$n_{H_2}=\dfrac{8,96}{22,4}=0,4mol$
$2Al+6HCl \rightarrow 2AlCl_3 + 3H_2$
x x/2 3/2x
$Fe+2HCl \rightarrow FeCl_2+H_2$
y. y y
hpt:
$\Leftrightarrow \left\{ \begin{array}{l}27x+56y=11\\\frac{3}{2}x+y=0,4\end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l}x=0,2\\x=0,1\end{array} \right.$
a. %$m_{Al}=\dfrac{0,2.27}{11}.100$%$=49,09$%
%$m_{Fe}=100$%$-49,09$%$=50,91$%
b. $n_{HCl}=3x+2y=0,8mol$
$V_{HCl}=\dfrac{0,8}{0,8}=1l$