Ta có: $n_{Cl2}$ = 8,96/22,4 = 0,4 mol
PT: Mg + Cl2 ---> MgCl2
x x (mol)
2Fe + 3Cl2 ---> 2FeCl3
2y 3y (mol)
Ta có HPT: $\left \{ {{x+3y=0,4} \atop {24x+112y=13,6}} \right.$
<=> $\left \{ {{x=0,1} \atop {x=0,1}} \right.$
=> $m_{Mg}$ = 24.0,1 = 2,4 g
=> %$m_{Mg}$ = $\frac{2,4}{13,6}$ .100% ≈ 17,65%