Bạn tham khảo:
$Fe+2HCl \to FeCl_2+H_2$
$2Al+6HCl \to 2AlCl_3+3H_2$
$n_{Fe}=a(mol)$
$n_{Al}=b(mol)$
$m_{hh}=56a+27b=11(1)$
$n_{H_2}=0,4(mol)$
$n_{H_2}=a+1,5b=0,4(2)$
$(1)(2)$
$a=0,1$
$b=0,2$
$\%m_{Fe}=\frac{0,1.56}{11}.100\%=50,9\%$
$\%m_{Al}=49,1\%$