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A cylinder of gas is assumed to contain 11.2 kg of butane `(C_(4)H_(10))`. If a normal family needs 20000 kJ of energy per day, the cylinder will last in (given that `DeltaH` for combustion of butane is -2658 kJ)A. 20 daysB. 25 daysC. 26 daysD. 24 days |
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Answer» Correct Answer - C Cylinder contains 11.2 kg or 193.10 mole of butane. [`because` moleclar mass of butane=58] `because`Energy released by 1 mole of butanne=-2658 kJ `therefore` Energy released by 193.10 mole of butane `=-2658xx193.10` `=5.13xx10^(5)kJ` `therefore`Cylinder will last in `(5.13xx10^(5))/(20000)=25.66` or 26 days. |
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