Gọi $n_{Fe} = a(mol) ; n_{Al} = b(mol)$
$\Rightarrow 56a + 27b = 11(1)$
$Fe + 2HCl \to FeCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$n_{H_2} = a + 1,5b = 8,96 : 22,4 = 0,4(2)$
Từ (1)(2) suy ra a = 0,1 ; b = 0,2
Vậy :
$\%m_{Fe} = \dfrac{0,1.56}{11}.100\% = 50,91\%$
$\%m_{Al} = 100\% -50,91\% = 49,09\%$