Question

A cylinder of gas is assumed to contains 14 kg of butane. If a normal family needs 20,000 kJ of energy per day for cooking, The cylinder will last:
(Given : The enthalpy of combustion, $-\Delta H=-$2658 kJ / mole for butane).

• A. 24.33 days.
• B. 28.44 days.
• C. 29.33 days.
• D. 32 days.

Answer 1

The correct option is D 32 days.

Molecular formula of butane = $C_4{H}_{10}$

Molecular mass of butane $=4\times 12+10\times 1=5^8$

Heat of combustion of butane = $2658KJmo{l}^{-1}$

$1$ mole of $5^8$ g of butane on complete combustion given heat $={26}_{5}8KJ$

$14\times {10}^{3}g$ of butane on complete combustion gives heat =

$\frac{\left(2658\right)\times \left(14\right)\times {10}^3}{58}=641586$

The family needs $20,000$ KJ of heat for cooking per day.

641586 KJ of heat will be used for cooking by a family in

$\frac{641586}{20000}=32$ days