Giải thích các bước giải:

\(\begin{array}{l}
CuO + {H_2}S{O_4} \to C{\rm{uS}}{O_4} + {H_2}O\\
C{\rm{uS}}{O_4} + 2NaOH \to Cu{(OH)_2} + N{a_2}S{O_4}\\
Cu{(OH)_2} \to CuO + {H_2}O\\
a)\\
{m_{{H_2}S{O_4}}} = \dfrac{{20 \times 9,8\% }}{{100\% }} = 1,96g\\
 \to {n_{{H_2}S{O_4}}} = 0,02mol\\
 \to {n_{{\rm{CuS}}{O_4}}} = {n_{CuO}} = {n_{{H_2}S{O_4}}} = 0,02mol\\
 \to {m_{{\rm{CuS}}{O_4}}} = 3,2g\\
 \to {m_{{\rm{CuO}}}} = 1,6g\\
 \to {m_{{\rm{dd}}A}} = {m_{CuO}} + {m_{{\rm{dd}}{H_2}S{O_4}}} = 1,6 + 1,96 = 3,56g\\
 \to C{\% _{{\rm{CuS}}{O_4}}} = \dfrac{{3,2}}{{3,56}} \times 100\%  = 89,89\% \\
b)\\
{n_{NaOH}} = 2{n_{{\rm{CuS}}{O_4}}} = 0,04mol\\
 \to {m_{NaOH}} = 1,6g\\
 \to {m_{{\rm{dd}}NaOH}} = \dfrac{{1,6}}{{8\% }} \times 100\%  = 20g\\
c)\\
{n_{Cu}}_O = {n_{{\rm{CuS}}{O_4}}} = 0,02mol\\
 \to {m_{Cu}}_O = 1,6g
\end{array}\)