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Gay Lussac's Law

When 180 gram...

Question

When 180 grams of glucose is subjected to combustion, the volume of $C O_{2}$ liberated at STP is

A

22.4 l

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B

67.2 l

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C

44 l

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D

134.4 l

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Solution

The correct option is D 134.4 l
The balanced equation for the combustion of glucose can be written as:
$C_{6} H_{12} O_{6} + 6 O_{2} \to 6 C O_{2} + 6 H_{2} O$
$∴$ 180g of $C_{6} H_{12} O_{6}$ produces = $6 \times 22.4$ = 134.4l of $C O_{2}$

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