A cylinder of gas supplied by Bharat Petroleum is assumed to contain `14 kg` of butane. 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. `15` days
B. `20` days
C. `50` days
D. `40` days

Answer

D

Explanation

Molar mass of butane, `C_(4)H_(10)=12xx4+10`
`=48+10=58 g mol^(-1)`
`58g` of butane gives `2658 kJ` of heat energy
`:. 14 kg` of butane will give heat energy
`=(2658kJxx(14xx10^(3)g))/(58g)=641.5862xx10^(3)kJ`
Daily energy requirement for cooking
`=20,000 kJ`
`=2xx10^(4) kJ "day"^(-1)`
`:. ` No. of days cylinder will last
`=(641.5862xx10^(3)kJ)/(2xx10^(4)kJ"day^(-1))=32.08"days"`