a)
\(2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ b)\\ n_{Al} = a(mol) ; n_{Fe} = b(mol) \Rightarrow 27a + 56b =16,6(1)\\ n_{H_2} = 1,5a + b = \dfrac{11,2}{22,4} = 0,5(2)\\ (1)(2)\Rightarrow a = 0,2 ; b = 0,2\\ \Rightarrow m_{Al} = 0,2.27 = 5,4(gam)\ ;\ m_{Fe} = 0,2.56 = 11,2(gam)\)