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Ways of Representing the Concentration of a Solution

10 moles of ...

Question

10 moles of $S O_{2}$ and 15 moles of $O_{2}$ were passed over a catalyst to produce 8 moles of $S O_{3}$ . The ratio of moles of $S O_{2}$ and $S O_{3}$ in mixture is:

A

\frac{5}{4}

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B

\frac{1}{4}

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C

\frac{1}{2}

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D

\frac{3}{4}

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Solution

The correct option is B $\frac{1}{4}$
Reaction is:
$2 S O_{2} + O_{2} \to 2 S O_{3}$
Initial concentration of $S O_{2}$ =10 mol and $O_{2}$ =15 mol
At equillibrium, say $x$ mol of $O_{2}$ are consumed, therefore, $2 x$ of $S O_{3}$ will form as
$2 S O_{2} 10 - 2 x + O_{2} 15 - x \to 2 S O_{3} 2 x$
Since moles of $S O_{3}$ formed = 8 mol
$2 x = 8 o r x = 4$
$\frac{[S O_{2}]}{[S O_{3}]} = \frac{10 - 8}{8} = \frac{1}{4}$
Therefore option B is correct.

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