Question

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 days
B. 25 days
C. 26 days
D. 24 days

Answer 1

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.