Em tham khảo nha:

\(\begin{array}{l}
1)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
hh:Al(a\,mol),Fe(b\,mol)\\
\left\{ \begin{array}{l}
27a + 56b = 8,3\\
1,5a + b = 0,25
\end{array} \right.\\
 \Rightarrow a = b = 0,1\\
\% {m_{Al}} = \dfrac{{0,1 \times 27}}{{8,3}} \times 100\%  = 32,53\% \\
\% {m_{Fe}} = 100 - 32,53 = 67,47\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,5\,mol\\
{V_{{\rm{dd}}HCl}} = \dfrac{{0,5}}{2} + 0,01 = 0,26l\\
2)\\
C{l_2} + 2NaBr \to 2NaCl + B{r_2}\\
hh:NaCl(a\,mol),NaBr(b\,mol)\\
\left\{ \begin{array}{l}
58,5a + 103b = 50\\
58,5a + 58,5b = 41,1
\end{array} \right.\\
 \Rightarrow a = \dfrac{{98}}{{195}};b = 0,2\\
{m_{NaCl}} = 50 - 0,2 \times 103 = 29,4g\\
\% {m_{NaCl}} = \dfrac{{29,4}}{{50}} \times 100\%  = 58,8\% 
\end{array}\)