Đáp án:
a) 0,038M
b) 1,456l
Giải thích các bước giải:
\(\begin{array}{l}
{n_{S{O_2}}} = 0,13mol\\
{n_{NaOH}} = 0,364mol\\
\to \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,354}}{{0,13}} = 2,8
\end{array}\)
-> Tạo 1 muối: \(N{a_2}S{O_3}\)
\(\begin{array}{l}
S{O_2} + 2NaOH \to N{a_2}S{O_3} + {H_2}O\\
{n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,13mol\\
\to C{M_{N{a_2}S{O_3}}} = \dfrac{{0,13}}{{2,912 + 0,5}} = 0,038M\\
N{a_2}S{O_3} + 2HCl \to 2NaCl + S{O_2} + {H_2}O\\
{n_{HCl}} = 0,175mol\\
{n_{N{a_2}S{O_3}}} = \dfrac{{0,13}}{2} = 0,065mol\\
\to {n_{S{O_2}}} = {n_{N{a_2}S{O_3}}} = 0,065mol\\
\to {V_{S{O_2}}} = 0,065 \times 22,4 = 1,456l
\end{array}\)