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At 300 K and 1 atm, 15 mL of a gaseous h...

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_2H_12`

B

`C_4H_8`

C

`C_4H_10`

D

`C_3 H_6`

Text Solution

Verified by Experts

The correct Answer is:
A
`C_xH_y(g) + (x + y/4) O_2(g) to xCO_2(g) + y/2 H_2O(l)`
15mL
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 - 300 = 30 mL`
`C_x H_y(g) + (x + y/4) O_2(g) to xCO_2 (g) + y/2H_2O(l)`

`rArr y = 12`
`rArr C_2H_12`.
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