Đáp án:

$\% {m_{Fe}} = 58,03\% $

Giải thích các bước giải:

 Sơ đồ phản ứng:

\[\left\{ \begin{gathered} Al \hfill \\ Fe \hfill \\ \end{gathered} \right.\xrightarrow{{ + HCl}}\left\{ \begin{gathered} AlC{l_3},FeC{l_2}\xrightarrow{{ + NaOH}}\left\{ \begin{gathered} Na\,Al{O_2},\,\,NaCl\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ \downarrow Fe{(OH)_2}\xrightarrow{{{t^o}}}F{e_2}{O_3} + {H_2}O \hfill \\ \end{gathered} \right. \hfill \\ {H_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ \end{gathered} \right.\]

Chất rắn: $F{e_2}{O_3}:\,\,8\,\,gam \to {n_{F{e_2}{O_3}}} = 0,05\,\,mol$

Bảo toàn nguyên tố $Fe$: ${n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,05.2 = 0,1\,\,mol$

$\begin{gathered} \to {m_{Fe}} = 0,1.56 = 5,6\,\,gam\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ \to \% {m_{Fe}} = \frac{{5,6}}{{9,65}}.100\% = 58,03\% \hfill \\ \end{gathered} $