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Enthalpy of Combustion

A cylinder of...

Question

A cylinder of gas is assumed to contains 11.2 kg of butane. If normal family needs 20,000 kJ of energy per day for cooking, how long will the cylinder last if the enthalpy of combustion, $Δ H = - 2658 k J / m o l e$ for butane.

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Solution

Everyday they are using 20,000 kJ of energy.
$∴$ 1 day $⟶ (\frac{20, 000}{2658}) \times$ weight of butane
Means 1 day consumes $(\frac{20, 000}{2658}) \times$ 58 grams of butane
Let 'x' be the number of days to complete 11.12 kg=11120 grams of butane.
1 day $⟶$ 436.42 grams
x days $⟶$ 11120 grams
$∴ x = \frac{11120 \times 1}{436.42} = 25.48 d a y s ≃ 25.5 d a y s$
$∴$ The cylinder will consume within 25.5 days.

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