\(n_{CuO}=\dfrac{16}{80}=0.2\left(mol\right)\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(0.2.........0.4..............0.2\)
\(V_{dd_{HCl}}=\dfrac{0.4}{0.5}=0.8\left(l\right)\)
\(m_{CuCl_2}=0.2\cdot151=30.2\left(g\right)\)
\(C_{M_{CuCl_2}}=\dfrac{0.2}{0.8}=0.25\left(M\right)\)