$Mg + 2HCl --> Mg$$Cl_{2}$ + $H_{2}$
x 2x x x
$Fe + 2HCl --> Fe$$Cl_{2}$ + $H_{2}$
y 2y y y
$n_{H_{2}}$ = $\frac{6,72}{22,4}$ = 0,3mol
Ta có HPT
$\left \{ {{x + y=0,3} \atop {24x + 56y = 13,6}} \right.$
<=>$\left \{ {{x=0,1} \atop {y= 0,2}} \right.$
=>$m_{Mg}$ = 0,1.24=2,4g
$m_{Fe}$ = 0,2.56=11,2g
=>%Mg = $\frac{2,4}{13,6}$ .100 = 17,6%
%Fe = $\frac{11,2}{13,6}$ .100 = 82,4%