Question

At `300 K` and `1 atm, 15 mL` of a gaseous hydrocarbon requires `375 mL` air containing `20% O_(2)` 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
A. `C_(3)H_(8)`
B. `C_(4)H_(8)`
C. `C_(4)H_(10)`
D. `C_(3)H_(6)`

Answer 1

`C_(x)H_(y)(g)+underset(75 mL)((x+(y)/(4)))O_(2)(g)rarrxCO_(2)underset(30 mL)((g))+(y)/(2)H_(2)O(l)`
`O_(2)` used `= 20%` of `375 = 75 mL`
Inert part of air `= 80%` of `375 = 300 mL`
Total volume of gases `= CO_(2)`+ Inert part of air
`= 30 + 300 = 330 mL`
`(x)/(1) = (30)/(15) rArr x = 2`
`(x+y)/(1) = (75)/(15) rArr x+(y)/(4) = 5`
`rArr x = 2, y = 12 rArr C_(2)H_(12)`