$n_{NaOH}=0,4.0,5=0,2(mol)$
$2NaOH+H_2SO_4\to Na_2SO_4+2H_2O$
$\Rightarrow n_{H_2SO_4\text{dư}}=0,1(mol)$
$n_{Al_2O_3}=\dfrac{20,4}{102}=0,2(mol)$
$n_{MgO}=\dfrac{8}{40}=0,2(mol)$
$Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O$
$MgO+H_2SO_4\to MgSO_4+H_2O$
$\Rightarrow n_{H_2SO_4\text{pứ}}=0,2.3+0,2=0,8(mol)$
$\sum n_{H_2SO_4}=0,8+0,1=0,9(mol)$
$\to C\%_{H_2SO_4}=\dfrac{0,9.98.100}{122,5}=72\%$