Đáp án:

\({m_{F{\text{e}}}} = 3,36{\text{ gam}}\)

\({m_{{O_2}}} = 1,28{\text{ gam}}\)

\( {m_{KCl{O_3}}} = 3,63{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(3F{\text{e}} + 2{{\text{O}}_2}\xrightarrow{{{t^o}}}F{{\text{e}}_3}{O_4}\)

Ta có:

\({n_{F{{\text{e}}_3}{O_4}}} = \frac{{4,64}}{{56.3 + 16.4}} = 0,02{\text{ mol}}\)

\( \to {n_{F{\text{e}}}} = 3{n_{F{{\text{e}}_3}{O_4}}} = 0,02.3 = 0,06{\text{ mol}}\)

\({n_{{O_2}}} = \frac{2}{3}{n_{F{\text{e}}}} = 0,04{\text{ mol}}\)

\( \to {m_{F{\text{e}}}} = 0,06.56 = 3,36{\text{ gam}}\)

\({m_{{O_2}}} = 0,04.32 = 1,28{\text{ gam}}\)

Điều  chế \(O_2\)

\(2KCl{O_3}\xrightarrow{{{t^o}}}2KCl + 3{{\text{O}}_2}\)

\( \to {n_{{O_2}{\text{ lt}}}} = \frac{{0,04}}{{100\%  - 10\% }} = \frac{2}{{45}} \to {n_{KCl{O_3}}} = \frac{2}{3}.\frac{2}{{45}} = \frac{4}{{135}}\)

\( \to {m_{KCl{O_3}}} = \frac{4}{{135}}.(39 + 35,5 + 16.3) = 3,63{\text{ gam}}\)