Đáp án:
\(\begin{array}{l}
\% {m_{Fe}} = 50,91\% \\
\% {m_{Al}} = 49,09\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
2Al + 6{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
hh:Fe(a\,mol),Al(b\,mol)\\
{n_{S{O_2}}} = \dfrac{{10,08}}{{22,4}} = 0,45\,mol\\
\left\{ \begin{array}{l}
56a + 27b = 11\\
1,5a + 1,5b = 0,45
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,2\\
\% {m_{Fe}} = \dfrac{{0,1 \times 56}}{{11}} \times 100\% = 50,91\% \\
\% {m_{Al}} = 100 - 50,91 = 49,09\%
\end{array}\)