`n_{CuO}=\frac{1,6}{80}=0,02\ (mol)`
`n_{H_2SO_4}=\frac{100.20%}{98.100%}=0,2\ (mol)`
a,
`CuO+H_2SO_4\to CuSO_4+H_2O`
b,
Do `n_{CuO}=0,02\ mol<n_{H_2SO_4}=0,2\ mol`
`=>CuO` hết, `H_2SO_4` dư
Theo PT: $n_{H_2SO_4\ pứ}=n_{CuSO_4}=n_{CuO}=0,02\ (mol)$
`=>`$n_{H_2SO_4\ dư}=0,2-0,02=0,18\ (mol)$
$m_{dd\ spứ}=m_{CuO}+m_{ddH_2SO_4}=1,6+100=101,6\ (g)$
`=>`$\left\{\begin{matrix}{C\%}_{ddH_2SO_4dư}=\dfrac{0,18.98}{101,6}.100\%=17,36\%\\{C\%}_{ddCuSO_4}=\dfrac{0,02.160}{101,6}.100\%=3,15\%\end{matrix}\right.$