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Q.

A gaseous hydrocarbon required 6 times its own volume of O2  for complete oxidation and produces 4 times its volume of  CO2.  Then the formula is, 
 

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a

Butane

b

Butene

c

Propene

d

Propane

answer is B.

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Detailed Solution

CxHy+(x+y/4)O2→xCO2+y/2H2O(ℓ)

1vo1 6=(x+y/4)  4=x

   6 times of (H.C)   4 times of (H.C)

x+y/4=6⇒4+y/4=6⇒y/4=2;y=8

∴C4H8

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