Em tham khảo nha :
\(\begin{array}{l}
a)\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
b)\\
{n_{CuO}} = \dfrac{{1,6}}{{80}} = 0,02mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{100 \times 20\% }}{{98}} = 0,204mol\\
\dfrac{{0,02}}{1} < \dfrac{{0,204}}{1} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{{H_2}S{O_4}pu}} = {n_{CuO}} = 0,02mol\\
{m_{{H_2}S{O_4}pu}} = 0,02 \times 98 = 1,96g\\
{m_{{H_2}S{O_4}d}} = 20 - 1,96 = 18,04g\\
{n_{CuS{O_4}}} = {n_{CuO}} = 0,02mol\\
{m_{CuS{O_4}}} = 0,02 \times 160 = 3,2g\\
C{\% _{{H_2}S{O_4}}} = \dfrac{{18,04}}{{1,6 + 100}} \times 100\% = 17,75\% \\
C{\% _{CuS{O_4}}} = \dfrac{{3,2}}{{1,6 + 100}} \times 100\% = 3,15\%
\end{array}\)
