Giải thích các bước giải:
\(\begin{array}{l}
a)V = 1,68l\\
{n_{S{O_2}}} = 0,075mol\\
{n_{NaOH}} = 0,16mol\\
\to \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,16}}{{0,075}} = 2,13
\end{array}\)
=> Tạo 1 muối: \(N{a_2}S{O_3}\)
\(\begin{array}{l}
S{O_2} + 2NaOH \to N{a_2}S{O_3} + {H_2}O\\
{n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,075mol\\
\to {m_{N{a_2}S{O_3}}} = 9,45g\\
b)V = 22,4l\\
{n_{S{O_2}}} = 1mol\\
{n_{NaOH}} = 0,16mol\\
\to \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,16}}{1} = 0,16
\end{array}\)
=> Tạo 1 muối: \(NaHS{O_3}\)
\(\begin{array}{l}
S{O_2} + NaOH \to NaHS{O_3}\\
{n_{NaHS{O_3}}} = {n_{S{O_2}}} = 1mol\\
\to {m_{NaHS{O_3}}} = 104g\\
c)V = 4,48l\\
{n_{S{O_2}}} = 0,2mol\\
{n_{NaOH}} = 0,16mol\\
\to \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,16}}{{0,2}} = 0,8
\end{array}\)
=> Tạo 1 muối: \(NaHS{O_3}\)
\(\begin{array}{l}
S{O_2} + 2NaOH \to NaHS{O_3}\\
{n_{NaHS{O_3}}} = {n_{S{O_2}}} = 0,2mol\\
\to {m_{NaHS{O_3}}} = 20,8g
\end{array}\)
