Em tham khảo nha:

\(\begin{array}{l}
1)\\
{n_{C{O_2}}} = \dfrac{{33,6}}{{22,4}} = 1,5\,mol\\
{n_{NaOH}} = \dfrac{{500 \times 20\% }}{{40}} = 2,5\,mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{C{O_2}}}}} = \dfrac{{2,5}}{{1,5}} = 1,67\\
1 < T < 2 \Rightarrow\text{ Tạo cả 2 muối} \\
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O(1)\\
N{a_2}C{O_3} + C{O_2} + {H_2}O \to 2NaHC{O_3}(2)\\
{n_{N{a_2}C{O_3}(1)}} = {n_{C{O_2}(1)}} = \dfrac{{2,5}}{2} = 1,25\,mol\\
{n_{N{a_2}C{O_3}(2)}} = {n_{C{O_2}(2)}} = 1,5 - 1,25 = 0,25\,mol\\
{n_{N{a_2}C{O_3}}} = 1,25 - 0,25 = 1\,mol\\
{n_{NaHC{O_3}}} = 0,25 \times 2 = 0,5\,mol\\
{m_{N{a_2}C{O_3}}} = 1 \times 106 = 106g\\
{m_{NaHC{O_3}}} = 0,5 \times 84 = 42g\\
2)\\
{n_{C{O_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_{Ca{{(OH)}_2}}} = 0,075 \times 1 = 0,075\,mol\\
T = \dfrac{{{n_{Ca{{(OH)}_2}}}}}{{{n_{C{O_2}}}}} = \dfrac{{0,075}}{{0,1}} = 0,75\\
0,5 < T < 1 \Rightarrow \text{ Tạo $CaCO_3$ và $Ca(HCO_3)_2$}
\end{array}\)