Ta có: [imath]n_{SO_2}=\frac{4,48}{22,4}=0,2(mol)[/imath]; [imath]n_{Ba^{2+}}=0,15.1.1=0,15(mol)[/imath]; [imath]n_{OH^{-}}=0,15.1.2=0,30(mol)[/imath]
Xét: [imath]{SO_2}+{BaOH_2}\rightarrow{BaCO_3}+{H_2O}[/imath]
[imath]\Rightarrow[/imath][imath]n_{BaSO_3}=0,15(mol)[/imath]; [imath]n_{{SO_2}_{\text{dư}}}=0,05(mol)[/imath]
Mặt khác: [imath]SO_2+BaSO_3+H_2O\rightarrow Ba(HSO3)_2[/imath]
[imath]\Rightarrow n_{{BaSO_3}_{\text{còn lại}}}=0,10(mol)\\ \Rightarrow m_{{BaSO_3}_{\text{còn lại}}}=21,7(gam)[/imath]