$m_{H_2SO_4}=19,6.20\%=3,92g \\⇒n_{H_2SO_4}=\dfrac{3,92}{98}=0,04mol \\n_{CuO}=\dfrac{16}{80}=0,2mol \\PTHH :$
$CuO + H_2SO_4\to CuSO_4+H_2↑$
$\text{Theo pt : 1 mol 1 mol}$
$\text{Theo đbài : 0,2 mol 0,04 mol}$
$\text{Tỷ lệ :}$ $\dfrac{0,2}{2}>\dfrac{0,04}{1}$
$\text{⇒Sau phản ứng CuO dư }$
$\text{Theo pt :}$
$n_{CuSO_4}=n_{CuO\ pư}=n_{H_2SO_4}=0,04mol \\⇒m_{CuSO_4}=0,04.160=6,4g \\m_{dd\ spu}=0,04.80+19,6=22,8g \\C\%_{CuSO_4}=\dfrac{6,4}{22,8}.100\%=28,07\%$
