Đáp án:

Giải thích các bước giải:

1)  

a. (1) 2KMnO4 → K2MnO4 + MnO2 + O2

(2) 2H2 + O2 → 2H2O

(3) 2H2O → 2H2 + O2

(4) H2 + PbO → Pb + H2O

b.(1) 2H2 + O2 → 2H2O

(2) 2H2O → 2H2 + O2

(3) 2Cu + O2 → 2CuO

(4) CuO + H2 → Cu + H2O

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

\( \to {n_{{H_2}}} = \frac{1}{2}{n_{HCl}} = 0,1{\text{ mol}}\)

\( \to {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)