 Let's return to the original example where we mixed two moles of hydrogen and two moles of oxygen. I'm going to represent each mole of gas by a balloon. The mole ratio is 2 to 1, so the two moles of hydrogen here are going to react with one mole of oxygen, and they're going to produce two moles of water. Well that accounts for one mole of oxygen. But the remaining mole of oxygen has nothing to react with, so it's unchanged. And what we're left with at the end is two moles of water and the one unreacted mole of original oxygen. Here the hydrogen is the limiting reactant because it was the one that ran out and determined when the reaction stopped. The oxygen was the excess reactant because we had more than was needed to exactly react with the limiting reactant. Okay, so let's ramp this up a bit. We'll use a different reaction and I'm going to ask if I react 1.8 moles of nitrogen with 2.5 moles of hydrogen, how many moles of ammonia will be produced? Well the first thing we have to do is to work out which is the limiting reactant, because it's that that determines how much product is formed. Remember, once the limiting reactant runs out the reaction stops and no more product is made. So if we're to calculate how much ammonia is made we first need to know which reactant is going to run out first. I've sketched the strategy for doing this here and I'm going to work through it. First, let's highlight how many moles of each reactant we actually have present. So that's 1.8 moles of nitrogen and 2.5 moles of hydrogen. Next we'll write out the relevant mole ratio from the equation. That's 1 mole of nitrogen reacts exactly with 3 moles of hydrogen. Now we choose one of the reactants and we write the number of moles that we actually have underneath the mole ratio. It doesn't matter which one we choose because it's a ratio we can use it in either direction. So let's take nitrogen for this example. We have 1.8 moles of nitrogen. Now the mole ratio is 1 to 3 so to find the amount of hydrogen that would be needed to react exactly with 1.8 moles of nitrogen we just multiply that by 3. That gives us 5.4 moles of hydrogen. What this tells us is that we need 5.4 moles of hydrogen to fully react with the 1.8 moles of nitrogen. But we only have 2.5 moles of hydrogen so that's not enough to fully react with the nitrogen and that means that the hydrogen is going to run out first. Hence hydrogen is the limiting reactant here. That means that nitrogen must be the excess reactant and that's because the hydrogen will run out before all of the nitrogen has reacted. So there'll be nitrogen left over. So to predict how much ammonia is going to form, that's the second part of this problem we should use the hydrogen amount as our basis. Since it is the limiting reactant, we assume that it all reacts to form ammonia. The ratio of hydrogen to ammonia is 3 to 2 from the equation. So we take the moles of hydrogen that we have available, that's 2.5, and we divide by 3 and multiply by 2 to give us 1.7 moles of ammonia produced. Now I just want to show you that it doesn't matter which reactant you choose when you're working out which one is the limiting reactant. Just now I started with nitrogen so let me now do it the other way round and start with hydrogen. I'll write out the mole ratio again and we know that we have 2.5 moles of hydrogen present. So I'll put that in as the amount that's present. Now the mole ratio is 3 hydrogens to 1 nitrogen so to work out how much nitrogen is needed to fully react with that amount of hydrogen to divide by 3. So that's 2.5 divided by 3 which equals 0.83 moles of nitrogen. So I need 0.83 moles of nitrogen to fully react with 2.5 moles of hydrogen and you'll see that we have more than 0.83 moles of nitrogen present. In fact, we have 1.8 moles. So that means that all the hydrogen will react and there will still be some nitrogen left over afterwards. So that leads us to the same conclusion as before. Then in this case, hydrogen is the limiting reactant and nitrogen is the excess reactant.