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Stoichiometric Calculations

5 moles of ...

Question

$5$ moles of ${S O}_{2}$ and $5$ moles of $O_{2}$ are allowed to react to form ${S O}_{3}$ in a closed vessel. At the equilibrium, $60$ % of ${S O}_{2}$ is used up. The total number of moles in the vessel at equilibrium is:

A

10.0

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B

8.5

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C

10.5

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D

3.9

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Solution

The correct option is D $8.5$
The reaction is depicted as:
${S O}_{2} + \frac{1}{2} O_{2} ⟷ {S O}_{3}$

We have initially:
${S O}_{2} = 5$ mol
$O_{2} = 5$ mol
${S O}_{3} = 0$ mol

At equilibrium :

60% $S O_{2}$ is used up-
${S O}_{2}$ = $5 - 5 (0.6)$ = $2$ mol

$O_{2} = 5 - 5 (0.3)$ = $3.5$ mol

${S O}_{3} = 5 (0.6) = 3$ mol

Therefore, total moles of ${S O}_{2}$ , $O_{2}$ and ${S O}_{3}$ at equilibrium $= 8.5$ mol

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