\(n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(n_{NaOH}=1.0,3=0,3\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{n_{NaOH}}{n_{SO_2}}=\dfrac{0,3}{0,3}=1\) => Tạo muối NaHSO3
PTHH: NaOH + SO2 --> NaHSO3
0,3--------------->0,3
=> \(m_{NaHSO_3}=0,3.104=31,2\left(g\right)\)