On treating a compound with warm dil.
Question:

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

  1. $\mathrm{X}=\mathrm{SO}_{2}, \mathrm{Y}=\mathrm{Cr}_{2} \mathrm{O}_{3}$

  2. $\mathrm{X}=\mathrm{SO}_{3}, \mathrm{Y}=\mathrm{Cr}_{2} \mathrm{O}_{3}$

  3. $\mathrm{X}=\mathrm{SO}_{2}, \mathrm{Y}=\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$

  4. $\mathrm{X}=\mathrm{SO}_{3}, \mathrm{Y}=\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$


Correct Option: , 3

Solution:

$\mathrm{SO}_{2}+\operatorname{dil} \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{SO}_{3}(\mathrm{~g})$

$\mathrm{SO}_{3}+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \frac{\text { dil. }}{\mathrm{H}_{2} \mathrm{SO}_{4}} \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$

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