$n_{H_2}=0,896/22,4=0,04mol$
$PTHH :$
$2Al+6HCl\to 2AlCl_3+3H_2↑$
$Fe+2HCl\to FeCl_2+H_2↑$
Gọi $n_{Al}=a;n_{Fe}=b(a,b>0)$
Ta có :
$m_{hh}=27a+56b=1,1g$
$n_{H_2}=1,5a+b=0,04mol$
Ta có hpt :
$\left\{\begin{matrix}
27a+56b=1,1g & \\
1,5a+b=0,04 &
\end{matrix}\right.$
$⇔\left\{\begin{matrix}
a=0,02 & \\
b=0,01 &
\end{matrix}\right.$
$⇒m_{Al}=0,02.27=0,54g$
$m_{Fe}=1,1-0,54=0,56g$