Presentation on theme: "Theoretical and Percent Yield"— Presentation transcript:
1 Theoretical and Percent Yield Limiting ReactantTheoretical and Percent Yield
2 Limiting Reactant: Cookies 1 cup butter1/2 cup white sugar1 cup packed brown sugar1 teaspoon vanilla extract2 eggs2 1/2 cups all-purpose flour1 teaspoon baking soda1 teaspoon salt2 cups semisweet chocolate chipsMakes 3 dozenLimiting Reactant - The reactant in a chemical reaction that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed.Excess Reactant - The reactant in a chemical reaction that remains when a reaction stops when the limiting reactant is completely consumed. The excess reactant remains because there is nothing with which it can react.
3 Limiting ReactantMost of the time in chemistry we have more of one reactant than we need to completely use up other reactant.That reactant is said to be in excess (there is too much).The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant.
4 Limiting ReactantTo find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one.The lower amount of a product is the correct answer.The reactant that makes the least amount of product is the limiting reactant.Be sure to pick the same product!
5 Limiting Reactant: Example 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced?2 Al + 3 Cl2 2 AlCl3Start with Al:Now Cl2:10.0 g Al 1 mol Al mol AlCl g AlCl327.0 g Al mol Al mol AlCl3= 49.4g AlCl335.0g Cl mol Cl mol AlCl g AlCl371.0 g Cl mol Cl mol AlCl3= 43.9g AlCl3
6 Theoretical YieldWe get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant.Theoretical Yield – The predicted amount of product is the theoretical yield g of AlCl3 is the calculated product, so that is the theoretical yield.
7 Limiting Reactant Practice 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.2 K + I2 2 KI15.0 g K x 1mole x 2 KI x 166 g = g KI40 g 2 K 1 mole15. 0 g I2 x 1 mole x 2 KI x 166g = 19.8 g KI252 g I2 1 mole
8 Finding the Amount of Excess By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess.Can we find the amount of excess potassium in the previous problem?
9 Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2 2 KIWe found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced.15.0 g I mol I mol K g K254 g I mol I mol K= 4.62 g K USED!15.0 g K – 4.62 g K = g K EXCESSGiven amount of excess reactantAmount of excess reactant actually usedNote that we started with the limiting reactant! Once you determine the LR, you should only start with it!
10 Percentage YieldPercentage yield is the percent of the theoretical yield that was made by our experiment. actual amount of product percentage yield = x 100 theoretical yield(the amount you expect to get from the reaction)
11 Percentage YieldExample: A student conducts a single displacement reaction that produces grams of copper. If grams of copper should have been produced what is the student's percentage yield?
12 Percentage Yield2.755g percentage yield = x 100 g percentage yield = %
13 Sample problemQ - What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2?Step 1: write the balanced chemical equation2H2 + O2 2H2OStep 2: determine actual and theoretical yield. Actual is given, theoretical is calculated:1 mol H22.02 g H2x2 mol H2O2 mol H2x18.02 g H2O1 mol H2Ox# g H2O=16 g H2143 g=Step 3: Calculate % yieldactualtheoretical138 g H2O143 g H2O=% yield =x 100%x 100%96.7%=
14 Practice problemQ - What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess N2?Step 1: write the balanced chemical equationN2 + 3H2 2NH3Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated:2 mol NH33 mol H2x17.04 g NH31 mol NH3x# g NH3=20.0 mol H2227 g=Step 3: Calculate % yieldactualtheoretical40.5 g NH3227 g NH3=% yield =x 100%x 100%17.8%=