Em tham khảo nha :
\(\begin{array}{l}
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
{n_{CuO}} = \dfrac{4}{{80}} = 0,05mol\\
{n_{HCl}} = 0,1 \times 1,5 = 0,15mol\\
\dfrac{{0,05}}{1} < \dfrac{{0,15}}{2} \Rightarrow HCl\text{ dư}\\
{n_{HC{l_d}}} = 0,15 - 0,1 = 0,05mol\\
{n_{CuC{l_2}}} = {n_{CuO}} = 0,05mol\\
{C_{{M_{HC{l_d}}}}} = \dfrac{{0,05}}{{0,1}} = 0,5M\\
{C_{{M_{CuC{l_2}}}}} = \dfrac{{0,05}}{{0,1}} = 0,5M
\end{array}\)