Giải thích các bước giải:
\(\begin{array}{l}
a)CuO + 2HCl \to CuC{l_2} + {H_2}O\\
b)\\
{n_{CuO}} = 0,02mol\\
\to {n_{CuC{l_2}}} = {n_{CuO}} = 0,02mol\\
\to {m_{CuC{l_2}}} = 2,7g\\
c)\\
{n_{HCl}} = 2{n_{CuO}} = 0,04mol\\
\to {m_{HCl}} = 1,46g\\
\to C{\% _{{\rm{dd}}HCl}} = \dfrac{{1,46}}{{200}} \times 100\% = 0,73\% \\
d)\\
{m_{{\rm{dd}}}} = {m_{CuO}} + {m_{{\rm{dd}}HCl}} = 201,6g\\
\to C{\% _{CuC{l_2}}} = \dfrac{{2,7}}{{201,6}} \times 100\% = 1,34\%
\end{array}\)