Em tham khảo nha:

\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{17,92}}{{22,4}} = 0,8\,mol\\
hh:Al(a\,mol),Fe(b\,mol)\\
\left\{ \begin{array}{l}
27a + 56b = 22\\
1,5a + b = 0,8
\end{array} \right.\\
 \Rightarrow a = 0,4;b = 0,2\\
\% {m_{Al}} = \dfrac{{0,4 \times 27}}{{22}} \times 100\%  = 49,09\% \\
\% {m_{Fe}} = 100 - 49,09 = 50,91\% \\
b)\\
{n_{KOH}} = 0,2 \times 0,15 = 0,03\,mol\\
KOH + HCl \to KCl + {H_2}O\\
{n_{HCl}}  \text{ dư }= {n_{KOH}} = 0,03\,mol\\
{n_{HCl}}  \text{ phản ứng }= 2{n_{{H_2}}} = 1,6\,mol\\
{V_{{\rm{dd}}HCl}} = \dfrac{{1,6 + 0,03}}{2} = 0,815l\\
c)\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,2\,mol\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,4\,mol\\
{C_M}FeC{l_2} = \dfrac{{0,2}}{{0,815}} = 0,245M\\
{C_M}AlC{l_3} = \dfrac{{0,4}}{{0,815}} = 0,491M
\end{array}\)