\(2NaOH+SO_2\rightarrow Na_2SO_3+H_2O\)
theo bài ra ta có:\(n_{NaOH}=0,1.0,2=0,02mol\)
theo PTHH: \(n_{SO_2}=2n_{NaOH}=2.0,02=0,04mol\)
\(\Rightarrow V_{SO_2}=0,04.22,4=0.896l\)
n NaOH=0,1.0,2=0,02(mol)
2NaOH+SO2--->Na2SO3+H2O
0,02-----0,01(mol)
VSO2=22,4.0,01=0,224(l)
SO2+2NaOH-->Na2SO3+H2O
0,01--0,02 mol
nNaOH=0,02 mol
=>VSO2=0,01.22,4=0,224 L