Em tham khảo nha:
\(\begin{array}{l}
1)\\
{n_{S{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{NaOH}} = \dfrac{{28}}{{40}} = 0,7\,mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,7}}{{0,3}} = 2,33\\
T > 2 \Rightarrow \text{ Chỉ tạo muối $Na_2SO_3$ }\\
2NaOH + S{O_2} \to N{a_2}S{O_3} + {H_2}O\\
{n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,3\,mol\\
{m_{N{a_2}S{O_3}}} = 0,3 \times 126 = 37,8g\\
2)\\
{n_{Ca{{(OH)}_2}}} = 0,075 \times 2 = 0,15\,mol\\
{n_{S{O_2}}} = \dfrac{{12,8}}{{64}} = 0,2\,mol\\
T = \dfrac{{{n_{Ca{{(OH)}_2}}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,15}}{{0,2}} = 0,75\\
0,5 < T < 1 \Rightarrow \text{ Tạo cả 2 muối }\\
Ca{(OH)_2} + S{O_2} \to CaS{O_3} + {H_2}O(1)\\
CaS{O_3} + S{O_2} + {H_2}O \to Ca{(HS{O_3})_2}(2)\\
{n_{CaS{O_3}(1)}} = {n_{S{O_2}(1)}} = {n_{Ca{{(OH)}_2}}} = 0,15\,mol\\
{n_{CaS{O_3}(2)}} = {n_{S{O_2}(2)}} = 0,2 - 0,15 = 0,05\,mol\\
{n_{CaS{O_3}}} = 0,15 - 0,05 = 0,1\,mol\\
{m_{CaS{O_3}}} = 0,1 \times 120 = 12g
\end{array}\)
