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When a certain amount of ethylene was combusted, $6226 kJ$ heat was evolved. If heat of combustion of ethylene is $1411 kJ,$ the volume of $O_{2}$ (at $STP$ ) that entered into the reaction is:

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296.5 litres

300.3 litres

6226 x 22.7 litres

22.7 litres

answer is A.

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Detailed Solution

$C_{2} H_{4} + 3 O_{2} \to 2 {CO}_{2} + 2 H_{2} OΔH = - 6226 kJ$

There are moles of $O_{2}$ used in the reaction.

$n_{ethylene} = \frac{6226}{1411} = 4.41$

$V_{O_{2}} = 4.41 \times 3 \times 22.7 L$

= $296.5 litres$

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