le
Thứ nhất: pt số 3 sẽ k xảy ra pứ
Thứ hai: pt số 5 sai
1) 2Cu + O2 -to-➢ 2CuO
CuO + 2HCl → CuCl2 + H2O
CuCl2 + 2NaOH → Cu(OH)2 + 2NaCl
2) S + H2 -to-➢ H2S
3) S + O2 -to-➢ SO2
2SO2 + O2 -to-➢ 2SO3
SO3 + H2O → H2SO4
H2SO4 + 2KOH → K2SO4 + 2H2O
4) S + Fe -to-➢ FeS
a)\(2Cu+O_2\underrightarrow{t^o}2CuO\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(CuCl_2+H_2SO_4\rightarrow CuSO_4+2HCl\)
\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
\(CuO+H_2\rightarrow Cu+H_2O\)
\(Cu+Cl_2\rightarrow CuCl_2\)
\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)
b)\(2Al+3Cl_2\rightarrow2AlCl_3\)
\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\)
\(2Al\left(OH\right)_3\underrightarrow{t^o}Al_2O_3+3H_2O\)
\(Al_2O_3+6HNO_3\rightarrow2Al\left(NO_3\right)_3+3H_2O\)
\(2Al\left(NO_3\right)_3+3Mg\rightarrow3Mg\left(NO_3\right)_2+2Al\)
a,
Cu+ \(\frac{1}{2}\)O2\(\underrightarrow{^{to}}\) CuO
CuO+ 2HCl \(\rightarrow\) CuCl2+ H2O
Phương trình 3 ko tồn tại
CuSO4+ 2NaOH \(\rightarrow\) Cu(OH)2+ Na2SO4
Cu(OH)2\(\underrightarrow{^{to}}\) CuO+ H2O
CuO+ CO \(\underrightarrow{^{to}}\)Cu+ CO2
Cu+ Cl2 \(\underrightarrow{^{to}}\)CuCl2
CuCl2+ 2NaOH\(\rightarrow\) Cu(OH)2+ 2NaCl
b,
Al+ 3/2Cl2 \(\underrightarrow{^{to}}\)AlCl3
AlCl3+ 3NH3+ 3H2O\(\rightarrow\)Al(OH)3+ 3NaCl
2Al(OH)3 \(\underrightarrow{^{to}}\) Al2O3+ 3H2O
Al2O3+ 6HNO3 \(\rightarrow\)2Al(NO3)3+ 3H2O
2Al(NO3)3+ 3Mg\(\rightarrow\) 3Mg(NO3)2+ 2Al
$Cu(OH)_2 \xrightarrow{t^o} CuO + H_2O$
$CuO + 2HCl \to CuCl_2 + H_2o$
$CuCl_2 + 2AgNO_3 \to Cu(NO_3)_2 + 2AgCl$
$Cu(NO_3)_2 + 2KOH \to Cu(OH)_2 + 2KNO_3$
giúp e câu này với ah
Câu 11: Nhiệt phân hoàn toàn 30,3 gam hỗn hợp 2 bazơ không tan Cu(OH)2 và Fe(OH)3, sau pư thu được 24 gam 2 oxit.
a) Tính % số mol mỗi bazơ trong hỗn hợp đầu?
Cho 2 oxit trên pư vừa đủ với 200 ml dung dịch HCl xM. Tìm x?