\(n_{SO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right);n_{NaOH}=0,05.1=0,05\left(mol\right)\)
Xét \(T=\dfrac{n_{NaOH}}{n_{SO_2}}=\dfrac{0,05}{0,05}=1\) => Tạo muối axit
PTHH: \(NaOH+SO_2\rightarrow NaHSO_3\)
0,05---------------->0,05
=> mmuối = 0,05.104 = 5,2 (g)