LPG is a mixture of n-butane & iso-butane. The volume of oxygen needed to burn 1 kg of LPG at NTP would be:

2240 L

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2690 L

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1000 L

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500 L

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Solution

The correct option is A 2690 L
reaction:
$C_{4} H_{10} + \frac{13}{2} O_{2} \to 4 C O_{2} + 5 H_{2} O$
$58 g m \frac{13}{2} \times 32$
$58 g C H_{4} r e q u i r e d v o l u m e o f O_{2} a t N T P = \frac{13}{2} \times 24 l i t$
so, 1kg of $C H_{4}$ requires $= \frac{13 \times 1000}{2 \times 58} \times 24 l i t = 2690 L$

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S O_{3}

S O_{3}

The rate of effusion of LPG (a mixture of n-butane and propane) is 1.25 times that of $S O_{3}$ . Hence, mass fraction of n-butane in LPG is :

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