Bạn tham khảo:
$a/$
$Mg+FeCl_2 \to MgCl_2+Fe$
$b/$
$n_{Mg}=a(mol)$
$m_{tăng}=56a-24a=36,8-24$
$ \to a=0,4(mol)$
$m_{Mg}=0,4.24=9,6(g)$
$m_{Fe}=0,4.56=22,4(g)$
$c/$
$\%m_{Mg}=\frac{24-9,6}{36,8}.100\%=39,13\%$
$\%m_{Fe}=60,87\%$
$d/$
$n_{Mg(dư)}=\frac{24-9,6}{24}=0,6(mol)$
$Mg+2HCl \to MgCl_2+H_2$
$Fe+2HCl \to FeCl_2+H_2$
$n_{H_2}=0,4+0,6=1(mol)$
$V=1.22,4=22,4(l)$
$e/$
$Mg+4HNO_3 \to Mg(NO_3)_2+2NO_2+2H_2O$
$Fe+6HNO_3 \to Fe(NO_3)_3+3NO_2+3H_2O$
$n_{NO_2}=1,2+1,2=2,4(mol)$
$V=2,4.22,4=53,76(l)$