a)
Gọi $n_{Fe} = a(mol) ; n_{Al} = b(mol) \Rightarrow 56a + 27b = 11(1)$
$Fe + 2HCl \to FeCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH :
$n_{HCl} = 2a + 3b = 0,4.2 = 0,8(2)$
Từ (1)(2) suy ra a = 0,1 ; b = 0,2
$\%m_{Fe} = \dfrac{0,1.56}{11}.100\% = 50,91\%$
$\%m_{Al} = 100\%- 50,91\% = 49,09\%$