$n_{H_2}=5,6/22,4=0,25mol$
$PTHH :$
$2Al+6HCl\to 2AlCl_3+3H_2↑(1)$
$Fe+2HCl\to FeCl_2+H_2↑(2)$
$Gọi\ n_{Al}=a,n_{Fe}=b(a,b>0)$
$\text{Ta có :}$
$m_{hh}=27a+56b=8,1g$
$n_{H_2}=1,5a+b=0,25mol$
$\text{Ta có hpt :}$
$\left\{\begin{matrix}
27a+56b=8,1 & \\
1,5a+b=0,25 &
\end{matrix}\right.$
$⇔\left\{\begin{matrix}
a≈0,1 & \\
b≈0,09 &
\end{matrix}\right.$
$⇒\%m_{Al}=\dfrac{0,1.27}{8,1}.100\%=33,33\%$
$\%m_{Fe}=100\%-33,33\%=66,67\%$
$\text{b.Theo pt (1) và (2) :}$
$n_{HCl}=3a+2b=0,48mol$
$⇒m_{HCl}=0,48.36,5=17,52g$
$⇒C\%_{HCl}=\dfrac{17,52}{100}.100\%=17,52\%$