\(n_{SO_2}=\frac{6,72}{22,4}=0,3\left(mol\right)\\ TL:\frac{n_{OH}}{n_{SO_2}}=\frac{0,2.2}{0,3}=1,\left(3\right)\)
→ Tạo ra hh 2 muối
\(n_{CaSO_3}=x;n_{Ca\left(HSO_3\right)_2}=y\)
\(PTHH:SO_2+Ca\left(OH\right)_2\rightarrow CaSO_3+H_2O\\ PTHH:Ca\left(OH\right)_2+2SO_2\rightarrow Ca\left(HSO_3\right)_2\\ hpt:\left\{{}\begin{matrix}x+2y=0,3\\x+y=0,2\end{matrix}\right.\Leftrightarrow x=y=0,1\\ m_M=0,1.120+202.0,1=32,2\left(g\right)\)