\(n_{SO_2}=0,45\left(mol\right)\\ n_{CaSO_3}=0,15\left(mol\right)< n_{SO_2}\)
Đặt \(n_{Ca\left(OH\right)_2}=a\left(mol\right)\)
=> Đã có pứ hòa tan ↓
Ca(OH)2 + SO2 -------> CaSO3 + H2O
a___________a____________a
CaSO3 + H2O + SO2 -------> Ca(HSO3)2
0,45-a_______________0,45-a
\(\Rightarrow n_{CaSO_3\left(con\right)}=2a-0,45=0,15\\ \Rightarrow a=0,3\\ \Rightarrow m_{Ca\left(OH\right)_2}=22,2\left(g\right)\Rightarrow x=14,8\)
\(n_{SO_2}=0,45\left(mol\right)\\ n_{CaSO_3}=0,15\left(mol\right)\)
TH1: ↓ chưa bị hòa tan
Ca(OH)2 + CO2 ------> CaSO3 + H2O
0,15____________________\(_{\leftarrow}\)0,15
\(\Rightarrow m_{Ca\left(OH\right)_2}=11,1\left(g\right)\Rightarrow x=7,4\)
TH2: ↓ đã bị hòa tan
Ca(OH)2 + SO2 ------> CaSO3 + H2O
a\(_{\rightarrow}\)__________a__________a
CaSO3 + SO2 + H2O ------> Ca(HSO3)2
\(0,45-a\)__\(_{\leftarrow}0,45-a\)
\(\Rightarrow a-\left(0,45-a\right)=0,15\\ \Rightarrow a=0,3\\ \Rightarrow m_{Ca\left(OH\right)_2}=22,2\left(g\right)\\ \Rightarrow x=14,8\)
