nSO2 = 0,005 mol
nCa(OH)2 = 0,003 mol
Ta có
\(\dfrac{n_{Ca\left(OH\right)2}}{n_{SO2}}\) = \(\dfrac{0,003}{0,005}\) = 0,6
\(\Rightarrow\) Tạo 2 muối
Ca(OH)2 + SO2 \(\rightarrow\) CaSO3 + H2O (1)
x..............x...................x
Ca(OH)2 + 2SO2 → Ca(HSO3)2 (2)
y..................2y............y
Từ (1)(2) ta có hệ
\(\left\{{}\begin{matrix}x+y=0,003\\x+2y=0,005\end{matrix}\right.\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}x=0,001\\y=0,002\end{matrix}\right.\)
\(\Rightarrow\) mCaSO3 = 0,001.120 = 0,12 (g)
\(\Rightarrow\) mCa(HSO3)2 = 0,002.202 = 0,404 (g)