Em tham khảo nha:

\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
\left\{ \begin{array}{l}
56a + 27b = 8,3\\
a + 1,5b = 0,25
\end{array} \right.\\
 \Rightarrow a = b = 0,1\\
\% {m_{Fe}} = \dfrac{{0,1 \times 56}}{{8,3}} \times 100\%  = 67,47\% \\
\% {m_{Al}} = 100 - 67,47 = 32,53\% \\
b)\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,1\,mol\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,1\,mol\\
m = 0,1 \times 127 + 0,1 \times 133,5 = 26,05g\\
c)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,5\,mol\\
{n_{HCl}} = 0,5 + 0,5 \times 10\%  = 0,55\,mol\\
{V_{{\rm{dd}}HCl}} = \dfrac{{0,55}}{2} = 0,275l\\
d)\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
2Al + 3C{l_2} \to 2AlC{l_3}\\
{n_{FeC{l_3}}} = {n_{Fe}} = 0,1\,mol\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,1\,mol\\
m = 0,1 \times 162,5 + 0,1 \times 133,5 = 29,6g
\end{array}\)