n Ba(OH)2=0,2.1,25=0,25 mol
n BaSO3=$\frac{32,55}{217}$=0,15 mol
TH1:Tạo 1 muối trung hòa
SO2+Ba(OH)2→BaSO3↓+H2O
0,15 ←0,15 mol
V=V SO2 (đktc)=0,15.22,4=3,36 l
TH2:Tạo 2 muối
SO2+Ba(OH)2→BaSO3↓+H2O (1)
0,15 0,15 ←0,15 mol
2SO2+Ba(OH)2→Ba(HSO3)2 (2)
0,2 ←0,1 mol
n Ba(OH)2 (2)=∑n Ba(OH)2-n Ba(OH)2 (1)=0,25-0,15=0,1 mol
⇒∑n SO2=0,15+0,2=0,35 mol
⇒V=V SO2 (đktc)=0,35.22,4=7,84 l
-------------------------Nguyễn Hoạt---------------------------
