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A cylinder of gas is assumed to contain 14 kg of butane. A normal family requires 20,000 kJ of energy per day for cooking. How many days will the butane gas in the cylinder last?


(a) 15 days
(b) 20 days
(c) 50 days
(d) 32 days
Can you answer this question?
 
 

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Answer: 32 days
Molar mass of butane, $C_4H_10 = 12 \times 4 + 10 = 58 g mol^{-1}$
58 g of butane gives 2658 kJ of heat energy.
$\therefore$ 14 kg of butane will give heat energy = $\frac{2658 kJ \times (14 \times 10^3 g)}{58 g} = 641.5862 \times 10^3 kJ$
Daily energy requirement for cooking = 20,000 kJ = $2 \times 10^4$ kJ $day^{-1}$
$\therefore$ Number of days cylinder will last = $\frac{641.5862 \times 10^3 \text{ kJ }}{2\times 10^4 \text{ kJ }day^{-1}}$ = 32.08 days
answered Feb 26, 2014 by mosymeow_1
 

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