2. Limiting Reagents Mrs. Kay Chem 11

3. Grilled Cheese Sandwich Bread + Cheese  ‘Cheese Melt’ 2 B + C  B 2 C 100 bread 30 slices ? sandwiches

4. Container 1 Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 269

5. Before and After Reaction 1 All the hydrogen and nitrogen atoms combine. Before the reaction After the reaction Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 269

6. Container 2 Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 270

7. Before and After Reaction 2 Before the reactionAfter the reaction Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 270

8. Real-World Stoichiometry: Limiting Reactants LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World, 1996, page 366 Ideal Stoichiometry Limiting Reactants

9. Limiting Reactants aluminum + chlorine gas  aluminum chloride Al(s) + Cl 2 (g)  AlCl 3 2 Al(s) + 3 Cl 2 (g)  2 AlCl 3 100 g 100 g ? g A. 200 gB. 125 gC. 667 gD. ???

10. 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 ingredients listed, could we make 4 dozen cookies? What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies? What if we only had one egg, could we make 3 dozen cookies?

11. Limiting Reactant Most 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.

12. Limiting Reactant To 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. 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!

13. 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 Cl 2  2 AlCl 3 Start with Al: Now Cl 2 : 10.0 g Al 1 mol Al 2 mol AlCl 3 133.5 g AlCl 3 27.0 g Al 2 mol Al 1 mol AlCl 3 = 49.4g AlCl 3 35.0g Cl 2 1 mol Cl 2 2 mol AlCl 3 133.5 g AlCl 3 71.0 g Cl 2 3 mol Cl 2 1 mol AlCl 3 = 43.9g AlCl 3 Limiting Reactant

14. LR Example Continued We 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. Once the 35.0g of chlorine is used up, the reaction comes to a complete.

15. 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.

16. 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?

17. Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I 2  2 KI We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced. 15.0 g I 2 1 mol I 2 2 mol K 39.1 g K 254 g I 2 1 mol I 2 1 mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = 10.38 g K 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 start with it!

18. 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!

19. Do the following: Pg 362 #11-23 Pg 367 # 3 and 4

20. Percent Yields Theoretical yield: max amount of a product that is formed in a reaction. Actual yield: amount of product that is actually obtained in a reaction Usually less than theoretical. Why?

21. Why? Theoretical has assumed that all of limiting reagent has completely reacted. Many reactions do not go to completion Unexpected competing side reactions limit the formation of products. Some reactants are lost during the separation process (remember in the lab, pouring off water, leaving silver behind!) Impure reactants Faulty measuring Poor experimental design or technique

22. How to calculate? Percent Yield = Actual Yield x 100% Theoretical yield

23. Practice together: 20g of HBrO 3 is reacted with excess HBr. 1.What is the theoretical yield of Br 2 ? 2.What is the percent yield, if 47.3g is produced? HBrO 3 + 5HBr  3H 2 O + 3 Br 2

24. Practice alone: When 35g of Ba(NO 3 ) 2 is reacted with excess Na 2 SO 4, 29.8g of BaSO 4 is recovered. Ba(NO 3 ) 2 + Na 2 SO 4  BaSO 4 + 2NaNO 3 1.Calculate the theoretical yield of BaSO 4 2.Calculate the percent yield of BaSO 4

25. Advanced Level Applying Percentage Purity

26. Percentage Purity Of a sample describes by mass the composition of a specific compound or element Ex: sample of gold that is 98% pure means that for every 100g sample of gold there is 98 g of pure gold and 2% of impurity

27. Finding percentage purity (from text pg 146) Suppose you have 13.9 g of impure iron pyrite (FeS 2 ). When the reaction proceeds you obtain 8.02 g of Fe 2 O 3. What was the percentage of iron pyrite in the original sample assuming the reaction goes to completion? 4FeS 2 + 11 O 2  2Fe 2 O 3 + 8 SO 2

28. Find the mass of FeS2 expected to have given you 8.02 g of Fe2O3 and then check to see its percent purity to the originally stated mass of 13.9 g.

29. 4FeS 2 + 11 O 2  2Fe 2 O 3 + 8 SO 2 8.02 g Fe2O3 x 1mol Fe2O3 x 4 mol FeS2 x 120 g FeS2 159.7 g 2 mol Fe2O3 1 mol FeS2 = 12.05 g (We were told that 13.9 g of FeS 2 reacted, yet according to our calculations only 12.05 g of FeS 2 would be capable of making 8.02 g of Fe 2 O 3. So there is impurity.)

30. Percentage = theoretical mass x 100 % purity sample size Perc. Purity = 12.05 g x 100% = 86.7 % 13.9 g

31. Applying percentage yield Ethylene oxide, C 2 H 4 O, is prepared by reacting C 2 H 5 OBr with NaOH. If this reaction proceeds with an 89% yield, what mass of C 2 H 4 O can be obtained when 15.0 g of C 2 H 5 OBr reacts with excess NaOH. C 2 H 5 OBr(aq) + NaOH  C 2 H 4 O(aq) + NaBr(aq) + H 2 O(l)

32. Work backwards, solve for 100%, then multiply by final answer by 0.89 to get 89% of the yield 15g x 1mol C2H5OBr x 1mol C2H4O x 44.06 g C2H4O = 5.29g 124.97 g 1 mol C2H5OBr mol 5.29g = 100% yield, but we only have 89% yield so the final answer is 5.59 g x 0.89 = 4.71 g

33. Practice Read Pg 137-148 Practice questions Pg 139 # 31 - 33 Pg 141 #34 and 35 Pg 147 # 38 – 40

34. Solution Stoichiometry Often reactions involve liquids that are related in volumes and concentration, instead of using mass. Molar concentration is the amount of moles of a solute in 1 litre of solution. Ex: 1.0 mol/L of HCl means that there is 1mol of HCl (aq) dissolved in 1 L of solution.

35. Example: 6HCl(aq) + 2Al  2AlCl 3 (aq) + 3H 2 (g) If 50.0 mL of 1.75 M HCl reacts with 25.0 g of aluminum, what is the maximum amount of AlCl 3 made? 1.75 M = 1.75 mol/L How much AlCl 3 is produced if all HCl is used? 3.89 g1.75 mol x 0.050L x 2mol AlCl 3 x 133.33 g = 3.89 g 1L 6 mol HCl mol AlCl 3 How much AlCl 3 is produced if all Al is used? 25.0 g x 1 mol x 2 molAlCl 3 x 133.33 g = 124 g 26.98 g 2 mol Al molAlCl 3 This is the amount made

36. Practice on your own. The scanned notes and handout online have additional solution questions to practice. This will be on our advanced Unit test.