Đáp án:
\(\begin{array}{l}
{C_{{M_{N{a_2}S{O_3}}}}} = 0,042M\\
{C_{{M_{NaHS{O_3}}}}} = 0,098M\\
{m_{N{a_2}S{O_3}}} = 2,646g\\
{m_{NaHS{O_3}}} = 5,096g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
{n_{S{O_2}}} = \dfrac{{1,568}}{{22,4}} = 0,07mol\\
{n_{NaOH}} = 0,5 \times 0,182 = 0,091mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,091}}{{0,07}} = 1,3\\
1 < T < 2 \Rightarrow\text{Phản ứng tạo ra 2 muối :} N{a_2}S{O_3},NaHS{O_3}\\
2NaOH + S{O_2} \to N{a_2}S{O_3} + {H_2}O\\
NaOH + S{O_2} \to NaHS{O_3}\\
hh:N{a_2}S{O_3}(a\,mol),NaHS{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
2a + b = 0,091\\
a + b = 0,07
\end{array} \right.\\
\Rightarrow a = 0,021;b = 0,049\\
{C_{{M_{N{a_2}S{O_3}}}}} = \dfrac{{0,021}}{{0,5}} = 0,042M\\
{C_{{M_{NaHS{O_3}}}}} = \dfrac{{0,049}}{{0,5}} = 0,098M\\
{m_{N{a_2}S{O_3}}} = 0,021 \times 126 = 2,646g\\
{m_{NaHS{O_3}}} = 0,049 \times 104 = 5,096g
\end{array}\)
