$\underset{\left(500\mathrm{ml}\right)}{{\mathrm{SO}}_2}+\underset{\left(200\mathrm{ml}\right)}{{\mathrm{O}}_2}\to \underset{\left(400\mathrm{ml}\right)}{{\mathrm{SO}}_3}$in this reaction

Considered the following Stoichiometric equation.

$\underset{500\mathrm{ml}}{2{{\mathrm{SO}}_2}_{\left(\mathrm{g}\right)}}+\underset{200\mathrm{ml}}{{\mathrm{O}}_{2\left(\mathrm{g}\right)}}\to \underset{400\mathrm{ml}}{2{{\mathrm{SO}}_3}_{\left(\mathrm{g}\right)}}$

200 mL of ${\mathrm{O}}_2$=400 mL of ${\mathrm{SO}}_3$

200 mL of ${\mathrm{O}}_2$ =400mL of ${\mathrm{SO}}_2$

400 mL of ${\mathrm{SO}}_2$=400 mL of ${\mathrm{SO}}_3$

Therefore, for the given reaction ${\mathrm{SO}}_2$ acts as excess reagent and ${\mathrm{O}}_2$ limiting reagent (500- 400 =100mL)