Question

15mL of a gaseous hydrocarbon required for complete combustion, 357mL air contianing 21% oxygen by volume. The gaseous product occupied 327mL. If all the volumes are measured at STP. The molecular formula of hydrocarbon will be:
A. `C_(2)H_(6)`
B. `C_(4)H_(10)`
C. `C_(3)H_(8)`
D. `C_(2)H_(4)`

Answer 1

Correct Answer - C
`"Volume of oxygen in air"=(21)/(100)xx357=74.97`
=75mL
Volume of air excluding oxygen=357-75=282mL
Let hydrocarbon be `C_(x)H_(y)`. The combustion may be represented as
`C_(x)H_(y)+(x+(y)/(4))O_(2) to xCO_(2)+(y)/(2)H_(2)O`
115mL of hydrocarbon uses `15mL "hydrocarbon uses " 15(x+(y)/(4))O_(2) "for combustion"`
`therefore 15(x+(y)/(4))=75`
`"Volume of " CO_(2) "after combustion "=327-282=45mL`
`therefore 15x=45`
Putting the value of x in eqn. (i) we get, y=8
Moleular fromula of hydrocarbon=`C_(3)H_(8)`