C\(_2\)H\(_{4(g)}\)  +  3O\(_{2(g)}\) → 2CO\(_{2(g)}\)  +  2H\(_2\)O\(_{(g)}\) The above equation represents the combustion of ethene.If 10cm\(^3\) of...

C\(_2\)H\(_{4(g)}\) + 3O\(_{2(g)}\) → 2CO\(_{2(g)}\) + 2H\(_2\)O\(_{(g)}\) The above equation represents the combustion of ethene.If 10cm\(^3\) of...

JAMB 2024

C\(_2\)H\(_{4(g)}\) + 3O\(_{2(g)}\) → 2CO\(_{2(g)}\) + 2H\(_2\)O\(_{(g)}\)

The above equation represents the combustion of ethene.If 10cm\(^3\) of ethene is burnt in 50cm\(^3\) of oxygen, what would be the volume of oxygen that would remain at the end of the reaction?

A. 20cm\(^3\)
B. 40cm\(^3\)
C. 60cm\(^3\)
D. 50cm\(^3\)

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Correct Answer: Option A

Explanation

Gay Lussac’s Law of Combining Volumes states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the product(s) formed if gaseous, provided the temperature and pressure remain constant.

C\(_2\)H\(_{4(g)}\) + 3O\(_{2(g)}\) → 2CO\(_{2(g)}\) + 2H\(_2\)O\(_{(g)}\)

1 mole : 3 moles

Total volume required: 10 cm\(^3\) 50 cm\(^3\)

Reacted Volume: 10 cm\(^3\) 30 cm\(^3\)

Residual volume: 0 (50 - 30) = 20 cm\(^3\)

There is an explanation video available below.

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