Đáp án:

\({m_{Fe}} = 14{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)

\(FeS{O_4} + 2KOH\xrightarrow{{}}Fe{(OH)_2} + {K_2}S{O_4}\)

\(2Fe{(OH)_2} + \frac{1}{2}{O_2} + {H_2}O\xrightarrow{{}}2Fe{(OH)_3}\)

\(2Fe{(OH)_3}\xrightarrow{{{t^o}}}F{e_2}{O_3} + 3{H_2}O\)

\({n_{F{e_2}{O_3}}} = \frac{{20}}{{160}} = 0,125{\text{ mol}}\)

\( \to {n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,25{\text{ mol}}\)

\( \to {m_{Fe}} = 0,25.56 = 14{\text{ gam}}\)

\({n_{Fe}} = {n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,25{\text{ mol}}\)

\(V = {V_{{H_2}}} = 0,25.22,4 = 5,6{\text{ lít}}\)

\({V_{dd{{\text{H}}_2}S{O_4}}} = \frac{{0,25}}{1} = 0,25{\text{ lít}}\)

\({n_{KOH}} = 2{n_{FeS{O_4}}} = 0,25.2 = 0,5{\text{ mol}}\)

\( \to {C_{M{\text{ KOH}}}} = \frac{{0,5}}{{0,25}} = 2M\)