Đáp án:

\(\begin{array}{l}
a)\\
{m_{Al}} = 5,4g\\
{m_{Fe}} = 5,6g\\
b)\\
{V_{HCl}} = 132,72ml
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4mol\\
hh:Al(a\,mol),Fe(b\,mol)\\
\left\{ \begin{array}{l}
27a + 56b = 11\\
\frac{3}{2}a + b = 0,4
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,1\\
{m_{Al}} = 0,2 \times 27 = 5,4g\\
{m_{Fe}} = 11 - 5,4 = 5,6g\\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,8mol\\
{m_{HCl}} = 0,8 \times 36,5 = 29,2g\\
{m_{{\rm{dd}}HCl}} = \frac{{29,2 \times 100}}{{20}} = 146g\\
{V_{HCl}} = \dfrac{{146}}{{1,1}} = 132,72ml
\end{array}\)