Write the balanced chemical equation for the complete combustion of each of the following compounds:
c. 2,3-dimethylhexane

Question

Write the balanced chemical equation for the complete combustion of each of the following compounds:

c. 2,3-dimethylhexane

Accepted Answer

Hi, everybody. Here's our next problem. Identify the products and provide the balanced chemical equation for the complete combustion of liquid five ethyl, two methyl heptane. And we have four different possibilities for our balanced equation. So we'll look at those after we've gone through our calculations to see which matches up correctly. So first of all, we need to look at the compound that we're combusting and figure out its molecular formula. So we're given its name, we have a heptane. So a seven carbon chain and it has a, an ethyl group on carbon five. So that will be two more carbons and then a methyl group on carbon tube. So that will be one more carbon. So we have seven carbons plus two carbons plus one carbon will equal a total of 10 carbon atoms. So to write our molecular formula, we remember that for alkanes, which is what we have. We see that ending A ne no double bonds are only substituent groups are alkyl groups. So for them, the formula is C subscript, NH script, subscript two N plus two. So we know our number of carbons. So N must equal 10. So therefore two N plus two will equal 10, multiplied by two is 20 plus two is 22. So our molecular formula of our five ethyl, two methyl hept will be C 10, age 22. So we have our proper molecular formula for the chemical that will be participating in this reaction. We recall that in a combustion reaction, we have 02 molecular oxygen as a reactant with the chemical, we are combusting and then carbon dioxide and water are the products. So let's write out the basic equation and then we'll work on balancing it. So we know we have C 10 H 22 and that would be liquid. We're told in the problem is going to react with and I'm going to leave a space because I'm going to be putting in coefficients as I balance my equation, it will react with 02 which is a gas. It's just a single arrow because combustion it's complete combustion. So everything will proceed to the products which will be CO2 gas and then it'll be water vapor, not liquid water since everything will be so hot. So H2O is also a gas. So recall that when we balance these combustion reactions, our first step number one is balance the carbon and we do that by adding a coefficient to the carbon dioxide. I'm going to do my coefficients in red as I add this. So you can see as I change them and add them. So we have 10 carbons on the left side in our alkane and then carbon dioxide has a single carbon atom per molecule. So we just put a coefficient of 10 in front of our carbon dioxide that will balance our carbons step two is to balance the hydrogens and we balance them by adding a coefficient to water, which is our hydrogen containing compound on the right side. So we have 22 hydrogens on the left water has two hydrogens per molecule. So our coefficient will be 11, since there will then be 11 multiplied by two hydrogens on the right side. So now we go to step number three, which is balance the oxygen. And this is a little trickier because on the right side, we have oxygen in both molecules. So we have to account for that. So we look at how much oxygen there is on the left. Well, there's just 02. So just two atoms of oxygen. And then on the right, we have from the carbon dioxide, we have 10 multiplied by 202, there's two oxygen atoms per molecule. So that would be 20. And then in the water, we have one oxygen atom per water molecule and there's 11 water molecules. So that's 11 oxygens on this side. So a total of 31 oxygen atoms on the right with two oxygen atoms on the left. So and here's where the balancing oxygen is sort of a two step process. First, we'll start with a fractional coefficient since we have the oxygen in this 02 form. So if we have 31 oxygens on the right side and two per molecule on the left, we can put the coefficient as 31 halves because multiply by two that will make leave you with 31 on the left side, balancing a 31 on the right. Obviously, we can't leave our equation in this form of a fractional coefficient. I'm just gonna erase my oxygen calculations here. So I just need to number four, get rid of fractional coefficients. In this case. Since my fractional coefficient is in the form of a half, 31 halves, I can multiply my whole equation by two. Remember that the entire equation must be multiplied by that. So I'll do two, multiplied by the whole equation I have so far it's balanced. I just don't want a fractional coefficient. So that will give me a final equation of two C 10 H 22 liquid plus and I now have 31 because 31 halves times two is 31 02, which is a gas. Now, my coefficient for carbon dioxide was 10. I multiply by that by two and get 20 CO2 gas. My coefficient for water was 11. So multiply that by two and I have 22 H2O gas, we can double check and bad. It is balanced. We have 20 carbons on the left, 20 carbons on the right. Terms of hydrogens, we have 44 on the left and 44 on the right. And then oxygens, we have 31 times two, so 62 on the left and then on the right, we have 40 plus 2262 on the right. So we have our balanced equation with no fractional coefficients. Let's look at our answer. Choices. Choice A does not have the two in front of C 10 H 22. So we can rule it out right away without even looking further. Choice B we have two C 10 H 22 plus 31 +02 becomes 20 co two plus 22 water. So Choice B matches what we've calculated to be thorough. We can take a quick look at choice C and D. Choice C has the correct coefficient in front of our heptane, but it has 32 oxygens instead of 31. So that's not correct. And choice D does not have the two in front of our Hep team. So that's also crossed out right away. So once again, we've identified the products as carbon dioxide and water and provide the balanced chemical equation for this complete combustion of liquid five ethyl, two methyl heptane. And that is Choice B two C 10 H 22 liquid plus 31 +02 gas becomes 20 co2 gas and 22 H2O gas. See you in the next video.

Write the balanced chemical equation for the complete combustion of each of the following compounds:
c. 2,3-dimethylhexane

Verified video answer for a similar problem:

Key Concepts

Combustion Reaction

Balanced Chemical Equation

Structural Formula of Organic Compounds

Write the balanced chemical equation for the complete combustion of each of the following compounds:c. 2,3-dimethylhexane

Verified video answer for a similar problem:

Key Concepts

Combustion Reaction

Balanced Chemical Equation

Structural Formula of Organic Compounds

Write the balanced chemical equation for the complete combustion of each of the following compounds:
c. 2,3-dimethylhexane