 So we are looking at reactions, combustion, oxidation. We've looked at how we handle air. Now what we're going to do, we're going to take a look at how we determine how much fuel is being combusted with the air. And so typically what we do is we look at kind of a simplified reaction equation as follows. So we have fuel plus dry air. Now, sometimes you can have water vapor in the air. That would make the problem a little more complex. For now, we'll just consider it to be dry air. And then what we would have are our products. So let's consider a hydrocarbon fuel under what we will call theoretical combustion. And we'll define that in a moment. And what that implies is that there is no oxygen in the products. So some generic hydrocarbon fuel. We have carbon and hydrogen, x moles of carbon, y of hydrogen. Plus some theoretical amount of air. And we talked about air being 21 percent oxygen, 79 percent nitrogen. And so we came up with a way of expressing it in terms of kilomoles in the following. And that will then react into, I'll say, x moles of carbon dioxide, water vapor and finally nitrogen. So what we're going to do, like we would do in any kind of stoichiometric balance, we're going to do a mass balance for each of the different components. So we'll do that for carbon, hydrogen, oxygen and nitrogen. So we're now going to do this for a simple fuel. We'll consider a hydrocarbon fuel of methane. So if we take methane and we write out our reaction equation, we have the theoretical amount of air. So we'll do the balance for each of the individual components, starting with carbon. So looking at carbon, we have carbon here on the left. There's no other places for carbon on the right. Carbon is there. Those are the two places where we have carbon. So with that, we can write one on the left is equal to x on the right. Next thing we have is hydrogen. So let's do the same thing. I'll use blue now and we have four hydrogen there. And then on the right, we had y over two for diatomic hydrogen. And that's it. So for hydrogen, what we can write is four on the left. And then y over two times two just comes out to be y oxygen. And I'm going to consider diatomic oxygen here. So in terms of diatomic oxygen, we have it there. With theoretical air, it appears there. It appears in the water as well. So oxygen appears a couple of places on the right. And what we end up with is a theoretical is x. So we're dealing with diatomics. We're taking the x moles plus. And then we have monatomic here and it's multiplied by y over two. We're balancing diatomic. So that should be y over four instead of y over two. And so with that, what we can say is that theoretical is equal to, while we know x from here, so we can substitute that in. And we know why from here is we can make that substitution. So we can then rewrite this as being one plus four over four is equal to two. So theoretical error in this case is two. And finally, for diatomic nitrogen, just for sake of completeness, what I'll do, I'll go to a dark green pan. Diatomic nitrogen is there and there. It's not doing anything in this reaction. So all we have is 3.76 a theoretical on the left equals 3.76 a theoretical on the right. And so with that, we can rewrite the oxidation reaction for methane as being the following. So that would be our reaction equation. And A-T-H is the stoichiometric or theoretical error. So that would be our combustion or reaction equation. If we had a stoichiometric balance between the air and the fuel coming in.