A cylinder of gas is assumed to contain ...

A cylinder of gas is assumed to contain 11.6 kg of butane `(C_4H_10)`. If a normal family needs 20000kJ of energy per day, then the cylinder will last for : (Given that ΔH for combustion of butane is – 2658kJ)

20 days

25 days

26 days

24 days

Text Solution

Verified by Experts

The correct Answer is:

1 mole of butane `(C_(4)H_(10)) = 58g`.
58g. of butane = 2658kJ
11200g. of butane `=(2658)/(58) xx 11200 -= 513269kJ`
20000 kJ of energy required =1 day
513269 kJ of energy required `=(513269)/(20000)= 25.66` day. `~~` 26 days (approx).

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