Em tham khảo nha:

\(\begin{array}{l} a)\\ nKOH = 0,4 \times 1,5 = 0,6mol\\ T = \dfrac{{nKOH}}{{nS{O_2}}} = \dfrac{{0,6}}{{0,25}} = 2,4\\ T > 2 \Rightarrow\text{ Tạo muối trung hòa} \\ 2KOH + S{O_2} \to {K_2}S{O_3} + {H_2}O\\ n{K_2}S{O_3} = nS{O_2} = 0,25\,mol\\ {C_M}{K_2}S{O_3} = \dfrac{{0,25}}{{0,4}} = 0,625M\\ n_{KOH}\ dư = 0,1\ mol \to C_M(KOH) = \dfrac{0,1}{0,4} = 0,25\ M\\ b)\\ nNaOH = 0,8 \times 0,25 = 0,2mol\\ T = \dfrac{{nNaOH}}{{nS{O_2}}} = \dfrac{{0,2}}{{0,25}} = 0,8\\ T < 1 \Rightarrow\text{ Tạo muối axit} \\ NaOH + S{O_2} \to NaHS{O_3}\\ nNaHS{O_3} = nNaOH = 0,2\,mol\\ {C_M}NaHS{O_3} = \dfrac{{0,2}}{{0,25}} = 0,8M\\ c)\\ nKOH = 0,2 \times 2 = 0,4mol\\ T = \dfrac{{nKOH}}{{nS{O_2}}} = \dfrac{{0,4}}{{0,25}} = 1,6\\ 1 < T < 2 \Rightarrow\text{ tạo cả 2 muối} \\ 2KOH + S{O_2} \to {K_2}S{O_3} + {H_2}O(1)\\ {K_2}S{O_3} + S{O_2} + {H_2}O \to 2KHS{O_3}(2)\\ nS{O_2}(1) = n{K_2}S{O_3}(1) = 0,2\,mol\\ nS{O_2}(2) = n{K_2}S{O_3}(2) = 0,25 - 0,2 = 0,05\,mol\\ n{K_2}S{O_3} = 0,2 - 0,05 = 0,15\,mol\\ nKHS{O_3} = 2 \times 0,05 = 0,1\,mol\\ {C_M}{K_2}S{O_3} = \dfrac{{0,15}}{{0,2}} = 0,75M\\ {C_M}KHS{O_3} = \dfrac{{0,1}}{{0,2}} = 0,5M \end{array}\)