Đáp án:
84%
Giải thích các bước giải:
\(\begin{array}{l}
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
FeO + {H_2}S{O_4} \to FeS{O_4} + {H_2}O\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
F{e_3}{O_4} + 4{H_2}O \to F{e_2}{(S{O_4})_3} + FeS{O_4} + 4{H_2}O\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,15\,mol\\
\% {m_{Fe}} = \dfrac{{0,15 \times 56}}{{10}} \times 100\% = 84\%
\end{array}\)