a) PTHH : Fe + 2HCl -> FeCl2 + H2
a -> a
Mg + 2HCl -> MgCl2 + H2
b -> b
b) Đặt nFe =a (mol) nMg = b(mol)
=> $\left \{ {{56a + 24b = 16} \atop {a + b = \frac{8,96}{22,4} }} \right.$ ⇔ $\left \{ {{a = 0,2} \atop {b=0,2}} \right.$
=> %mFe = 70%
=> %mMg = 30%