a. A cylinder of gas is assumed to contain `11.2 kg` of butane. If a normal family needs `20000kJ` of energy per day for cooking, how long will the cylinder last? Given that the heat of combustion of butane is `2658 kJ mol^(-1)`.
b. If the air supply of the burner is insufficient (i.e. you have a yellow instead of a blue flame), a portion of the gas escape without combustion. Assuming that `33%` of the gas is wasted due to this inefficiency, how long would the cylinder last?

Explanation

(a) Molecular formula of butane = `C_(4)H_(10)`
Molecular mass of butane = `4 xx 12 + 10 xx 1 = 58`
Heat of combustion of butane = `2658 kJ mol^(-1)`
1 mole of 58 g of butane on complete combustion give haet = 2658 kJ
`therefore 14 xx 10^(3)g` of butane on complete combustion give heat = `(2658 xx 14 xx 10^(3))/(58)`
` = 641586`
The family needs 20,000 kJ of heat per day 20,000 kJ of heat is used for cooking by a family in = 1 day 641586 kJ of heat will be used for cooking by a family in = `(641586)/(20000) = 32 days`
The cylinder will last for 32 days.
(b) 25 per cent of the gas is wasted due to inefficicency.This means that only 75% of butane gets combusted.Therefore, The energy produced by 75% combustion of butane = `(641586 xx 75)/(100) = 481190 kJ`
`therefore` the number of days the cylinder will last =`(481190)/(20000) = 24` days.