Em tham khảo nha :
\(\begin{array}{l}
a)\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
{n_{CuO}} = \dfrac{{16}}{{80}} = 0,2mol\\
{n_{HCl}} = \dfrac{{284 \times 12,85\% }}{{36,5}} = 1mol\\
\dfrac{{0,2}}{1} < \dfrac{1}{2} \Rightarrow HCl\text{ dư}\\
{n_{CuC{l_2}}} = {n_{CuO}} = 0,2mol\\
{m_{CuC{l_2}}} = 0,2 \times 135 = 27g\\
b)\\
{n_{HC{l_d}}} = 1 - 0,4 = 0,6mol\\
{m_{HC{l_d}}} = 0,6 \times 36,5 = 21,9g\\
C{\% _{HCl}} = \dfrac{{21,9}}{{16 + 284}} \times 100\% = 7,3\% \\
C{\% _{CuC{l_2}}} = \dfrac{{27}}{{300}} \times 100\% = 9\%
\end{array}\)