Kerosene, a common space-heater fuel, is a mixture of hydrocarbons whose average
ID: 943229 • Letter: K
Question
Kerosene, a common space-heater fuel, is a mixture of hydrocarbons whose average formula is C12H26.
a) Determine the balanced equation, using the simplest whole-number coefficients, for the complete combustion of kerosene (C12H26) to gases. Include physical states of each reactant and product.
b) If Detla H rxn= -1.50 * 10^4 kJ for the combustion equation as written in part (a). determine delta H t of kerosene in kJ/mol.
c) Calculate the heat produced by the combustion of 0.82 gal of kerosene (density of kerosene= .749 g/mL)
d) How many gallons of kerosene must be burned for a kerosene furnace to produce 1,143 Btu (1 Btu= 1.005 kJ)?
Explanation / Answer
a) combustion will be
2C12H26 (l) + 37 O2 (g) -> 24CO2 (g) + 26H2O (l)
b) DeltaH rxn = Delta H of formation of products - DeltaH of formation of reactanst
we know that
Delat H of formation of CO2 = 393.51kJ/mol and
Delta H of formation of H2O = 241.82 kJ/mol
Delta H of formation of O2 = 0
-1.50 * 10^4 kJ = 24 X Delta H of formation of CO2 + 26 X Delta H of formation of H2O - 2 X Delta h of formation of C12H26
2 X Delta h of formation of C12H26 = -15731.56 + 15000 = -731.56 KJ
Delta h of formation of C12H26 =-365.78 KJ / mole
c) Heat produced by combustion of 2 moles of kerosene = -15000KJ
so heat produced by 340 (2 X 170) grams = -15000 KJ
so heat produced by 1 gram = 44.12 KJ
Now 1 gallon = 3785.41 mL
0.82 gallons = 3104 mL now 3104 mL = Volume X density grams = 3104 X 0.749 grams = 2324.89 grams
so heat produced by 2324.89 grams = 44.12 X 2324.89 KJ =102574 KJ
d) To produce 1,143 Btu of heat
We know that 1 BtU = 1.005 KJ
So 1143 Btu = 1148.72 KJ
102574 KJ Heat produced by 0.82 gallons
so 1KJ will be produced by 0.82 / 102574 gallons
so 1148.72 KJ will be produced by 0.82 X 1148.72 / 102574 gallons = 0.00918 gallons
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