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» <html><body><p>15 days<br/><a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a> dyas<br/>50 days<br/>40 days</p>Solution :Molar mass of <a href="https://interviewquestions.tuteehub.com/tag/butane-906337" style="font-weight:bold;" target="_blank" title="Click to know more about BUTANE">BUTANE</a>, `C_(4)H_(10) = <a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a> xx 4 + 1 xx 10 = 58 g"mol"^(-1)` <br/> 58 g of butane gives 2658 kJ of heat energy . <br/> 14 kg of butane will give heat energy <br/> `= (2658)/(58) xx 14 xx 10^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` <br/> `= 641.5862 xx 10^(3)` kJ <br/> <a href="https://interviewquestions.tuteehub.com/tag/daily-430694" style="font-weight:bold;" target="_blank" title="Click to know more about DAILY">DAILY</a> energy requirement for cooking = 20, 000 kJ= `2 xx 10^(4) kJ` <br/> No. of days cylinder will last = `(641.5862 xx 10^(3)kJ) /(2 xx 10^(4) kJ"day^(-1))` = 32.08 days.</body></html>