In this module we're going to look at the calculations associated with limiting reagent problems. By the end of this module, you should be able to calculate the quantity produced and identify the limiting and excess reagents in a particular reaction. When we do a limiting reagent problem we really approach the problem no differently then when we do any other stoichiometry problem. The main difference is, is that we have the amounts of two different substances. Here we're told we have 25 grams of iron and 25 grams of oxygen. So the first thing we have to figure out is which one is our limiting reagent. Because the limiting reagent is the substance that will limit the amount of product that can be formed. So for our limiting reagent problem we're actually going to do two calculations using stoichiometry. The first thing we're going to do is we're going to assume that iron is the limiting reagent. If we assume iron is the limiting reagent, we can figure out how much of our product or iron oxide will be produced, if we have 25 grams of iron with excess oxygen. So what we can do is we have our grams, we want to convert to moles. The molar mass of iron is 55.85 grams per mole. We know that for four moles of iron, we get two moles of Fe2O3. Then we can find the mass of Fe2O3 using it's molar mass, which is 159.7 grams per mole. So now, we see that grams of iron cancels with grams of iron. Moles of iron, cancels with moles of iron. Moles of Fe2O3, cancels with moles of Fe2O3. And what we're left with be grams of Fe2O3. And when I do the calculation, I find that I get 35.7 grams of Fe2O3. Now remember, we made the assumption that iron was the limiting reagent. Next, we're going to assume that oxygen is limiting. That would mean that we assume that the iron is the excess reagent. We still start the problem looking at 25 grams of oxygen. But now we have to actually do a different molar mass because we're looking at oxygen instead of iron, so we have 32 grams per mole of O2. We set up our mole to mole ratio. We see that based on our balanced equation we have 3 moles, of O2 for every two moles of Fe2O3, and again I use the molar mass of Fe2O3 to convert from moles to grams of Fe2O3. Again, grams of oxygen cancel with grams of oxygen. Moles of oxygen cancel with moles of oxygen. Moles of Fe2O3 cancel with moles of Fe2O3. And I end up with getting 83.2 grams of Fe2O3. Now, what I want to look at is to determine which one is the limiting reagent and which one is the excess reagent. When I assumed iron was the limited reagent, I produced the least amount of product, Iron is limiting. [SOUND] Oxygen is excess. And the maximum amount of product I can produce is 35.7 grams of Fe2O3. At that point all of my iron supply has been exhausted and so I cannot produce any more of our product. Just like with our snowmen example, once we've used up all of one part, we were done, we were limited in the amount of product that we can make. One thing to note is, in this problem, the amounts of both reactants were the same. However, even if they're different amounts, we can't just look at the grams of a substance to determine the limiting reagent. Just because there's a lower mass of one substance, doesn't necessarily mean it will limit the amount of the product formed. It will depend on the molar mass of that substance, as well as the molar ratio, between that reactant and the amount of product it can form. Now, let's let you try one. If we assume that magnesium is a limiting reagent, how many grams of magnesium nitride can be made through the reaction of 35 grams of magnesium and 15 grams of nitrogen? When we do our calculation we find that we can make 48.42 grams of magnesium nitride. Just like in our previous example, we're assuming that one substance is limiting, we're going to figure out how much product can be made if that is in fact the limiting reagent. So we're assuming magnesium is limiting. So we're going to start with the mass of magnesium given which is 35.00 grams of magnesium. I'm going to use the molar mass of magnesium which 24.31 grams per mole of magnesium. Then I could use my mole to mole ratio given from the balanced chemical equation, and what I see is that for every three moles of magnesium, I can produce one mole of magnesium nitride. And then I can use the molar mass of magnesium nitride, which is 100.95 grams per mole, and some that I can calculate the grams of magnesium nitride that can be produced if magnesium is in fact the limiting reagent. After we do the calculation, we end up with 48.42 grams of magnesium nitride. Now we need to assume that nitrogen is the limiting reactant, and determine how many grams of magnesium nitride can be made if the nitrogen is limiting. We take the same approach as we did before. We know we have 15 grams of nitrogen. We use our molar mass of nitrogen 28.02 grams per mole of N2. Then we use our mole ratio from our balance equation, which is one mole of N2 for every one mole of magnesium nitride. So even though it's a one to one ratio, I'm going to go ahead and put it in there so that my units work out correctly. And then I look at the molar mass of magnesium nitride and it is 100.95 grams per mole of Mg3N2. After I do the calculation, I find that I have 54.01 of magnesium nitride. So when I assume that nitrogen is a limiting reagent, I produce, can produce 54.01 grams of my product. Now we've done the two preliminary calculations assuming that magnesium was limiting, then assuming that nitrogen was limiting. Now I have to decide how much of my product can I actually make. Because 48.42 is the lesser amount, that tells me that that's the maximum amount of product I can produce, and therefore that magnesium is the limiting reagent and nitrogen is in excess. Note here that we started with a larger mass of magnesium than we did nitrogen, but magnesium is limited because we have to look at molar amounts and molar ratios of reactants to products. In the next module we'll look at theoretical and percent yields.