Đáp án:
\(\begin{array}{l}
{C_\% }{K_2}S = 4,14\% \\
{C_\% }KHS = 0,9\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
{n_{{H_2}S}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
{n_{KOH}} = 0,7 \times 1 = 0,7\,mol\\
2KOH + {H_2}S \to {K_2}S + 2{H_2}O\\
\dfrac{{0,7}}{2} < \dfrac{{0,4}}{1} \Rightarrow {H_2}S \text{ dư}\\
{n_{{H_2}S}} \text{ dư}= 0,4 - \dfrac{{0,7}}{2} = 0,05\,mol\\
{n_{{K_2}S}} = \dfrac{{0,7}}{2} = 0,35\,mol\\
{K_2}S + {H_2}S \to 2KHS\\
\dfrac{{0,35}}{1} > \dfrac{{0,05}}{1} \Rightarrow {K_2}S\text{ dư}\\
{n_{{K_2}S}}\text{ dư} = 0,35 - 0,05 = 0,3\,mol\\
{n_{KHS}} = 0,05 \times 2 = 0,1\,mol\\
{m_{{\rm{dd}}spu}} = 0,4 \times 34 + 700 \times 1,12 = 797,6g\\
{C_\% }{K_2}S = \dfrac{{0,3 \times 110}}{{797,6}} \times 100\% = 4,14\% \\
{C_\% }KHS = \dfrac{{0,1 \times 72}}{{797,6}} \times 100\% = 0,9\%
\end{array}\)
