Question

A cylinder of cooking gas supplied by Indian Oil Corporation is assumed to contain 14 kg of butane (ΔHc for
C4H10 = - 2600 kJ mol-1). If a small family of three persons, requires 10,000 J of heat energy per day for
cooking, the gas in the cylinder would last for
A) 44 days
B) 54days
C) 72 days
D) 63 days

Answer 1

Molecular formula of Butane = C_{4}H_{10}
Molecular mass of butane = 4 X12 +10 X1 = 58
Heat of combustion of butane = 2600 kJ/mol
1 mole or 58 g of butane on complete combustion gives heat = 2600kJ
14 x {{10}^{3}} g of butane on complete combustion gives heat = 2600 x 14 x {{10}^{3}} /58 = 627586.206
The family needs 0000 j of heat per day.
10000 j of heat is used for cooking by a family in = 1 day
627586.206 kj of heat will be used for cooking by a family in =627586.206 / 10000 = 62.7 days






simple steps

Molar mass of butane (C4H10) = 12x4 + 1x10 = 58g/mol

14kg = (14x1000)g = 14000g

Moles of butane = Given mass/molar mass

= 14000/58

= 241.38 moles


∆HC for C4H10 = - 2600 kj mol-1

This means 1 mole of butane gives 2600kJ of energy

241.38 moles will give = 241.38 x 2600 = 627588 J

= 627,588 J

The number of days = 627,588/10,000

= 62.75 days
= 63

The cylinder will last for 63 days