Em tham khảo nha :
\(\begin{array}{l}
{n_{S{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
a)\\
{n_{NaOH}} = 0,3 \times 1 = 0,3mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,3}}{{0,3}} = 1\\
\Rightarrow\text{Phản ứng tạo ra muối } NaHS{O_3}\\
NaOH + S{O_2} \to NaHS{O_3}\\
{n_{NaHS{O_3}}} = {n_{NaOH}} = 0,3mol\\
{C_{{M_{NaHS{O_3}}}}} = \dfrac{{0,3}}{{0,3}} = 1M\\
b)\\
{n_{NaOH}} = 0,5 \times 1 = 0,5mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,5}}{{0,3}} = 1,67\\
\Rightarrow\text{Phản ứng tạo ra muối } NaHS{O_3},N{a_2}S{O_3}\\
NaOH + S{O_2} \to NaHS{O_3}\\
2NaOH + S{O_2} \to N{a_2}S{O_3} + {H_2}O\\
hh:NaHS{O_3}(a\,mol),N{a_2}S{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,3\\
a + 2b = 0,5
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,2\\
{C_{{M_{NaHS{O_3}}}}} = \dfrac{{0,1}}{{0,5}} = 0,2M\\
{C_{{M_{N{a_2}S{O_3}}}}} = \dfrac{{0,2}}{{0,5}} = 0,4M\\
c)\\
{n_{NaOH}} = 0,8 \times 1 = 0,8mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,8}}{{0,3}} = 2,67\\
\Rightarrow\text{Phản ứng tạo ra muối } N{a_2}S{O_3}\\
2NaOH + S{O_2} \to N{a_2}S{O_3} + {H_2}O\\
{n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,3mol\\
{C_{{M_{N{a_2}S{O_3}}}}} = \dfrac{{0,3}}{{0,8}} = 0,375M\\
{n_{NaO{H_d}}} = 0,8 - 0,3 \times 2 = 0,2mol\\
{C_{{M_{NaOH}}}} = \dfrac{{0,2}}{{0,8}} = 0,25M
\end{array}\)
