\(n_{SO_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)
\(T=\dfrac{0.25}{0.3}=0.83\)
=> Tạo ra CaSO3
\(Ca\left(OH\right)_2+SO_2\rightarrow CaSO_3+H_2O\)
\(0.25..........0.25...........0.25\)
\(m_{CaSO_3}=0.25\cdot120=30\left(g\right)\)
\(m_{Ca\left(OH\right)_2\left(dư\right)}=\left(0.3-0.25\right)\cdot74=3.7\left(g\right)\)