\(n_{SO_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
\(n_{KOH}=0.2\cdot2=0.4\left(mol\right)\)
\(T=\dfrac{0.4}{0.15}=2.67\)
=> Tạo muối trung hòa
\(2KOH+SO_2\rightarrow K_2SO_3+H_2O\)
\(0.3.........0.15...........0.15\)
\(m_{K_2SO_3}=0.15\cdot158=23.7\left(g\right)\)