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Hydrocarbons

Question
CBSEENCH11008044

At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is:

  • C3H8

  • C4H8

  • C4H12

  • C3H6

Solution

C.

C4H12

straight C subscript straight x straight H subscript straight y space left parenthesis straight g right parenthesis space plus space stack open parentheses straight x space plus straight y over 4 close parentheses with 75 space ml below straight O subscript 2 space left parenthesis straight g right parenthesis space rightwards arrow with space on top space stack xCO subscript 2 space left parenthesis straight g right parenthesis with 30 space ml below space plus space straight y over 2 straight H subscript 2 straight O space left parenthesis straight l right parenthesis
straight O subscript 2 space used space equals space 20 percent sign space of space 375 space equals space 75 space mL
Inert space part space of space air space equals space 80 percent sign space of space 375 space equals space 300 space ml
Total space volume space of space gases
equals space CO subscript 2 space plus space inert space part space of space air
equals space 30 plus 300 space equals space 330 space mL
straight x over 1 space equals space 30 over 15 rightwards double arrow space straight x space equals space 2

fraction numerator straight x space plus begin display style straight y over 4 end style over denominator 1 end fraction space equals 75 over 15
rightwards double arrow space straight x space plus straight y over 4 space equals 5
rightwards double arrow space straight x space equals space 2 comma space straight y space equals 12 space rightwards double arrow space straight C subscript 2 straight H subscript 12