Em tham khảo nha:

\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
hh:Al(a\,mol),Fe(b\,mol)\\
\left\{ \begin{array}{l}
27a + 56b = 8,3\\
1,5a + b = 0,25
\end{array} \right.\\
 \Rightarrow a = b = 0,1\\
{m_{Al}} = 0,1 \times 27 = 2,7g\\
{m_{Fe}} = 8,3 - 2,7 = 5,6g\\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,5\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,5 \times 36,5}}{{9,8\% }} = 186,22g\\
{m_{{\rm{dd}}spu}} = 8,3 + 186,22 - 0,25 \times 2 = 194,02g\\
{C_\% }AlC{l_3} = \dfrac{{0,1 \times 133,5}}{{194,02}} \times 100\%  = 6,88\% \\
{C_\% }FeC{l_2} = \dfrac{{0,1 \times 127}}{{194,02}} \times 100\%  = 6,55\% 
\end{array}\)