\(n_{SO_2}=\dfrac{1,2395}{24,79}=0,05\left(mol\right)\)
\(m_{NaOH}=100.4\%=4\left(g\right)\Rightarrow n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)
\(\Rightarrow\dfrac{n_{NaOH}}{n_{SO_2}}=\dfrac{0,1}{0,05}=2\) → pư tạo muối Na2SO3
Chất tan trong X: Na2SO3.