Limiting Reactant and Theoretical Yield For our examples, let’s assume the following recipe for making a pizza: 1 crust + 2 cups cheese + 5 oz. of sauce à 1 pizza Limiting Reactant (Reagent) – The reactant that limits the amount of product in a chemical reaction. For example, if we have 4 crusts, 10 cups of cheese and 15 oz. of sauce, we have enough crust for 4 pizzas, enough cheese for 5 pizzas but only enough sauce for 3 pizzas. Therefore, our limiting reactant is sauce. Theoretical Yield – The amount of product that can be made in a chemical reaction based on the amount of the limiting reactant. Using our pizza example, the theoretical yield is 3 pizzas since that is all we can make with the limited amount of sauce we have. Actual Yield – The amount of product actually produced. Using our pizza example, suppose during our pizza making we burned one pizza. We would then have only made 2 pizzas. Our actual yield would be 2 pizzas. Percent Yield – The percentage of the theoretical yield that was attained based upon the actual amount of product produced. Percent yield can be calculated as (actual yield/theoretical yield) x 100. Using our pizza example, we theoretically could have made 3 pizzas but only actually made 2 pizzas. Our percentage yield would then be (2 pizzas/3 pizzas) x 100 = 67%. Limiting Reactant and Theoretical Yield Example Problem: A reaction mixture contains 42.5 g Mg and 33.8 g O2 and yields 55.9 g MgO. Identify the limiting reactant, theoretical yield (in g) and percent yield. 2 Mg (s) + O2 (g) à 2 MgO (s) Solution: Determine how much product (MgO) can be made considering each reactant separately. 42.5 g Mg x (1 mol Mg/24.31 g Mg) x (2 mol MgO/2 mol Mg) = 1.75 mol MgO 33.8 g O2 x (1 mol O2/32.00 g O2) x (2 mol MgO/1 mol O2) = 2.11 mol MgO Since 42.5 g Mg results in the least amount of MgO, it is the limiting reactant. Now determine the theoretical yield based upon the limiting reactant. Since you can only produce 1.75 mol of MgO with 42.5 g Mg, the theoretical yield is: 1.75 mol MgO x (40.31 g MgO/1 mol MgO) = 70.5 g MgO The percent yield is: (55.9 g MgO/70.5 g MgO) x 100 = 79.3 %