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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...

Chemistry
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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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Explanation Video

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