Question

5 moles of $S{O}_2$ and 5 moles of $O_2$ react in a closed vessel. At equilibrium 60% of the $S{O}_2$ is consumed. The total number of gaseous moles ($S{O}_2$, $O_2$ and $S{O}_3$) in the vessel is:

• A. 5.1
• B. 3.9
• C. 10.5
• D. 8.5

Answer 1

The correct option is D 8.5

Solution:-

${S{O}_2}_{\left(g\right)}+\frac{1}{2}{O_2}_{\left(g\right)}\rightleftharpoons {S{O}_3}_{\left(g\right)}$

From the above reaction-

No. of moles of $S{O}_3$ formed $=$ No. of moles of $S{O}_2$ consumed.

No. of moles of $O_2$ consumed $=$ half of no. of moles of $S{O}_2$ consumed.

Given that $60\%$ of $5$ moles of $S{O}_2$ is consumed.

No. of moles of $S{O}_3$ formed $=\frac{60}{100}\times 5=3\text{mol}$

No. of moles of $O_2$ consumed $=\frac{1}{2}\times 3=1.5\text{mol}$

Therefore, at equilibrium,

No. of moles of $S{O}_2$ present $=5-3=2$

No. of moles of $O_2$ present $=5-1.5=3.5$

No. of moles of $S{O}_3$ present $=3$

$\therefore$ Total no. of gaseous moles in the vessel $=2+3.5+3=8.5$

Hence the total no. of gaseous moles in the vessel is $8.5$.