# A cylinder of gas supplied by a company is assumed to contain 14 kg

(i) A cylinder of gas supplied by a company is assumed to contain 14 kg of butane. If a normal family requires 20000 kJ of energy per day for cooking, how long will the cylinder last?
(ii) If the air supplied to the burner is insufficient, a portion of gets escapes without combustion. Assuming that 25% of the gas is wasted due to this inefficiency, how long will the cylinder last? (Heat of combustion of butane = 2658 kJ/mol.)

(i)Molecular formula of Butane = \$C_{4}H_{10}\$
Molecular mass of butane = 4 X12 +10 X1 = 58
Heat of combustion of butane = 2568 kJ/mol
1 mole or 58 g of butane on complete combustion gives heat = 2568kJ
14 x \${{10}^{3}}\$ g of butane on complete combustion gives heat = 2658 x 14 x \${{10}^{3}}\$ /58 = 641586
The family needs 20000 kj of heat per day.
20000 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
(ii)25 per cent of the gas is wasted due to inefficiency. This means that only 75% of butane gets combusted.Therefore,
the energy produced by 75% combustion of butane = 641586 x 75 /100 = 481190 kJ
the number of the cylinder will last = 481190 /20000 = 24 days.