Ta có :\(n_{SO_2}=\dfrac{1,12}{22,4}=0,05mol;n_{Ca\left(OH\right)_2}=0,7.0,05=0,035mol\)
Nhận xét :\(1< \dfrac{n_{CO_2}}{n_{Ca\left(OH\right)_2}}=\dfrac{0,05}{0,035}=1,42< 2\)
\(\rightarrow\) Xảy ra 2 phương trình :
\(SO_2+Ca\left(OH\right)_2\rightarrow CaSO_3\downarrow+2H_2O.\)
x..............x........................x
\(2SO_2+Ca\left(OH\right)_2\rightarrow Ca(HSO_3)_2\)
2(0,035-x)....(0,035-x).......(0,035-x)
\(\rightarrow n_{SO_2}=x+2\left(0,035-x\right)=0,05\rightarrow x=0,02\)
Sau phản ứng :
\(\left\{{}\begin{matrix}n_{CaSO_3}=x=0,02mol\rightarrow m_{CaSO_3}=2,4g\\n_{Ca\left(HSO_3\right)_2}=0,035-x=0,015mol\rightarrow m_{Ca\left(HSO_3\right)_2}=3,03g\end{matrix}\right.\)
