On treating a compound with warm dil. H _ 2 SO _ 4 , gas X is...

Sulfur dioxide gas turns acidified potassium dichromate (VI) solution from orange to green. Chromium is reduced from (VI) to (III) after reaction.

${SO}_{2} + dil{H}_{2}{SO}_{4} ⟶ {SO}_{3}\left(g\right)$

${SO}_{3} + {K}_{2}{Cr}_{2}{O}_{7}\mathop{\to}_{{H}_{2}{SO}_{4}}^{dil}{Cr}_{2}{\left({SO}_{4}\right)}_{3}$

Hence, On treating a compound with warm dil. ${H}_{2}{SO}_{4}$, gas $X$ is evolved which turns ${K}_{2}{Cr}_{2}{O}_{7}$ paper acidified with dil. ${H}_{2}{SO}_{4}$ to a green compound $Y$. $X$ and $Y$ respectively are ${SO}_{3}$ and ${Cr}_{2}{( {SO}_{4} )}_{3}$.