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A 40 mL of a mixture of $H_2O$ and $O_2$ was placed in a gas burette at $18^{\large\circ}C$ and 1 atm P. A spark was applied so that the form a pure gas had a volume of 10mL at $18^{\large\circ}C$ and 1 atm P . If the remaining gas was $H_2$, what was the initial mole $\%$ of $H_2$ in mixture?


1 Answer

$2H_2 + O_2\rightarrow 2H_2O(l)$
Volume before reaction of $H_2O$ = a
Volume before reaction of $O_2$ = b
Volume before reaction of $2H_2O$ = O
Volume after reaction of $H_2O$ = (a-2b)
Volume after reaction of $O_2$ = O
Since at constnt P and T gases react in their volume ratio
a + b = 40
a - 2b = 10
$\therefore$ a = 30mL
b = 10 mL
Mole $\% of H_2 = Volume \% of H_2 = \large\frac{30}{40}\times100$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= 75$
Hence answer is (b)
answered Feb 23, 2014 by sharmaaparna1

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