Đáp án:

 \(\begin{array}{l}
a)\\
{V_{S{O_2}}} = 4,48l\\
b)\\
{C_M}NaCl = {C_M}HCl = 0,5M\\
c)\\
{C_M}KHS{O_3} = {C_M}{K_2}S{O_3} = 0,5M
\end{array}\)

Giải thích các bước giải:

 \(\begin{array}{l}
a)\\
N{a_2}S{O_3} + 2HCl \to 2NaCl + S{O_2} + {H_2}O\\
{n_{N{a_2}S{O_3}}} = \frac{{25,2}}{{126}} = 0,2\,mol\\
{n_{HCl}} = 0,8 \times 1 = 0,8mol\\
\dfrac{{0,2}}{1} < \dfrac{{0,8}}{2} \Rightarrow \text{ HCl dư}\\
{n_{S{O_2}}} = {n_{N{a_2}S{O_3}}} = 0,2\,mol\\
{V_{S{O_2}}} = 0,2 \times 22,4 = 4,48l\\
b)\\
{n_{NaCl}} = 2{n_{N{a_2}S{O_3}}} = 0,4\,mol\\
{n_{HCl}}\text{ dư} = 0,8 - 0,2 \times 2 = 0,4\,mol\\
{C_M}NaCl = {C_M}HCl \text{ dư}= \dfrac{{0,4}}{{0,8}} = 0,5M\\
c)\\
{n_{KOH}} = 0,2 \times 1,5 = 0,3\,mol\\
T = \dfrac{{{n_{KOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,3}}{{0,2}} = 1,5\\
1 < T < 2 \Rightarrow\text{ Tạo cả 2 muối} \\
2KOH + S{O_2} \to {K_2}S{O_3} + {H_2}O(1)\\
{K_2}S{O_3} + S{O_2} + {H_2}O \to 2KHS{O_3}(2)\\
{n_{S{O_2}(1)}} = {n_{{K_2}S{O_3}(1)}} = 0,15\,mol\\
{n_{S{O_2}(2)}} = {n_{{K_2}S{O_3}(2)}} = 0,2 - 0,15 = 0,05\,mol\backslash \\
{n_{{K_2}S{O_3}}} = 0,15 - 0,05 = 0,1\,mol\\
{n_{KHS{O_3}}} = 0,05 \times 2 = 0,1\,mol\\
{C_M}KHS{O_3} = {C_M}{K_2}S{O_3} = \dfrac{{0,1}}{{0,2}} = 0,5M
\end{array}\)