$n_{CuO}=\dfrac{1,6}{80}=0,02mol \\m_{H_2SO_4}=20\%.100=20g \\⇒n_{H_2SO_4}=\dfrac{20}{98}≈0,2mol \\PTHH :$
$CuO + H_2SO_4\to CuSO_4+H_2O$
$\text{Theo pt : 1 mol 1 mol}$
$\text{Theo đbài : 0,02 mol 0,2 mol}$
Tỷ lệ : $\dfrac{0,02}{1}<\dfrac{0,1}{1}$
$\text{⇒Sau pư H2SO4 dư}$
$\text{Theo pt :}$
$n_{H_2SO_4\ pư}=n_{CuO}=0,02mol \\⇔n_{H_2SO_4\ dư}=0,2-0,02=0,18mol \\⇒m_{H_2SO_4\ dư}=0,18.98=17,64g \\n_{CuSO_4}=n_{CuO}=0,02mol \\⇒m_{CuSO_4}=0,02.160=3,2g \\m_{dd\ spu}=1,6+100=101,6g \\⇒C\%_{H_2SO_4\ dư}=\dfrac{17,64}{101,6}.100\%=17,36\% \\C\%_{CuSO_4}=\dfrac{3,2}{101,6}.100\%=3,15\%$
