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300 ml of a gaseous hydrocarbon when burnt in excess of O 2 gave 2.4 lit. of CO 2 and 2.7 lit. of water vapour under same condition. The molecular formula of hydrocarbon is :– (1) C 4 H 8 (2) C 8 H 18 (3) C 6 H 14 (4) C 8 H 16

300 ml of a gaseous hydrocarbon when burnt in excess of O2 gave 2.4 lit. of CO2 and 2.7 lit. of water vapour under same condition. The molecular formula of hydrocarbon is :–
(1) C4H8
(2) C8H18
(3) C6H14
(4) C8H16

Grade:12

1 Answers

Naveen Kumar
askIITians Faculty 60 Points
6 years ago
The balenced equation can be given as:
CnHm...+(n+m/4)O2............>nCO2.....+(m/2)H2O
300ml..........excess.................0..............0.....
0...............................................2.4L.........2.7L.......
As in same condition, the ratio of moles or the ratio of the volumes of gases would be same.
so the molar ratio of CO2 and water will be=2.4/2.7=n/(m/2)=2*(n/m) =8/9
Hence n/m=4/9
So empirical formula= C4H9
From equation we can see that the volume ratio of CO2 and hydrocarbon=2400/300=molar ratio=n/1=8
hence m=9*8/4=18
Hence (2) is the option.

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