Đáp án:
\(\begin{array}{l}
a)\\
{C_M}{H_2}S{O_4} = 0,833M\\
{C_M}{K_2}S{O_4} = 0,167M\\
b)\\
m = 69,9g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{n_{KOH}} = \dfrac{{5,6}}{{56}} = 0,1\,mol\\
{n_{{H_2}S{O_4}}} = 0,3 \times 1 = 0,3\,mol\\
{n_{{H_2}S{O_4}}} > \dfrac{{{n_{KOH}}}}{2} \Rightarrow {H_2}S{O_4} \text{ dư } \\
{n_{{H_2}S{O_4}}} \text{ dư }= 0,3 - \frac{{0,1}}{2} = 0,25\,mol\\
{n_{{K_2}S{O_4}}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
{C_M}{H_2}S{O_4} \text{ dư }= \dfrac{{0,25}}{{0,3}} = 0,833M\\
{C_M}{K_2}S{O_4} = \dfrac{{0,05}}{{0,3}} = 0,167M\\
b)\\
{H_2}S{O_4} + Ba{(OH)_2} \to BaS{O_4} + 2{H_2}O\\
{K_2}S{O_4} + Ba{(OH)_2} \to BaS{O_4} + 2KOH\\
{n_{BaS{O_4}}} = 0,25 + 0,05 = 0,3\,mol\\
m = {m_{BaS{O_4}}} = 0,3 \times 233 = 69,9g
\end{array}\)
