Em tham khảo nha :
\(\begin{array}{l}
a)\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
b)\\
{n_{CuO}} = \dfrac{{16}}{{80}} = 0,2mol\\
{m_{{H_2}S{O_4}}} = \dfrac{{100 \times 20}}{{100}} = 20g\\
{n_{{H_2}S{O_4}}} = \dfrac{{20}}{{98}} = 0,204mol\\
\dfrac{{0,2}}{1} < \dfrac{{0,204}}{1} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{CuS{O_4}}} = {n_{Cu}} = 0,2mol\\
{m_{CuS{O_4}}} = 0,2 \times 160 = 32g\\
{n_{{H_2}S{O_4}d}} = 0,204 - 0,2 = 0,004mol\\
{m_{{H_2}S{O_4}d}} = 0,004 \times 98 = 0,392g\\
C{\% _{{H_2}S{O_4}}} = \dfrac{{0,392}}{{16 + 100}} \times 100\% = 0,0245\% \\
C{\% _{CuS{O_4}}} = \dfrac{{32}}{{16 + 100}} \times 100\% = 27,6\%
\end{array}\)
