$n_{H_2SO_4}=0,5.0,6=0,3(mol)$
$n_{HCl}=0,5.0,4=0,2(mol)$
Bảo toàn H:
$2n_{H_2}=2n_{H_2SO_4}+n_{HCl}$
$\Rightarrow n_{H_2}=0,4(mol)$
Gọi x, y là số mol $Al$, $Mg$
$\Rightarrow 27x+24y=8,25$ (1)
Bảo toàn e:
$3n_{Al}+2n_{Mg}=2n_{H_2}$
$\Rightarrow 3x+2y=0,8$ (2)
$(1)(2)\Rightarrow x=0,15; y=0,175$
$\%m_{Al}=\dfrac{0,15.27.100}{8,25}=49\%$
$\%m_{Mg}=51\%$