$\text{$m_{H_{2}SO_{4}}$ = 420*40% = 168 g ⇒ $n_{H_{2}SO_{4}}$ = 168/98 = 12/7 mol}$
$\text{$n_{CuO}$ = m/80 mol}$
$CuO_{ }$ + $H_{2}SO_{4}$ → $CuSO_{4}$ + $H_{2}O$
$\text{$n_{H_{2}SO_{4}}$ phản ứng = $n_{CuO}$ = m/80 mol}$
$\text{$n_{H_{2}SO_{4}}$ dư = $n_{H_{2}SO_{4}}$ - $n_{H_{2}SO_{4}}$ phản ứng = 12/7 - m/80 mol}$
$\text{Ta có: ($m_{H_{2}SO_{4}}$ dư/$m_{dd}$)*100% = 14%}$
$\text{⇔ [(12/7 - m/80)*98/m+420]*100% = 14% ⇔ m = 80}$
$\text{$n_{CuSO_{4}}$ = m/80 = 80/80 = 1 mol}$
$\text{⇒ x = C%$_{CuSO_{4}}$ = (160/80+420)*100% = 32%}$