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An LPG cylinder is assumed to contain 11...

An LPG cylinder is assumed to contain 11.2 kg of butane `(C_(4)H_(10))`. If a family needs 20000 kJ of energy per day, then cylinder will last for nearly how many days ? (Given that `Delta H` for combustion of butane is - 2658 kJ/mole)

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`C_(4)H_(10)(g) + 13/2 O_(2)(g) to 4CO_(2)(g) + 5H_(2)O(g), DeltaH = -2658` kJ
Molecular weight of `C_(4)H_(10) = 58` g/moles.
`therefore 11.2` kJ of butane on combustion produces `(2658 xx 10^(3) xx 11.2)/58` J
Family needs 20,000 kJ of energy per day.
`therefore` No. of days `=(2658 xx 10^(3) xx 11.2)/(58 xx 20,000) = 25.66` days.
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