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?
-
20cm\(^3\)
- 40cm\(^3\)
- 60cm\(^3\)
- 50cm\(^3\)
The correct answer is: 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 .