Write the balanced chemical equation for the complete combustion of each of the following hydrocarbons found in gasoline:
c. 3-ethyltoluene

Question

Write the balanced chemical equation for the complete combustion of each of the following hydrocarbons found in gasoline:

c. 3-ethyltoluene

Accepted Answer

Welcome back. Let's take a look at our next question. What is the correct balanced chemical equation that represents the complete combustion of 135 tri methyl benzene, a gasoline additive hydrocarbon. Then we have four different balanced equations which we can evaluate as we figure out what this equation should be. So we need to start with figuring out the molecular formula of our given compound. So we have its name 135 tri methyl benzene. So we know we have a benzene ring, draw that here six carbon ring with its three double bonds and we have methyl groups on carbons, one three and five. So now we want to figure out we have only carbon and hydrogen. How many of each do we have? Well, we know we have a six carbon ring and then three carbons from our methyl groups. So we have c subscript nine of nine carbons, hydrogens could be a little trickier. But in this case, we know on a benzene ring, any unsubs carbon has just one hydrogen due to those double bonds. So we have only three uns substituted carbons in that ring. So that's three hydrogens. And then each of the metal groups has three hydrogens, they're all ch three groups. So if we count up our hydrogens, we have 123 methyl groups times three is nine and then three more from the ring. So we have a total of 12 hydrogens. So our molecular formula is C nine H 12. So we can actually eliminate all but one of our answer choices just by having this molecular formula. So on a test that would get me out of this question way faster. No balancing equations, anything like that. But we will go through this more thoroughly since this is a practice question. But just so we look at this and see how we would do it on a test choice. A has the molecular formula C six H 16 liquid. But there both the carbon and hydrogen numbers are incorrect. Choice B says C nine H 12. So that is correct. Choice C says C six H 12. So it doesn't have the right number of carbons. So both A and C the molecular formula of our product that's being combusted is incorrect. And then choice D has C nine H 16. So the right number of carbons, wrong number of hydrogens. So a choice D is not correct either. So again, on a test, I would choose B and move on. But we'll go through the whole thing here in the interest of being thorough. So we start our balanced equation with our hydrocarbon C nine H 12. We see from our answer choices, they all list our hydrocarbon as being liquid. And so we don't need to figure that out. But just as a matter of interest in general, a hydrocarbon with anywhere from 5 to 16 carbons is liquid at room temperature, then that will be combusting with our other reactant. Always in a case of combustion is 02 gas or oxygen gas, 02 and the gas state, I'm leaving a space in front of it for us to put the coefficient as we balance our equation, it's a complete combustion. So an arrow going only in one direction will be used. Then we would have, I often list it in the other order, but I'm going to match it with my answer choices which have water first, but our products of the combustion will be water. So I'll write h2o with a space in front of it that will be in a gas state, its combustion, it will be very hot and then carbon dioxide and I'll leave a space CO2 and put it in a gas state. So what are our steps for balancing a combustion reaction? Number one, balance the number of carbon atoms. So balance C. So in the equation we've written, we have nine carbons on the left side and one carbon on the right. So we will use the coefficient nine in front of our carbon dioxide. So that balances the carbons step two is to balance the hydrogens. So we have hydrogens in our hydrocarbon, 12 of them on the left, on the right, it's in our water and there's two of them. So if we multiply our water by six, that's six water molecules, then we have six times two hydrogens per molecule and we have 12. So that's balanced. Step. Number three is to balance the oxygen. And this is last of these three as it's a little trickier since you have oxygen in the products in both the water and the carbon dioxide. So we need to take them both into account. So the left side, there's just those two atoms of oxygen in our oxygen gas. On the right side with the new coefficients, we now have six atoms from the water less nine times 2, 18 from the carbon dioxide should equal 24 oxygen atoms total on the right side. Keeping in mind that oxygen in the form of 02, if we have 12 molecules of 02, we have 12 times 2, 24 atoms of oxygen on the left, balancing those on the right going to erase my little mouth from my oxygen. So now our step four, if it were necessary would be to get rid of fractional coefficients, but we don't have any because our oxygen, we could use just a coefficient of 12 in front of it. So our equation is balanced as is. So there is our balanced equation. So let's double check that it matches with choice BC nine H 12 liquid plus 1202 gas goes to 6 H2O gas plus 9 Co2 gas. So that is indeed correct and will be our answer for our balanced equation for a complete combustion of 135 tri methyl benzene. See you in the next video.

Write the balanced chemical equation for the complete combustion of each of the following hydrocarbons found in gasoline:
c. 3-ethyltoluene

Verified video answer for a similar problem:

Key Concepts

Combustion Reaction

Hydrocarbons

Balancing Chemical Equations

Write the balanced chemical equation for the complete combustion of each of the following hydrocarbons found in gasoline:c. 3-ethyltoluene

Verified video answer for a similar problem:

Key Concepts

Combustion Reaction

Hydrocarbons

Balancing Chemical Equations

Write the balanced chemical equation for the complete combustion of each of the following hydrocarbons found in gasoline:
c. 3-ethyltoluene