Gọi số mol Al, Fe là a, b
=> 27a + 56b = 16,6
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
______a------------------------>1,5a
Fe + 2HCl --> FeCl2 + H2
b------------------------>b
=> mtăng = 16,6 - mH2 = 15,6
=> mH2 = 1 (g)
=> n\(n_{H_2}=\dfrac{1}{2}=0,5\left(mol\right)\)
=> 1,5a + b = 0,5
=> a = 0,2; b = 0,2
=> \(\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Fe}=0,2.56=11,2\left(g\right)\end{matrix}\right.\)