Em tham khảo nha :
\(\begin{array}{l}
a)\\
{n_{S{O_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
{n_{NaOH}} = 0,2 \times 2 = 0,4mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,4}}{{0,25}} = 1,6\\
\Rightarrow\text{Phản ứng tạo ra 2 muối } N{a_2}S{O_3},NaHS{O_3}\\
2NaOH + S{O_2} \to N{a_2}S{O_3} + {H_2}O\\
NaOH + S{O_2} \to NaHS{O_3}\\
hh:N{a_2}S{O_3}(a\,mol),NaHS{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,25\\
2a + b = 0,4
\end{array} \right.\\
\Rightarrow a = 0,15;b = 0,1\\
{m_{N{a_2}S{O_3}}} = 0,15 \times 126 = 18,9g\\
{m_{NaHS{O_3}}} = 0,1 \times 104 = 10,4g\\
m = 18,9 + 10,4 = 29,3g\\
b)\\
{n_{C{O_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Ca{{(OH)}_2}}} = 0,2 \times 0,5 = 0,1mol\\
T = \dfrac{{{n_{Ca{{(OH)}_2}}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,1}}{{0,15}} = \frac{2}{3}\\
\Rightarrow\text{Phản ứng tạo ra 2 muối } CaC{O_3},Ca{(HC{O_3})_2}\\
Ca{(OH)_2} + C{O_2} \to CaC{O_3} + {H_2}O\\
Ca{(OH)_2} + 2C{O_2} \to Ca{(HC{O_3})_2}\\
hh:CaC{O_3}(a\,mol);Ca{(HC{O_3})_2}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,1\\
a + 2b = 0,15
\end{array} \right.\\
\Rightarrow a = 0,05;b = 0,05\\
{m_{CaC{O_3}}} = 0,05 \times 100 = 5g\\
{m_{Ca{{(HC{O_3})}_2}}} = 0,05 \times 162 = 8,1g\\
m = 8,1 + 5 = 13,1g
\end{array}\)
