Chem 400 Chapter 3 Lecture Notes Part 3 <a href="/keyword/limiting-reactant/" >limiting reactant</a> Stoichiometry You now know how to use a balanced equation to calculate a theoretical yield of products. But, we made an assumption: we had exactly the correct amount of ALL reactants. But in reality, we often just throw in a lot of at least one reactant, or we have some reactants in excess. This means that when the reaction is over, whatever was in excess will be leftover. Although this seems strange, it often helps increase the %-yield by reducing the amount of side reactions. For example, combustion reactions are typically run with a huge excess of oxygen, to lower the amount of carbon monoxide which is produced. And if some reactants are in excess, there is usually 1 reactant which is the <a href="/keyword/limiting-reactant/" >limiting reactant</a> , that is, the reactant which limits the amount of product which may be made. The <a href="/keyword/limiting-reactant/" >limiting reactant</a> is used up in the reaction. The reaction STOPS when the <a href="/keyword/limiting-reactant/" >limiting reactant</a> runs out. So how do we calculate the theoretical yield if there are excess reactants and a <a href="/keyword/limiting-reactant/" >limiting reactant</a> ? There are 2 overall goals: 1) determine which reactant is limiting; and then 2) calculate the theoretical yield. There are several ways of doing <a href="/keyword/limiting-reactant/" >limiting reactant</a> problems, so I will show you a longer, but very logical, simple way; and then I will show you a short way. Here's an example problem using cetyl palmitate, a major constituent of beewax: C32H64O2 + 47 O2 32 CO2 + 32 H2O If you combust 91.57 g of cetyl palmitate with 24.07 g oxygen gas, which is the <a href="/keyword/limiting-reactant/" >limiting reactant</a> , how many g of water can be made, and how many g of the excess reactant are leftover? Process: Convert both reactants to mol or g of product. As you have 2 reactants, you will have 2 lines of work. Compare the 2 product amounts. Obviously, you can only make the smaller amount. So the <a href="/keyword/limiting-reactant/" >limiting reactant</a> is the reactant which makes the least amount of product. Again, the <a href="/keyword/limiting-reactant/" >limiting reactant</a> runs out, and when it runs out, the rxn dies. So in this process, you got both the <a href="/keyword/limiting-reactant/" >limiting reactant</a> identity and the theoretical yield. Now, you have to find the amount of excess reactant leftover (if problem asks for it). Here, you know the theoretical yield and you also know how many g the excess reactant could have made. So take the ratio of the theoretical yield over the g product from excess reactant. This is the fraction of excess product which is used. Then: theoretical yield g leftover excess reactant = 1 - (g excess reactant ) g product from excess reactant calculation OK, that's the longer way. What's the shorter way? Process: Find moles of the reactants. 1. Divide the number of moles of reactants by its coefficient in the balanced equation. This way you can directly compare the number of moles. This is a way of using the mole ratios. 2. Compare the number of adjusted moles of reactants (after dividing by the coefficient). Whichever is smaller is the <a href="/keyword/limiting-reactant/" >limiting reactant</a> . 3. Convert mol <a href="/keyword/limiting-reactant/" >limiting reactant</a> from step 1 to g of product (or mol if the question asked this). Now to find g of excess reactant. Convert number of mol of <a href="/keyword/limiting-reactant/" >limiting reactant</a> from step 1. to g of the excess reactant. This is how much of the excess reactant will be used. Subtract this from the original g of the excess reactant.