Đáp án:

\({m_{CaS{O_3}}} = 16,8{\text{ gam}}\)

\({C_{M{\text{ Ca(HS}}{{\text{O}}_3}{)_2}}} = 0,3M\)

\({{\text{C}}_{M{\text{ }}{{\text{K}}_2}S{O_3}}} = 0,1875M\)

Giải thích các bước giải:

 Ta có: \({n_{S{O_2}}} = \frac{{5,824}}{{22,4}} = 0,26{\text{ mol;}}{{\text{n}}_{Ca{{(OH)}_2}}} = 0,2.1 = 0,2{\text{ mol}} \to \frac{{{n_{S{O_2}}}}}{{{n_{Ca{{(OH)}_2}}}}} = \frac{{0,26}}{{0,2}} = 1,3\)

Do vậy tạo hỗn hợp 2 muối \(CaS{O_3};Ca{(HS{O_3})_2}\) với số mol lần lượt là x, y.

\( \to x + y = 0,2{\text{ mol;x + 2y = 0}}{\text{,26}} \to {\text{x = 0}}{\text{,14;y = 0}}{\text{,06}}\)

\({m_{CaS{O_3}}} = 0,14.120 = 16,8{\text{ gam}}\)

\({C_{M{\text{ Ca(HS}}{{\text{O}}_3}{)_2}}} = \frac{{0,06}}{{0,2}} = 0,3M\)

Ta có: 

\({n_{KOH}} = 0,12.1 = 0,12{\text{ mol = 2}}{{\text{n}}_{Ca{{(HS{O_3})}_2}}}\)

\(Ca{(HS{O_3})_2} + 2KOH\xrightarrow{{}}CaS{O_3} + {K_2}S{O_3} + 2{H_2}O\)

\( \to {n_{{K_2}S{O_3}}} = \frac{{0,12}}{2} = 0,06{\text{ mol;}}{{\text{V}}_{dd}} = 200 + 120 = 320{\text{ ml = 0}}{\text{,32 lít}} \to {{\text{C}}_{M{\text{ }}{{\text{K}}_2}S{O_3}}} = \frac{{0,06}}{{0,32}} = 0,1875M\)