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?

Options

A. 20 days
B. 21 days
C. 26 days
D. 30 days

Answer

C

Explanation

`Delta H_(c)` for butane `(C_(4)H_(10))" per mole "=2658 kJ`
`Delta H_(c)" for butane per "g=(2658)/("Molecular weight")=(2658)/(58)`
`=45.8 kJ`
[Molecular weight for butane `(C_(4)H_(10))=58`]
`therefore` Energy released from 11.2 kg of butane
`=45.8xx11.2xx1000=5,12,960 kJ`
20,000 kJ energy is consumed in one day `therefore 5,12,960 kJ` energy will be consumed `(5,12,960)/(20,000)=25.6 ~~ 26` days
`therefore 26` days