Check my work Be sure to answer all parts. Kerosene, a common space-heater fuel,
ID: 1035350 • Letter: C
Question
Check my work Be sure to answer all parts. Kerosene, a common space-heater fuel, is a mixture of hydrocarbons whose average formala is CyHze (a) Determine the balanced equation, using the simplest whole-aumber coefficients, for the complete combustion of kerosene to gases. Include physical states of each reactant and product. (b) If AH".-1.50 101 kJ for the combustion equation as written in part (a) determine AH?of kerosene k.J/mol (c) Calculate the heat produced by the combustion of 0.64 gal of kerosene (d of kerosene 0.749 g/mL). kJ (d) How many gallons of kerosene must be burned for a kerosene furnace to produce 1,075 Bru (1 Btu 1.055 kJ)? gallons K Prey20 of 30Explanation / Answer
Ans. #a. Balanced reaction for Kerosene combustion:
C12H26 + 18.5 O2 -----------> 12 CO2 + 13 H2O - per mol kerosene
2 C12H26 + 37 O2 -----------> 24 CO2 + 26 H2O - Reaction 1
Reaction 1 is the balanced reaction with whole number coefficients.
#b. In part a, the reaction involved combustion of 2 mol kerosene.
Now,
dH0f = (-dH0rxn) / moles of kerosene reacting
Or, dH0f = -(-1.50 x 104 kJ) / 2 mol
Hence, dH0f = 7.5 x 103 kJ/mol
#c. Amount of kerosene combusted = Volume x Density
= 0.64 gallon
= (0.64 gallon x 3785.41 mL gallon-1) x (0.749 g/ mol)
= 1814.5741376 g
= 1814.5741376 g / 170.33844 g mol-1
= 10.653 mol
Now,
Amount of heat produced = dH0rxn per mol x Moles of kerosene
= (-1.50 x 104 kJ / 2 mol) x 10.653 mol
= 7.99 x 104 kJ
#d. Required moles of kerosene = Total heat to be released / dH0rxn per mol
= (-1075 x 1.055 kJ) / (-1.50 x 104 kJ / 2 mol)
= 0.15122 mol
Note: The –ve sign of dH indicates release of heat.
# Required mass of kerosene = Required moles x MW
= 0.15122 mol x 170.33844 g mol-1
= 25.758 g
Now,
Required volume = Mass / Density
= 25.758 g / 0.749 g mL-1
= 34.390 mL
= 34.390 mL / 3785.41 mL gallon-1
= 9.085 x 10-3 gallons.