1. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 1 Chapter 6 Chemical Reactions and Quantities 6.8 Percent Yield and Limiting Reactants

2. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 2 Theoretical, Actual, and Percent Yield Theoretical yield:  the maximum amount of product, which is calculated using the balanced equation. Actual yield:  the amount of product obtained when the reaction takes place Percent yield:  the ratio of actual yield to theoretical yield Percent yield = actual yield (g) x 100 theoretical yield (g)

3. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 3 Guide to Calculations for Percent Yield

4. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 4 Suppose you have prepared cookie dough to make 5 dozen cookies. The phone rings and you answer. While you talk, a sheet of 12 cookies burns, and you have to throw them out. The rest of the cookies you make are okay. What is the percent yield of edible cookies? Theoretical yield: 60 cookies possible Actual yield: 48 cookies to eat Percent yield: 48 cookies x 100% = 80.% yield 60 cookies Calculating Percent Yield

5. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 5 Learning Check With a limited amount of oxygen, the reaction of carbon and oxygen produces carbon monoxide. 2C(g) + O 2 (g) 2CO(g) What is the percent yield if 40.0 g of CO are produced when 30.0 g of O 2 are used? 1) 25.0% 2) 75.0% 3) 76.2%

6. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 6 Solution 3) 76.2% yield STEP 1 Given: 40.0 g of CO produced (actual) 30.0 g of O 2 used Need: percent yield of CO STEP 2 Write a plan to calculate % yield of CO: g of O 2 moles of moles of g of CO O 2 CO (theoretical) Percent yield of CO = g of CO (actual) x 100% g of CO (theoretical)

7. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 7 Solution (continued) STEP 3 Write conversion factors: 1 mole of O 2 = 32.0 g of O 2 1 mole O 2 and 32.0 g O 2 32.0 g O 2 1 mole O 2 1 mole of O 2 = 2 moles of CO 1 mole O 2 and 2 moles CO 2 moles CO 1 mole O 2 1 mole of CO = 28.0 g of CO 1 mole CO and 28.0 g CO 28.0 g CO 1 mole CO

8. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 8 Solution (continued) STEP 4 Setup to calculate theoretical yield in g of O 2 : 30.0 g O 2 x 1 mole O 2 x 2 moles CO x 28.0 g CO 32.0 g O 2 1 mole O 2 1 mole CO = 52.5 g of CO (theoretical) Setup to calculate percent yield: 40.0 g CO (actual) x 100 = 76.2% yield (3) 52.5 g CO (theoretical)

9. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 9 Learning Check When N 2 and 5.00 g of H 2 are mixed, the reaction produces 16.0 g of NH 3. What is the percent yield for the reaction? N 2 (g) + 3H 2 (g) 2NH 3 (g) 1) 31.3% of NH 3 2) 56.9% of NH 3 3) 80.0% of NH 3

10. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 10 Solution 2) 56.9% STEP 1 Given: 16.0 g of NH 3 produced (actual) 5.00 g of H 2 used Need: percent yield of NH 3 STEP 2 Write a plan to calculate % yield of NH 3 : g of H 2 moles of moles of g of NH 3 H 2 NH 3 (theoretical) Percent yield of NH 3 = g of NH 3 (actual) x 100% g of NH 3 (theoretical)

11. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. Solution (continued) STEP 3 Write conversion factors: 1 mole of H 2 = 2.02 g of H 2 1 mole H 2 and 2.02 g H 2 2.02 g H 2 1 mole H 2 1 mole of H 2 = 2 moles of NH 3 1 mole H 2 and 2 moles NH 3 2 moles NH 3 1 mole H 2 1 mole of NH 3 = 17.0 g of NH 3 1 mole NH 3 and 17.0 g NH 3 17.0 g NH 3 1 mole NH 3

12. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 12 Solution (continued) STEP 4 Setup to calculate theoretical yield of g of NH 3 : 5.00 g H 2 x 1 mole H 2 x 2 moles NH 3 x 17.0 g NH 3 2.02 g H 2 3 moles H 2 1 mole NH 3 = 28.1 g of NH 3 (theoretical) Setup to calculate percent yield: Percent yield = 16.0 g NH 3 x 100 = 56.9% yield (2) 28.1 g NH 3

13. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 13 Limiting Reactant A limiting reactant in a chemical reaction is the substance that  is used up  limits the amount of product that can form and stops the reaction

14. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 14 Reacting Amounts In a table setting, there is 1plate, 1 fork, 1 knife, and 1 spoon. How many table settings are possible from 5 plates, 6 forks, 4 spoons, and 7 knives? What is the limiting item?

15. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 15 Reacting Amounts (continued) Only 4 place settings are possible. Initially Used Left over Plates 5 41 Forks 6 42 Spoons 4 4 0 Knives 7 4 3 The limiting item is the spoon.

16. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 16 Example 1 of an Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices of bread and 1 jar of peanut butter? With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item.

17. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 17 Example 2 of an Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices bread and 1 tablespoon of peanut butter? With 1 tablespoon of peanut butter, only 1 sandwich could be made. The peanut butter is the limiting item.

18. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 18 Guide to Calculating Product from a Limiting Reactant

19. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 19 Limiting Reactant When 4.00 moles of H 2 is mixed with 2.00 moles of Cl 2, how many moles of HCl can form? H 2 (g) + Cl 2 (g) 2HCl(g) 4.00 moles 2.00 moles ??? Moles  Calculate the moles of product that each reactant, H 2 and Cl 2, could produce.  The limiting reactant is the one that produces the smaller number of moles of product.

20. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 20 Limiting Reactant (continued) HCl from H 2 4.00 moles H 2 x 2 moles HCl = 8.00 moles of HCl 1 moles H 2 HCl from Cl 2 2.00 moles Cl 2 x 2 moles HCl = 4.00 moles of HCl 1 mole Cl 2 4.00 moles of HCl is the smaller number of moles produced. Thus, Cl 2 will be used up. The limiting reactant is Cl 2.

21. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 21 Check Calculations Equation Initially H 2 4.00 moles Cl 2 2.00 moles 2HCl 0 mole Reacted/ Formed –2.00 moles +4.00 moles Left after reaction 2.00 moles (4.00 – 2.00) Excess 0 moles (2.00 – 2.00) Limiting 4.00 moles (0 + 4.00) Product possible

22. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 22 Limiting Reactants Using Mass If 4.80 moles Ca are mixed with 2.00 moles N 2, which is the limiting reactant? 3Ca(s) + N 2 (g) Ca 3 N 2 (s) Moles of Ca 3 N 2 from Ca 4.80 moles Ca x 1 mole Ca 3 N 2 = 1.60 moles of Ca 3 N 2 3 moles Ca (Ca is used up) Moles of Ca 3 N 2 from N 2 2.00 moles N 2 x 1 mole Ca 3 N 2 = 2.00 moles of Ca 3 N 2 1 mole N 2 Ca is used up. Thus, Ca is the limiting reactant.

23. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 23 Learning Check What is the mass of water that can be produced when 8.00 g of H 2 and 24.0 g of O 2 react? 2H 2 (g) + O 2 (g) 2H 2 O(l) 1) 8.0 g of H 2 O 2) 27.0 g of H 2 O 3) 72 g of H 2 O

24. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 24 Solution 3) 72 g of H 2 O Moles of H 2 O from H 2 : 8.00 g H 2 x 1 mole H 2 x 2 moles H 2 O = 4.0 moles of H 2 O 2.0 g H 2 2 moles H 2 Moles of H 2 O from O 2 : 24.0 g O 2 x 1 mole O 2 x 2 moles H 2 O = 1.50 moles of H 2 O 32.0 g O 2 1 mole O 2 Smaller number of moles of H 2 O 1.50 moles H 2 O x 18.0 g H 2 O = 27.0 g of H 2 O 1 mole H 2 O