20 gms of sulphur on burning in air produced 11.2 its of ${\mathrm{SO}}_2$ at STP. The percentage of unreacted sulphur

$\mathrm{S}+{\mathrm{O}}_2\to {\mathrm{SO}}_2$

1mol sulphate = 1 mole ${\mathrm{SO}}_2=22.4 \mathrm{Lt} {\mathrm{SO}}_2$

$11.2 \mathrm{Lt} {\mathrm{SO}}_2=\frac{1}{2}\mathrm{mole} \mathrm{sulphure}=16 \mathrm{grams} \mathrm{sulphur}$

$\% \mathrm{unreacted}=\frac{4}{20}\times 100=20\%$