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\(SO_3+H_2O\rightarrow H_2SO_4\\ H_2SO_4+Na_2SO_3\rightarrow Na_2SO_4+SO_2+H_2O\\ SO_2+Na_2O\rightarrow Na_2SO_3\\ Na_2SO_3+2HCl\rightarrow2NaCl+SO_2+H_2O\)
1, SO3+H2O->H2SO4
2, H2SO4+Na2SO3->Na2SO4+H2O+SO2
3, SO2+2NaOH->Na2SO3+H2O
4, Na2SO3+H2SO4->Na2SO4+H2O+SO2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1\right)3Fe+2O_2\xrightarrow[]{t^0}Fe_3O_4\\ \left(2\right)Fe_3O_4+8HCl\rightarrow2FeCl_3+FeCl_2+4H_2O\\ \left(3\right)FeCl_2+Ag_2SO_4\rightarrow FeSO_4+2AgCl\\ \left(4\right)FeSO_4+Ba\left(NO_3\right)_2\rightarrow Fe\left(NO_3\right)_2+BaSO_4\\ \left(5\right)Fe\left(NO_3\right)_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaNO_3\\ \left(6\right)Fe\left(OH\right)_2\xrightarrow[]{t^0}FeO+H_2O\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
Ta có: \(n_{Al_2O_3}=\dfrac{15,3}{102}=0,15\left(mol\right)\)
b, \(n_{AlCl_3}=2n_{Al_2O_3}=0,3\left(mol\right)\Rightarrow m_{AlCl_3}=0,3.133,5=40,05\left(g\right)\)
c, \(n_{HCl}=6n_{Al_2O_3}=0,9\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,9.36,5}{14,6\%}=225\left(g\right)\)
d, m dd sau pư = 15,3 + 225 = 240,3 (g)
\(\Rightarrow C\%_{AlCl_3}=\dfrac{40,05}{240,3}.100\%\approx16,67\%\)
e, nH2SO4 = 0,3.1 = 0,3 (mol)
PT: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
Xét tỉ lệ: \(\dfrac{0,15}{1}>\dfrac{0,3}{3}\), ta được Al2O3 dư.
→ Al2O3 không tan hết.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: nCuCl2 (4M) = 0,22.4 = 0,88 (mol)
Gọi: VCuCl2 (1,2M) = a (l)
⇒ nCuCl2 (1,2M) = 1,2a (mol)
\(\Rightarrow\dfrac{0,88+1,2a}{0,22+a}=2\)
⇒ a = 0,55 (l) = 550 (ml)
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A: MgO, CuO
B: MgCl2, CuCl2
C: Mg(OH)2, Cu(OH)2
PT: \(2Mg+O_2\underrightarrow{t^o}2MgO\)
\(2Cu+O_2\underrightarrow{t^o}2CuO\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)
\(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_{2\downarrow}\)
\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)