(a) A cylinder of gas is assumed to contain 11.2 kg of butane. If a normal family needs 20,000 kJ of energyper dayfor cooking , how long will the cylinder last ? Given that the heat of combustion of butane is 2658 kJ mol^(-1) (b) If the air supply of the burner is sufficient (i.e., you havea yellow instead of a blue flame), a portion ofthe gas escapes without combustion. Assuming that 33% of the gas is wasted due to this inefficiency , how long would the cylinder last ?

(a) A cylinder of gas is assumed to contain 11.2 kg of butane. If a no

1.	(a) A cylinder of gas is assumed to contain 11.2 kg of butane. If a normal family needs 20,000 kJ of energyper dayfor cooking , how long will the cylinder last ? Given that the heat of combustion of butane is 2658 kJ mol^(-1) (b) If the air supply of the burner is sufficient (i.e., you havea yellow instead of a blue flame), a portion ofthe gas escapes without combustion. Assuming that 33% of the gas is wasted due to this inefficiency , how long would the cylinder last ?
Answer» Solution :(a) 1 Mole `C_(4)H_(10) = 58 g ``:. ` Heat produced from `11200 g = ( 2658) /( 58) xx 11200 = 513268.9kJ` No.of days which it will LAST` = 513268.9 // 20,000= 25.7 `days `= 26 ` days (b) After wastage, heat available `= ( 67)/( 100) xx 513268.9 = 343890J` No. of days for which it will last `= 343890 // 20000 = 17` days