TH1: Ca(OH)2 dư
\(\begin{array}{l}
CaO + {H_2}O \to Ca{(OH)_2}\\
S{O_2} + Ca{(OH)_2} \to CaS{O_3}\\
nCaS{O_3} = \frac{{1,2}}{{120}} = 0,01mol\\
= > nS{O_3} = 0,01mol\\
VS{O_2} = 0,01 \times 22,4 = 0,224l
\end{array}\)
TH2: SO2 dư
\(\begin{array}{l}
nCaO = \frac{{5,6}}{{56}} = 0,1mol\\
CaO + {H_2}O \to Ca{(OH)_2}\\
S{O_2} + Ca{(OH)_2} \to CaS{O_3}(1)\\
CaS{O_3} + S{O_2} + {H_2}O \to Ca{(HS{O_3})_2}(2)\\
nCaS{O_3} = \frac{{1,2}}{{120}} = 0,01mol\\
nCaS{O_3}(1) = nCa{(OH)_2} = 0,1mol\\
nCaS{O_3}(2) = 0,1 - 0,01 = 0,09mol\\
= > nS{O_3} = 0,1 + 0,09 = 0,19mol\\
VS{O_2} = 0,19 \times 22,4 = 4,256l
\end{array}\)
