Đáp án:

\(\begin{array}{l}
a)\\
{V_{S{O_2}}} = 5,6l\\
b)\\
{C_\% }{K_2}S{O_4} = 19,82\% \\
c)\\
{m_{Ca{{(HS{O_3})}_2}}} = 20,2g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
{K_2}S{O_3} + {H_2}S{O_4} \to {K_2}S{O_4} + S{O_2} + {H_2}O\\
{n_{{K_2}S{O_3}}} = \dfrac{{39,5}}{{158}} = 0,25\,mol\\
{n_{S{O_2}}} = {n_{{K_2}S{O_3}}} = 0,25\,mol\\
{V_{S{O_2}}} = 0,25 \times 22,4 = 5,6l\\
b)\\
{m_{{\rm{dd}}spu}} = 39,5 + 196 - 0,25 \times 64 = 219,5g\\
{n_{{K_2}S{O_4}}} = {n_{{K_2}S{O_3}}} = 0,25\,mol\\
{C_\% }{K_2}S{O_4} = \dfrac{{0,25 \times 174}}{{219,5}} \times 100\%  = 19,82\% \\
c)\\
{n_{Ca{{(OH)}_2}}} = 0,05 \times 2 = 0,1\,mol\\
T = \dfrac{{{n_{Ca{{(OH)}_2}}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,1}}{{0,25}} = 0,4\\
T < 0,5 \Rightarrow\text{ Tạo muối $Ca(HSO_3)_2$} \\
Ca{(OH)_2} + 2S{O_2} \to Ca{(HS{O_3})_2}\\
{n_{Ca{{(HS{O_3})}_2}}} = {n_{Ca{{(OH)}_2}}} = 0,1\,mol\\
{m_{Ca{{(HS{O_3})}_2}}} = 0,1 \times 202 = 20,2g
\end{array}\)

\begin{array}{l} a)\\ {K_2}S{O_3} + {H_2}S{O_4} \to {K_2}S{O_4} + S{O_2} + {H_2}O\\ {n_{{K_2}S{O_3}}} = \dfrac{{39,5}}{{158}} = 0,25\,mol\\ {n_{S{O_2}}} = {n_{{K_2}S{O_3}}} = 0,25\,mol\\ {V_{S{O_2}}} = 0,25 \times 22,4 = 5,6l\\ b)\\ {m_{{\rm{dd}}spu}} = 39,5 + 196 - 0,25 \times 64 = 219,5g\\ {n_{{K_2}S{O_4}}} = {n_{{K_2}S{O_3}}} = 0,25\,mol\\ {C_\% }{K_2}S{O_4} = \dfrac{{0,25 \times 174}}{{219,5}} \times 100\%  = 19,82\% \\ c)\\ {n_{Ca{{(OH)}_2}}} = 0,05 \times 2 = 0,1\,mol\\ T = \dfrac{{{n_{Ca{{(OH)}_2}}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,1}}{{0,25}} = 0,4\\ T < 0,5 \Rightarrow\text{ Tạo muối $Ca(HSO_3)_2$} \\ Ca{(OH)_2} + 2S{O_2} \to Ca{(HS{O_3})_2}\\ {n_{Ca{{(HS{O_3})}_2}}} = {n_{Ca{{(OH)}_2}}} = 0,1\,mol\\ {m_{Ca{{(HS{O_3})}_2}}} = 0,1 \times 202 = 20,2g \end{array}