Bạn tham khảo:
$Fe+H_2SO_4 \to FeSO_4+H_2$
$Mg+H_2SO_4 \to MgSO_4+H_2$
$a/$
$n_{Fe}=a(mol)$
$n_{Mg}=b(mol)$
$m_{hh}=56a+24b=4,8(1)$
$n_{H_2}=a+b=0,12(2)$
$(1)(2)$
$a=b=0,06$
$\%m_{Fe}=\frac{0,06.56}{4,8}.100\%=70\%$
$\%m_{Al}=30%$
$b/$
$n_{H_2SO_4}=a+b=0,12(mol)$
$m_{ddH_2SO_4}=\frac{0,12.98.100}{10}=117,6(g)$
$n_{FeSO_4}=0,06(mol)$
$n_{MgSO_4}=0,06(mol)$
$C\%_{FeSO_4}=\frac{0,06.152}{4,8+117,6-0,12.2}.100\%=7,5\%$
$C\%_{MgSO_4}=\frac{0,06.120}{4,8+117,6-0,12.2}=5,9\%$
