$n_{NaOH}=0,5(mol)$
$NaOH+SO_2\to NaHSO_3$
$\to n_{SO_2}=0,5(mol)$
$\to n_{SO_4^{2-}\text{muối}}=\dfrac{n_e}{2}=n_{SO_2}=0,5(mol)$
Bảo toàn $S$:
$n_{H_2SO_4\text{pứ}}=n_{SO_4^{2-}\text{muối}}+n_{SO_2}=1(mol)$
$n_{H_2SO_4\text{bđ}}=\dfrac{140.98\%}{98}=1,4(mol)$
$\to$ dư $1,4-1=0,4$ mol $H_2SO_4$
$m_X=14,2+140-0,5.64=122,2g$
Vậy $C\%_{H_2SO_4\text{dư}}=\dfrac{0,4.98.100}{122,2}=32,08\%$