1.

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)

Answer»

20 days
25 days
26 days
24 days

Solution :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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