$n_{CuO}=\dfrac{8}{80}=0,1mol \\m_{H_2SO_4}=100.20\%=20g \\⇒n_{H_2SO_4}=\dfrac{20}{98}=0,2mol \\PTHH :$
$CuO + H_2SO_4\to CuSO_4+H_2O$
$\text{Theo pt : 1 mol 1 mol}$
$\text{Theo đbài : 0,1 mol 0,2 mol}$
$\text{Tỷ lệ :}$ $\dfrac{0,1}{1}<\dfrac{0,2}{1}$
$\text{⇒Sau phản ứng H2SO4 dư}$
T$\text{heo pt :}$
$n_{H_2SO_4\ pu}=n_{CuO}=0,1mol \\⇒m_{H_2SO_4\ dư}=20-0,1.98=10,2g \\n_{CuSO_4}=n_{CuO}=0,1mol \\⇒m_{CuSO_4}=0,1.160=16g \\m_{dd\ spu}=8+100=108g \\⇒C\%_{H_2SO_4\ dư}=\dfrac{10,2}{108}.100\%=9,44\% \\C\%_{CuSO_4}=\dfrac{16}{108}.100\%=14,81\%$
