Giải thích các bước giải:
a,
\(\begin{array}{l}
{n_{S{O_2}}} = 0,8mol\\
{n_{NaOH}} = 1mol\\
\to \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = 1,25
\end{array}\)
Tạo 2 muối
Gọi a và b là số mol của \(S{O_2}\)
\(\begin{array}{l}
S{O_2} + NaOH \to NaHS{O_3}\\
S{O_2} + 2NaOH \to N{a_2}S{O_3} + {H_2}O\\
\left\{ \begin{array}{l}
a + b = 0,8\\
a + 2b = 1
\end{array} \right.\\
\to a = 0,6 \to b = 0,2\\
\to {m_{Muối}} = {m_{NaHS{O_3}}} + {m_{N{a_2}S{O_3}}} = 87,6g\\
\to C{M_{NaHS{O_3}}} = \dfrac{{0,6}}{{17,92 + 0,5}} = 0,03M\\
\to C{M_{N{a_2}S{O_3}}} = \dfrac{{0,2}}{{17,92 + 0,5}} = 0,01M
\end{array}\)
b,
\(\begin{array}{l}
{n_{S{O_2}}} = 0,5mol\\
{n_{NaOH}} = 1mol\\
\to \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = 2
\end{array}\)
Tạo muối trung hòa
\(\begin{array}{l}
S{O_2} + 2NaOH \to N{a_2}S{O_3} + {H_2}O\\
{n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,5mol\\
\to {m_{N{a_2}S{O_3}}} = 63g\\
\to C{M_{N{a_2}S{O_3}}} = \dfrac{n}{V} = \dfrac{{0,5}}{{11,2 + 0,5}} = 0,04M
\end{array}\)
c,
\(\begin{array}{l}
{n_{S{O_2}}} = 0,4mol\\
{n_{NaOH}} = 1mol\\
\to \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = 2,5
\end{array}\)
Tạo muối trung hòa, NaOH dư
\(\begin{array}{l}
S{O_2} + 2NaOH \to N{a_2}S{O_3} + {H_2}O\\
{n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,4mol\\
{n_{NaOH}}dư= 1 - 0,4 \times 2 = 0,2mol\\
\to {m_{N{a_2}S{O_3}}} = 50,4g\\
\to C{M_{N{a_2}S{O_3}}} = \dfrac{n}{V} = \dfrac{{0,4}}{{8,96 + 0,5}} = 0,04M\\
\to C{M_{NaOH}}dư = \dfrac{n}{V} = \dfrac{{0,2}}{{8,96 + 0,5}} = 0,02M
\end{array}\)
