Ta có: \(\left\{{}\begin{matrix}n_{SO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\n_{Ca\left(OH\right)_2}=1\cdot0,2=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo muối trung hòa
PTHH: \(Ca\left(OH\right)_2+SO_2\rightarrow CaSO_3\downarrow+H_2O\)
Vì Ca(OH)2 dư nên tính theo SO2
\(\Rightarrow n_{CaCO_3}=0,15\left(mol\right)\) \(\Rightarrow m_{CaSO_3}=0,15\cdot120=18\left(g\right)\)