Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Fe}} = 67,47\% \\
\% {m_{Al}} = 32,53\% \\
b)\\
a = 2,5
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
hh:Fe(a\,mol),Al(b\,mol)\\
\left\{ \begin{array}{l}
56a + 27b = 8,3\\
a + 1,5b = 0,25
\end{array} \right.\\
\Rightarrow a = b = 0,1\,mol\\
\% {m_{Fe}} = \dfrac{{0,1 \times 56}}{{8,3}} \times 100\% = 67,47\% \\
\% {m_{Al}} = 100 - 67,47 = 32,53\% \\
b)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,25\,mol\\
{C_M}{H_2}S{O_4} = \dfrac{{0,25}}{{0,1}} = 2,5M \Rightarrow a = 2,5
\end{array}\)
