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Fifteen milliliters of gaseous hydrocarb...

Fifteen milliliters of gaseous hydrocarbo (A) was required for complete combustion 357 ml of air (21 % oxygen by volume) and gaseous products occupied 327 ml (all volumes being measured at STP).

The molecular formula of the hydrocarbon (A) is:

A

`C_(2)H_(6)`

B

`C_(2)H_(4)`

C

`C_(3)H_(6)`

D

`C_(3)H_(8)`

Text Solution

Verified by Experts

The correct Answer is:
D
Let the formula of hydrocarbon A is CxHy.
`C_(x)H_(y)(g) +(x+(y)/(4))O_(2) (g) rarr xCO_(2)(g) + (y)/(2)H_(2)O(I)`
1 ml `(x + (y)/(4))ml` x ml ----
15 ml `(x + (y)/(4))ml` 15x ml ----
Volulme of `O_(2) =(357xx21)/(100) =75 ml` , volume of `N_(2)` `=357 - 75 =282 ml`
Volume of `CO_(2) =327 - 282 =45 ml`
`:. 15x=45, x=3`
`:. ({:(15(x+y/4)=75),(x+y/4=5):})`
`3 + (y)/(4) =5` , On solving, we get y = 8.
Formula of (A) `= C_(3)H_(8)`
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