Đáp án:

\(\begin{array}{l}
a)\\
\% {m_{Fe}} = 75,68\% \\
\% {m_{Al}} = 24,32\% \\
b)\\
{C_\% }FeC{l_2} = 11,87\% \\
{C_\% }AlC{l_3} = 8,32\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
hh:Fe(a\,mol),Al(b\,mol)\\
\left\{ \begin{array}{l}
a + 1,5b = 0,3\\
56a + 27b = 11,1
\end{array} \right.\\
 \Rightarrow a = 0,15;b = 0,1\\
\% {m_{Fe}} = \dfrac{{0,15 \times 56}}{{11,1}} \times 100\%  = 75,68\% \\
\% {m_{Al}} = 100 - 75,68 = 24,32\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,6\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,6 \times 36,5}}{{14,6\% }} = 150g\\
{m_{{\rm{dd}}spu}} = 11,1 + 150 - 0,3 \times 2 = 160,5g\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,15\,mol\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,1\,mol\\
{C_\% }FeC{l_2} = \dfrac{{0,15 \times 127}}{{160,5}} \times 100\%  = 11,87\% \\
{C_\% }AlC{l_3} = \dfrac{{0,1 \times 133,5}}{{160,5}} \times 100\%  = 8,32\% 
\end{array}\)