\(\begin{array}{l}
3)\\
a)CuO+H_2SO_4\to CuSO_4+H_2O\\
b)\\
n_{H_2SO_4}=n_{CuO}=n_{CuSO_4}=\frac{50.19,6\%}{98}=0,1(mol)\\
C\%_{CuSOO_4}=\frac{0,1.160}{0,1.80+50}.100\%\approx 27,59\%\\
4)\\
a)\\
n_{MgO}=\frac{4}{40}=0,1(mol)\\
MgO+2HCl\to MgCl_2+H_2O\\
n_{HCl}=2n_{MgO}=0,2(mol)\\
V_{dd\,HCl}=\frac{0,2}{1}=0,2(l)\\
b)\\
n_{MgCl_2}=n_{MgO}=0,1(mol)\\
C_{M\,MgCl_2}=\frac{0,1}{0,2}=0,5M
\end{array}\)