Giải thích các bước giải:
\(\begin{array}{l}
a)CuO + {H_2}S{O_4} \to C{\rm{uS}}{O_4} + {H_2}O\\
{n_{CuO}} = 0,02mol\\
{m_{{H_2}S{O_4}}} = \dfrac{{100 \times 20\% }}{{100\% }} = 20g\\
\to {n_{{H_2}S{O_4}}} = 0,2mol\\
\to {n_{{H_2}S{O_4}}} > {n_{CuO}} \to {n_{{H_2}S{O_4}}}dư\\
\to {n_{{H_2}S{O_4}(pt)}} = {n_{CuO}} = 0,02mol\\
\to {n_{{H_2}S{O_4}}}dư = 0,18mol\\
b)\\
{n_{{\rm{CuS}}{O_4}}} = {n_{CuO}} = 0,02mol \to {m_{{\rm{CuS}}{O_4}}} = 3,2g\\
{m_{{H_2}S{O_4}}}dư = 17,64g\\
{m_{{\rm{dd}}}} = {m_{CuO}} + {m_{{\rm{dd}}{{\rm{H}}_2}S{O_4}}} = 21,6g\\
\to C{\% _{CuS{O_4}}} = \dfrac{{3,2}}{{21,6}} \times 100\% = 14,8\% \\
\to C{\% _{{H_2}S{O_4}}}dư = \dfrac{{17,64}}{{21,6}} \times 100\% = 81,7\%
\end{array}\)
