Em tham khảo nha :
\(\begin{array}{l}
a)\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
b)\\
{n_{CuO}} = \dfrac{{1,6}}{{80}} = 0,02mol\\
{m_{{H_2}S{O_4}}} = \dfrac{{100 \times 20}}{{100}} = 20g\\
{n_{{H_2}S{O_4}}} = \dfrac{{20}}{{98}} = 0,2mol\\
\dfrac{{0,02}}{1} < \dfrac{{0,2}}{1} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{{H_2}S{O_4}d}} = 0,2 - 0,02 = 0,18mol\\
{m_{{H_2}S{O_4}d}} = 0,18 \times 98 = 17,64g\\
{n_{CuS{O_4}}} = {n_{CuO}} = 0,02mol\\
{m_{CuS{O_4}}} = 0,02 \times 160 = 3,2g\\
C{\% _{CuS{O_4}}} = \dfrac{{3,2}}{{1,6 + 100}} \times 100\% = 3,15\% \\
C{\% _{{H_2}S{O_4}}} = \dfrac{{17,64}}{{1,6 + 100}} \times 100\% = 17,36\%
\end{array}\)