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_(2)H_(12)`
B. `C_(4)H_(8)`
C. `C_(4)H_(10)`
D. `C_(3)H_(6)`

Answer 1

Correct Answer - A
`underset(15 ml)(C_(x)H_(y)+(x+(y)/(4)))O_(2)(g) rarr xCO_(2) (g)+(y)/(2)H_(2)O(l)`
Volume of `O_(2)` used `= (20)/(100) xx 375 = 75 ml`
Volume of air remaining = 300 ml
Total volume of gas left after combustion = 330 ml
Volume of `CO_(2)` gases after combustion = 330 - 300 = 30 ml
`{:(C_(x)H_(y)+(x+(y)/(4))O_(2)(g) rarr xCO_(2) (g)+(y)/(2)H_(2)O(l)),("15 ml 75 ml 30 ml"):}`
`(x)/(1) = (30)/(15) = rArr x = 2`
`(x+(y)/(4))/(1) = (75)/(15) rArr x + (y)/(4) = 5 rArr y = 12 rArr C_(2)H_(12)` Confirmed