Đáp án:

a,

\(\begin{array}{l}
{n_{S{O_2}}} = 0,05mol\\
{n_{NaOH}} = 0,05mol\\
\dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = 1
\end{array}\)

Chỉ tạo muối axit: \(NaOH + S{O_2} \to NaHS{O_3}\)

\({n_{NaHS{O_3}}} = {n_{NaOH}} = 0,05mol \to m = 5,2g\)

b,\(\begin{array}{l}
{n_{S{O_2}}} = 0,05mol\\
{n_{NaOH}} = 0,08mol\\
\dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = 1,6
\end{array}\)

Tạo 2 muối,

\(\begin{array}{l}
NaOH + S{O_2} \to NaHS{O_3}\\
2NaOH + S{O_2} \to N{a_2}S{O_3} + {H_2}O
\end{array}\)

\(\left\{ \begin{array}{l}
a + 2b = 0,08\\
a + b = 0,05
\end{array} \right.\)

a=0,02 b=0,03

\(\begin{array}{l}
{m_{NaHS{O_3}}} = 2,08g\\
{m_{N{a_2}S{O_3}}} = 3,78g
\end{array}\)

c,

\(\begin{array}{l}
{n_{S{O_2}}} = 0,05mol\\
{n_{NaOH}} = 0,1mol\\
\dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = 2\\
2NaOH + S{O_2} \to N{a_2}S{O_3} + {H_2}O\\
{n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,05mol\\
{m_{N{a_2}S{O_3}}} = 6,3g
\end{array}\)