The automobile fuel called E85 consists of 85 % ethanol and 15 % gasoline. E85 c

Question

The automobile fuel called E85 consists of 85 % ethanol and 15 % gasoline. E85 can be used in so-called "flex-fuel" vehicles (FFVs), which can use gasoline, ethanol, or a mix as fuels. Assume that gasoline consists of a mixture of octanes (different isomers of C8H18), that the average heat of combustion of C8H18(l) is 5400 kJ/mol, and that gasoline has an average density of 0.70 g/mL. The density of ethanol is 0.79 g/mL.

A note from your instructor:

For Part A, start with the given L of gasoline and use the density of the gasoline as well as the molar mass of  C8H18 to determine the moles of gasoline. Once you have found the moles of gasoline you will use the provided heat of combustion to determine the energy produced. You do not need to use Appendix C because you are using the provided heat of combustion.

For Part B, do the same with the ethanol. You will use the heat of combustion for ethanol from the extra credit homework problem 5.79. If you did not do the that homework problem, you will need to write the balanced equation for the combustion of ethanol paying attention to the phases. Use the phase of gas for the water. Then calculate the heat of the combustion reaction from the heats of formation in Table C.

For Part C, you will need to use a weighted average. This is the same formula we used to determine the average atomic mass.

For Part D, you are starting with the given gallons of gasoline and will need to use the heat of combustion produced for the given liters of gasoline as a conversion factor to determine the energy produced from the given gallons. The heat of combustion produced from the liters of gasoline is from Part A. Now take the energy produced from the gallons of gas and use that as the given. Using the heat of combustion produced for the liters of E85 from Part C, determine the gallons of E85 necessary to produce that much energy.

For Part E, you will use the gallons of gasoline from part D and the gallons of E 85 from part D.

1. By using the information given as well as data in Appendix C, calculate the energy produced by combustion of 3.5 L of gasoline.

Explanation / Answer

ANSWER

C8H18 (l) + 25/2O2 ------> 8CO2 (g) + 9H2O (g)    H = 5400 kJ/mol54

5400 kJ of heat is released when one mole or 114g of C8H18 undergoes combustion.

3.5L of gasoline = 3500mL because 1L = 1000mL

mass of 3500mL gasoline = density X volume

mass of 3500mL gasoline = 0.70 X 3500 = 2450g

Energy released by one gram of gaoline = 5400 / 114 = 47.37KJ

Energy released by 2765g of gaoline = 47.37 X 2450 = 116056.5KJ

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