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:

Options

A. `C_(2)H_(6)`
B. `C_(2)H_(4)`
C. `C_(3)H_(6)`
D. `C_(3)H_(8)`

Answer

D

Explanation

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)`