1)
a)
\(2KMn{O_4}\xrightarrow{{{t^o}}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)
\(2{H_2} + {O_2}\xrightarrow{{{t^o}}}2{H_2}O\)
\(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)
\(PbO + {H_2}\xrightarrow{{{t^o}}}Pb + {H_2}O\)
b)
\(2{H_2} + {O_2}\xrightarrow{{{t^o}}}2{H_2}O\)
\(2{H_2}O\xrightarrow{{đp}}2{H_2} + {O_2}\)
\(2Cu + {O_2}\xrightarrow{{{t^o}}}2CuO\)
\(CuO + {H_2}\xrightarrow{{{t^o}}}Cu + {H_2}O\)
2)
Phản ứng xảy ra:
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
Ta có:
\({n_{Fe}} = \frac{{8,4}}{{56}} = 0,15{\text{ mol = }}{{\text{n}}_{{H_2}}}\)
\({n_{HCl}} = 2{n_{{H_2}}} = 0,3{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,15.22,4 = 3,36{\text{ lít}}\)
\({C_{M{\text{ HCl}}}} = \frac{{0,3}}{{0,2}} = 1,5M\)
Khi dùng 200 ml \(HCl\) 1M hòa tan lượng \(Fe\) trên
\({n_{HCl}} = 0,2.1 = 0,2{\text{ mol < 0}}{\text{,3 mol}}\)
Vậy axit thiếu
\( \to {n_{{H_2}}} = \frac{1}{2}{n_{HCl}} = 0,1{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)