The automobile fuel called E85 consists of 85% ethanol and 15% gasoline. E85 can be used in the 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) By using the information given as well as data in Appendix C, compare the energy produced by combustion of 1.0 L of gasoline and of 1.0 L of ethanol.

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Hi everyone for this problem, we're told to assume that the average heat produced by the combustion of diesel is 44.80 kg jules program, compare the energy produced by five liters of diesel and five liters of methanol. If the heat of combustion of methanol is negative, 726 kg joules per mole. The density of diesel and methanol is 0.85 g per milliliter and 0.79 g per milliliter respectively. So for this problem, we're comparing two values. We're comparing diesel and methanol and we're starting off with the volume. So we're told that we have five liters of each and we need to compare the Energy produced by both. Okay, so our goal is to go from 5l of both to energy produced by both. So let's go ahead and get started. So let's start off with methanol. Okay, so methanol is c H 30. H. And we need to know this because we're going to need to know the molar mass of methanol. So let's start off with what they tell us. They tell us we have five leaders of methanol and we want to go from leaders of methanol to kill a jewels produced energy produced. So let's go ahead and we're going to need the density of methanol and they tell us that density is 0.79 g per mil leader. So we need to first go from leaders of methanol to mill leaders. So we say that in one middle leader, We have 10 to the negative three leaders. Okay, so now we were in units of mill leaders and we can use the density that was given in the problem, they tell us that The density is .79 g per mil leader and that's our Unit conversion here. So in one millimeter There is 0.79 g of methanol. Okay, so now we're in grams of methanol and remember we want to go to energy. And what we can do now is use our molar mass of methanol to go from grams to moles. So in one mole of methanol, our molar mass of methanol Is 32.05 g. Okay, so let's just make sure our units are canceling properly here. So our leaders of methanol cancel. Mill leaders cancel grams of methanol cancel. Now we're in molds of methanol and we want to go to kill a jules. And what we can use here is we're told in the problem that methanol produces 726 kg jewels per mole. Okay, so that is the conversion that we can use to go from moles to kill a jewels. So in one mole of methanol we have 726. Kill a jewels of energy produced. So now our molds of methanol cancel and we're left with killer jewels. So let's go ahead and calculate that number. So we have eight 0.95 times 10 to the four killer jewels of energy produced by this five liters of methanol. And the question is asking us to compare the energy produced between methanol and diesel. So now we need to do the same thing. But for diesel and then we can compare the two. Okay, so for diesel, Our formula is C12 h 23. So we're told that we have five leaders and we want to do the same thing. We want to go from leaders of diesel to energy of diesel produced. So we need to convert this to mill leaders in one millimeter. We have 10 to the negative three leaders and were also given the density of diesel. So now we can use that here And one militia leader of diesel, we have 0.85 grams of diesel. This is our density. Okay, and For diesel, the problem tells us that the combustion of diesel is 44.80 killer jewels per gram. So we can use that here because if you take a look at our units are leaders cancel our middle leaders canceled. And now we're in grams for methanol, we were in moles and we used the combustion of methanol to solve. But here for diesel we have grams. So we're going to use this combustion of diesel at 44.80 kg joules per gram to figure out our energy produced. So when we plug that unit conversion in for every one g Of diesel, we have 44.80 kila jewels of energy produced. So if you see here are grams cancel and we're left with kila jewels of heat. And so this gives us 1.90 times 10 to the 5th killer jewels. Okay, so these are our two energies, it says to compare the the energy produced by both, and these are our respective energies produced by both. That's the end of this problem. I hope this was helpful.

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Key Concepts

Heat of Combustion

Density and Volume Relationships

Combustion Reaction Stoichiometry