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 to gases. Include physical states of each reactant and product.

(b) If ΔHorxn = −1.50 ×10⁴ kJ for the combustion equation as written in part (a),determine ΔHof of kerosene.

(c)  Calculate the heat produced by the combustion of 0.38 gal of kerosene (d of kerosene = 0.749 g/mL).

(d)  How many gallons of kerosene must be burned for a kerosene furnace to produce 1,329 Btu (1 Btu = 1.055 kJ)?

Answer 1

(a) The combustion of a hydrocarbon in the presence of oxygen yields carbon dioxide and water. So, the balanced chemical equation for the combustion of kerosene (CH26) is shown below: 2СН(1) +370, (g) >24C0, (8)+26H,0(€) (b) The heat of formation of a reaction can be expressed as дн - Хн; (products) - Z н; (reactants) (1) Here = AH Standard heat of formation of a reaction 2 Total standard heat of formation of products = >H2 (products) Total standard heat of formation of reactants = H (reactants) So, the standard heat of the combustion of kerosene can be given by using equation (1): дна - 24 molx H; (СО,) + 26 mol xн; (Н,О)-2 molx н; (C,Н;) -37 molxH2 (02)

Substitute all the standard values and the given value of AH in the above equation -1.50 x10 kJ 24 molx(-393.5 kJ/mol)+ 26 mol x(-241.8 kJ/mol)- X 2 molxH (CH6)-37 molx(0 kJ /mol) -1.50 x104 kJ= -9444 kJ-6286.8 kJ-2 molx H2 (C,H)+0 X 2 mol x H (C,н,) — -15730.8 kJ +15000 kJ -730.8 kJ н; (С,Н) = 2 mol н (С,Н) %- -365.4 kJ/ mol Hence, the heat of formation of the combustion of kerosene is -365.4kJ/mol (c) Density of kerosene =0.749 g /mL Volume of kerosene = 0.38 gal So, the volume of kerosene in milliliters will be: Volume of kerosene = 0 38 galy 3785.41 mL 1 gal {1 gal 3785.41 mL} X - 1438.46 mL

be determined by using the density formula The mass of kerosene can Mass Density Volume Mass of kerosene 0.749 g/mL 1438.46 mL Mass of kerosene 0.749 g /mL x1438.46 mL =1077.4 g According -1.50 x104kJ ofheat to the reaction stoichiometry, 2x170 g of kerosene will produce So, the heat produced by 1077.4 g ofkerosene is calculated as: Heat produced=-1.50 -104KJ-1077.4 g 340 g =-4.75x10 kJ Hence, the heat produced by the combustion of kerosene is -4.75x104 kJ (d) Since 1 Btu 1.055 kJ

Therefore 1329 Btu 1329 Btu x.055 kJ 1 Btu =1402.095 kJ According to the reaction stoichiometry, 340 g of kerosene will produce 1.50 X104KJ of heat So, the mass of kerosene that will produce -1402.095 kJ of heat will be determined as: 340 gx-1402.095 kJ) -1.50 x104kJ Mass of kerosene =31.78 g Substitute the above value in density formula to calculate the volume of kerosene: 31.78 g 0.749 g/mL = Volume of kerosene 31.78 g Volume of kerosene = 0.749 g/mL 1 gal =42.43 mLx {1 gal 3785.41 mL} X 3785.41 mL =0.01064 gal -1.06x10 gal

Hence, 1.06x102 gallons must be burmed to produce 1329 Btu ofheat.