Đáp án:
\(\begin{array}{l}
\% Fe = 67,47\% \\
\% Al = 32,53\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
2Al + 6{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
{n_{S{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
hh:Fe(a\,mol);Al(b\,mol)\\
\left\{ \begin{array}{l}
56a + 27b = 8,3\\
\dfrac{3}{2}a + \dfrac{3}{2}b = 0,3
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,1\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
\% Fe = \dfrac{{5,6}}{{8,3}} \times 100\% = 67,47\% \\
\% Al = 100 - 67,47 = 32,53\%
\end{array}\)