1.

A cylinder of gas supplied by Bharat Petroleum is assumed to contain 14 kg of butane. If a normal family requires 20,000 kJ of energy per day for cooking, butane gas in the cylinder last …….. days (Delta H_(C) of C_(4)H_(10) = - 2658 kJ per mole)

Answer»

15 days
20 days
50 days
32 days

Solution :MOLAR mass of butane, `C_(4)H_(10) = 12 xx 4 + 1 xx 10 = 58 g mol^(-1)`
58 g of butane gives 2658 KJ of HEAT energy.
14 kg of butane will give heat energy
`= (2658)/(58) xx 14 xx 10^(3)`
`= 641.5862 xx 10^(3) kJ`
Daily energy requirement for cocking = 20.000 kJ
`= 2 xx 10^(4) kJ`
No. of days CYLINDER will last `= (641.5862 xx 10^(3) kJ)/(2 xx 10^(4) kJ "day"^(-1))`
= 32.08 days.


Discussion

No Comment Found

Related InterviewSolutions