$m_{H_2SO_4} = 420.40\% = 168(gam)$
Gọi $n_{CuO} = a(mol)$
$CuO + H_2SO_4 \to CuSO_4 + H_2O$
Sau phản ứng :
$m_{H_2SO_4\ dư} = 168 - 98a(gam)$
$m_{dd} = 80a + 420(gam)$
Ta có :
$C\%_{H_2SO_4\ dư} = \dfrac{168 - 98a}{80a + 420}.100\% = 14\%$
$\Rightarrow a =1(mol)$
$\Rightarrow a = 1.80 = 80(gam)$
$C\%_{CuSO_4} = \dfrac{1.160}{80.1+420}.100\% = 32\%$