Đáp án:
\(\begin{array}{l}
{C_\% }{H_2}S{O_4} = 17,36\% \\
{C_\% }CuS{O_4} = 3,15\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
b)\\
{n_{CuO}} = \dfrac{{1,6}}{{80}} = 0,02\,mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{100 \times 20\% }}{{98}} \approx 0,2\,mol\\
{n_{CuO}} < {n_{{H_2}S{O_4}}} \Rightarrow {H_2}S{O_4} \text{ dư } \\
{n_{{H_2}S{O_4}}} \text{ dư }= 0,2 - 0,02 = 0,18\,mol\\
{n_{CuS{O_4}}} = {n_{CuO}} = 0,02\,mol\\
{m_{{\rm{dd}}spu}} = 1,6 + 100 = 101,6g\\
{C_\% }{H_2}S{O_4}\text{ dư } = \dfrac{{0,18 \times 98}}{{101,6}} \times 100\% = 17,36\% \\
{C_\% }CuS{O_4} = \dfrac{{0,02 \times 160}}{{101,6}} \times 100\% = 3,15\%
\end{array}\)