1)

Phản ứng xảy ra:

\(2KCl{O_3}\xrightarrow{{{t^o}}}2KCl + 3{O_2}\)

Ta có: 

\({n_{{O_2}}} = \frac{{5,6}}{{22,4}} = 0,25{\text{ mol}} \to {{\text{n}}_{{O_2}{\text{ lt}}}} = \frac{{0,25}}{{100\%  - 1,2\% }} = \frac{{125}}{{494}}{\text{ mol}} \to {{\text{n}}_{KCl{O_3}}} = \frac{2}{3}{n_{{O_2}}} = \frac{{125}}{{494}}.\frac{2}{3} = \frac{{125}}{{741}} \to {m_{KCl{O_3}}} = \frac{{125}}{{741}}.(39 + 35,5 + 16.3) = 20,665{\text{ gam}}\)

2)

Phản ứng xảy ra:

\(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)

\(N{a_2}O + {H_2}O\xrightarrow{{}}2NaOH\)

Khí B là hidro

Ta có: \({n_{Na}} = \frac{{0,23}}{{23}} = 0,01{\text{ mol}} \to {{\text{n}}_{{H_2}}} = \frac{1}{2}{n_{Na}} = 0,005{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,005.22,4 = 0,112{\text{ lít}}\)

 Dẫn khí B qua CuO

Phản ứng xảy ra:

\(CuO + {H_2}\xrightarrow{{{t^o}}}Cu + {H_2}O\)

Ta có: \({n_{CuO}} = \frac{{3,2}}{{64 + 16}} = 0,04{\text{ mol > }}{{\text{n}}_{{H_2}}}\) do đó CuO dư, hiệu suất tính theo hidro.

\( \to {n_{Cu}} = {n_{{H_2}}}.70\%  = 0,005.70\%  = 0,0035{\text{ mol}} \to {{\text{m}}_{Cu}} = 0,0035.64 = 0,224{\text{ gam}}\)