00:01
This problem is a little long, but let's get started.
00:03
The automobile fuel called e85 is 85 % ethanol, which i'll abbreviate e -t -o -h from now on, and 15 % octane, which we will put as c -8h -18.
00:28
You can use that in flex fuel vehicles.
00:31
It is not as chalkful of energy as pure gasoline.
00:39
So gasoline is a consistency of octanes that has an average of 5 ,400 kilojoules per mole.
00:51
And gasoline, this is gasoline, and gasoline has a density of 0 .70 grams per milliliter.
01:01
The density of ethanol is 0 .79 grams per millimeter.
01:10
And we're going to calculate, from the balanced chemical equation, how much ethanol.
01:16
Energy is produced by combustion, breathing energy by the letter's energy of one liter of each fuel.
01:35
We're already given the kilojoules per mole for gasoline or octane.
01:40
We need to find the ethanol.
01:43
And i found this on a different problem.
01:45
I found this, if you want to go look, i've also solved problem 29.
01:50
And i'm just going to write this one down quickly.
01:53
So i'm not going to be talking on this.
01:59
I'll talk a little bit.
02:01
This is the chemical formula for ethane, undergoing combustion.
02:07
This is the balanced chemical equation.
02:12
I'm going to apply hessis law to find my heat for the reaction.
02:32
I have a coefficient of two in front of my co2, and that's negative 393 .5 kilojoules per mole.
02:43
Add to that the coefficient of 3 times gas h2o is negative 241 .8 kilojoules.
02:56
And we'll subtract from that coefficient of 1 times ethanol is negative 27 .7 .0.
03:10
And of course, water, or excuse me, oxygen is zero.
03:20
For the reaction, once i do my math on this, i get, negative 1, 2, 3, 4 .7 kilojoules.
03:33
So here are my two values that i'm going to be using.
03:38
Okay.
03:42
Next, i'm going to do two calculations.
03:44
I'm going to switch colors for this.
03:45
I'll do my gasoline.
03:48
And for gasoline, i've got one liter of each.
03:50
So i'm just going to start with 1 ,000 milliliters.
03:55
And i'm going to use the density of gasoline with 0 .70 grams per milliliter.
04:03
And i was given 5 ,400 kilojoules per mole, and i've got to get rid my grams here, which is 114 .21 grams per mole.
04:28
So my grams cancel, my milliliters cancel, and my moles cancel.
04:34
So here i'm going to get 3 .31 times 10 to the 4 kilojoules per liter.
04:45
And now let's do ethanol, 1 ,000 milliliters.
04:54
We were given 0 .79 grams per milliliter, a little higher density, and we just calculate negative 1, 2, 3, 4 .8 kilojoules per mole, and the molar mass is 46 .07 grams per mole...