\(\begin{array}{l}
{{\rm{n}}_{{\rm{CuO}}}}{\rm{ = 0,05}}\\
{n_{HCl}} = \frac{{200\,.\,\,1,46\% }}{{36,5}} = 0,08\\
\mathop {CuO}\limits_{0,05} + \mathop {2HCl}\limits_{0,08} \to CuC{l_2} + {H_2}O\\
{m_{dd\,\,sau\,pu}} = 4 + 200 = 204\\
{n_{CuC{l_2}}} = \frac{1}{2}{n_{HCl}} = 0,04 => C\% = \frac{{0,04.135}}{{204}}.100\% = 2,65\%
\end{array}\)