\(a.n_{SO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ n_{NaOH}=0,15\left(mol\right)\\ LậpT:\dfrac{n_{NaOH}}{n_{SO_2}}=\dfrac{0,15}{0,1}=1,5\\ \Rightarrow Tạo2muối\\ NaOH+SO_2\rightarrow NaHSO_3\\ 2NaOH+SO_2\rightarrow Na_2SO_3+H_2O\\ Đặt\left\{{}\begin{matrix}NaHSO_3:x\left(mol\right)\\Na _2SO_3:y\left(mol\right)\end{matrix}\right.\\ Tacó:\left\{{}\begin{matrix}x+y=0,1\\x+2y=0,15\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\\ \Rightarrow m_{NaHSO_3}=5,2\left(g\right);m_{Na_2SO_3}=6,3\left(g\right)\\ b.CM_{NaHSO_3}=\dfrac{0,05}{0,15}=\dfrac{1}{3}M;CM_{Na_2SO_3}=\dfrac{0,05}{0,15}=\dfrac{1}{3}M\)
