Đáp án:
\(\begin{array}{l}
{C_\% }{H_2}S{O_4} = 9,074\% \\
{C_\% }CuS{O_4} = 14,815\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
b)\\
{n_{CuO}} = \dfrac{m}{M} = \dfrac{8}{{80}} = 0,1\,mol\\
{m_{{H_2}S{O_4}}} = \dfrac{{{m_{{\rm{dd}}}} \times {C_\% }}}{{100}} = \dfrac{{100 \times 19,6}}{{100}} = 19,6g\\
{n_{{H_2}S{O_4}}} = \dfrac{m}{M} = \dfrac{{19,6}}{{98}} = 0,2\,mol\\
\dfrac{{0,1}}{1} < \dfrac{{0,2}}{1} \Rightarrow \text{ NaOH dư}\\
{n_{{H_2}S{O_4}}} \text{ dư}= 0,2 - 0,1 \times 1 = 0,1\,mol\\
{n_{CuS{O_4}}} = {n_{CuO}} = 0,1\,mol\\
{m_{{\rm{dd}}spu}} = 8 + 100 = 108g\\
{C_\% }{H_2}S{O_4} \text{ dư}= \dfrac{{0,1 \times 98}}{{108}} \times 100 = 9,074\% \\
{C_\% }CuS{O_4} = \dfrac{{0,1 \times 160}}{{108}} \times 100 = 14,815\%
\end{array}\)
