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Ideal Gas Equation

A 40 mL of a ...

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A 40 mL of a mixture of $H_{2}$ and $O_{2}$ was placed in a gas burette at $18^{o} C$ and 1 atm pressure. A spark was applied so that the formation of water was complete. The remaining pure gas had a volume of 10 mL at $18^{o} C$ and 1 atm P. If the remaining gas was $H_{2}$ , what was the initial mole % of $O_{2}$ in the mixture?

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Solution

$H_{2} + (\frac{1}{2}) O_{2} \to H_{2} O (l)$
Volume before reaction a b 0
Volume after reaction a-2b 0 2b
Since the reaction is studied at constant P and T
$a + b = 40; a - 2 b = 10$
$a = 30 m L ∴ b = 10 m L$
Mole % of $O_{2} =$ volume % of $O_{2}$
$= \frac{10}{(30 + 10)} \times 100 = 25$ %

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