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3. Stoichiometry: Limiting Reagents At the conclusion of our time together, you should be able to: 1. Determine the limiting reagent 2. Use the limiting reagent to determine the amount of product produced in a reaction 3. Determine the excess amount(s) in the reaction

4. Limiting Reactant: Cookies 1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen If we had the specified amount of all ingredients listed, could we make 4 dozen cookies? If not, what limits us?? What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies? If not, what limits us??? What if we only had one egg, could we make 3 dozen cookies?

5. Limiting Reactant Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant. excess That reactant is said to be in excess (there is too much). limiting reactant The other reactant limits how much product we get. Once it runs out, the reaction ’s. This is called the limiting reactant.

6. Limiting Reactant all a each To find the correct amount of product produced, 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 lower amount of a product is the correct answer. limiting reactant The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same! Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!

7. Limiting Reactant: Example 10.0 g 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 Cl 2  2 AlCl 3 Start with Al: Now Cl 2 : 10.0 g Al 1 mol Al 2 mol AlCl 3 133.33 g AlCl 3 26.98 g Al 2 mol Al 1 mol AlCl 3 = 49.42 g AlCl 3 35.0 g Cl 2 1 mol Cl 2 2 mol AlCl 3 133.33 g AlCl 3 70.90 g Cl 2 3 mol Cl 2 1 mol AlCl 3 = 43.88 g AlCl 3 Limiting Reactant

8. LR Example Continued We get 49.42 g of aluminum chloride from the given amount of aluminum, but only 43.88 g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0 g of chlorine is used up, the reaction comes to a complete.

9. Limiting Reactant Practice 15.0 g of aluminum reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made. A little quicker way to do this is to pick one reactant and determine what amount of the other reactant is needed to completely react the first.

10. Calculate the mass in grams of iodine required to react completely with 15.0 g of aluminum. Al + I 2  AlI 3 2 Al + 3 I 2  2 AlI 3 = 212 g I 2 x 3 mol I 2 2 mol Al x 253.80 g I 2 1 mol I 2 15.0 g Al x 1 mol Al 26.98 g Al

11. Finding the Amount of Excess To completely react 15.0 g of aluminum, we would need 212 g of iodine. Since we only have 15.0 g of iodine, the iodine limits the amount of product we can make. 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 aluminum in the previous problem?

12. Finding Excess Practice 15.0 g of aluminum reacts with 15.0 g of iodine. Calculate the excess of aluminum. 2 Al + 3 I 2  2 AlI 3 Always start with the limiting reactant: 15.0 g I 2 1 mol I 2 2 mol Al 26.98 g Al 253.80 g I 2 1 mol I 2 1 mol Al = 3.19 g Al USED! 15.0 g Al – 3.19 g Al = 11.81 g Al EXCESS Given amount of excess reactant Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only use it!

13. Limiting Reactant: Recap 1. You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. 2. Convert ALL of the reactants to the SAME product (pick any product you choose.) 3. The lowest answer is the correct answer. 4. The reactant that gave you the lowest answer is the LIMITING REACTANT. 5. The other reactant(s) are in EXCESS. 6. To find the amount of excess, subtract the amount used from the given amount. 7. If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!

14. Stoichiometry: Limiting Reagents Let’s see if you can: 1. Determine the limiting reagent 2. Use the limiting reagent to determine the amount of product produced in a reaction 3. Determine the excess amount(s) in the reaction

15. In the lab where you combined iron (II) with sulfur, if 5.00 g of iron is combined with 5.00 g of sulfur, what is the limiting reagent and how much excess reagent is left? 5.00 g Fex 1 mol Fe 55.85 g Fe = 0.0895 mol Fe Fe + S  FeS 5.00 g Sx 1 mol S 32.07 g S = 0.156 mol S

16. 0.0895 mol Fe + 0.156 mol S  FeS The ratio must be 1:1, therefore Fe is limiting and will combine with 0.0895 mol S to produce 0.0895 mol of FeS 0.0895 mol Sx 32.07 g S 1 mol S = 2.87 g S used 5.00 g S – 2.87 g S used = 2.13 g excess S