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 for....days(DeltaH_(c) of C_(4)H_(10) = -2658 kJ per mole)

A cylinder of gas supplied by Bharat Petroleum is assumed to contain 1

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 for....days(DeltaH_(c) of C_(4)H_(10) = -2658 kJ per mole)
Answer» 15 days 20 dyas 50 days 40 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 cooking = 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.