Giải thích các bước giải:
\(\begin{array}{l}
a)S{O_2} + 2NaOH \to N{a_2}S{O_3} + {H_2}O\\
b)\\
{n_{NaOH}} = 0,15mol\\
\to {n_{S{O_2}}} = \dfrac{1}{2}{n_{NaOH}} = 0,075mol \to {V_{S{O_2}}} = 1,68l\\
c)\\
{n_{N{a_2}S{O_3}}} = \dfrac{1}{2}{n_{NaOH}} = 0,075mol\\
\to C{M_{N{a_2}S{O_3}}} = \dfrac{{0,075}}{{1,68 + 0,3}} = 0,038M\\
d)\\
N{a_2}S{O_3} + 2HCl \to 2NaCl + S{O_2} + {H_2}O\\
{n_{HCl}} = 0,12mol\\
\to {n_{HCl}} > {n_{N{a_2}S{O_3}}} \to {n_{HCl}}dư\\
\to {n_{NaCl}} = 2{n_{N{a_2}S{O_3}}} = 0,15mol \to {m_{NaCl}} = 8,775g
\end{array}\)
