\(Ba\left(OH\right)2+SO2-->BaSO3+H2O\)(1)
\(n_{SO2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{BaSO3}=\frac{21,7}{217}=0,1\left(mol\right)\)
\(n_{SO2}>n_{BaSO3}\Rightarrow\) SO2 dư .Tạo 2 muối
\(CO2+BaSO3+H2O-->Ba\left(HSO3\right)2\)(2)
Theo pthh1
\(n_{SO2}=n_{BaSO3}=0,1\left(mol\right)\)
\(n_{SO2}\left(2\right)=0,15-0,1=0,05\left(mol\right)\)
\(n_{BaSO3}\left(2\right)=n_{SO2}=0,05\left(mol\right)\)
\(n_{BaSO3}\left(1\right)=0,1-0,95=0,05\left(mol\right)\)
\(n_{Ba\left(OH\right)2}=n_{BaSO3}=0,05\left(mol\right)\)
\(a=C_{M\left(Ba\left(OH\right)2\right)}=\frac{0,05}{0,3}=\frac{1}{6}\left(M\right)\)
