A cylinder of gas supplied by Bharat Petroleum is assumed to contain `14 kg` of butance. If a normal family requires 20,000 kJ of energy per day for cooking, butane gas in the cylinder lasts
`(Delta_(C ) H^(@)` of `C_(4) H_(10) = - 2658 kJ mol^(-1))`

Options

A. 20
B. 50
C. 40
D. 32

Answer

D

Explanation

Number of days cylinder will last
`("Energy available from 14 kg butane")/("Energy requirement per day")`
Energy released out by the combustion of 14 kg of butane is
= (Number of moles of butane)
`((14 xx 10^(3)g)/(58 g mol^(-1))) (2658 kJ mol^(-1))`
`= 641.6 xx 10^(3) kJ`
Note that `n_(C_(4)H_(10)) = Mass_(C_(4)H_(10))//"Molar mass"_(C_(4)H_(10))` and molar
mass of `C_(4) H_(10` is `58 g mol^(-1)`.
Thus, the number of days cylinder will last
`= (641.6 xx 10^(3) kJ)/(20 xx 10^(3) kJ day^(-1))`
`=32.08 days`