1. Chapter 9 Chemical Quantities in Reactions

2. Mole Relationships in Chemical Equations Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

3. The equation can be read in “moles” by placing the word “______” or “______” after each coefficient. 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) 4 mol Fe + 3 mol O 2 2 mol Fe 2 O 3 Moles in Equations

4. 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) _____ molecules NH 3 react with _____ molecules O 2 to produce _____ molecules NO and ____ molecules H 2 O and _____ mol NH 3 react with ______ mol O 2 to produce ______ mol NO and ______ mol H 2 O Quantities in A Chemical Reaction

5. Take another look at this equation 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) 4 mol Fe + 3 mol O 2 2 mol Fe 2 O 3 Moles in Equations

6. A mole-mole factor is a ______ of the moles for two substances in an equation. 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Fe and O 2 4 mol Fe and 3 mol O 2 3 mol O 2 4 mol Fe Fe and Fe 2 O 3 4 mol Fe and 2 mol Fe 2 O 3 2 mol Fe 2 O 3 4 mol Fe O 2 and Fe 2 O 3 3 mol O 2 and 2 mol Fe 2 O 3 2 mol Fe 2 O 3 3 mol O 2 Writing Mole-Mole Factors

7. Consider the following equation: 3H 2 (g) + N 2 (g) 2NH 3 (g) 1. A mole factor for H 2 and N 2 is 2. A mole factor for NH 3 and H 2 is Learning Check

8. How many moles of Fe 2 O 3 can form from 6.0 mol O 2 ? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) STEP 1 Given 6.0 mol O 2 Need: moles of Fe 2 O 3. STEP 2 moles O 2 moles Fe 2 O 3 STEP 3 3 mol O 2 = 2 mol Fe 2 O 3 3 mol O 2 and 2 mol Fe 2 O 3 2 mol Fe 2 O 3 3 mol O 2 STEP 4 Set up problem using the mol factor. 6.0 mol O 2 x 2 mol Fe 2 O 3 = 4.0 mol Fe 2 O 3 3 mol O 2 Calculations with Mole Factors

9. How many moles of Fe are needed to react with 12.0 mol O 2 ? 4Fe(s) + 3O 2 (g) 2 Fe 2 O 3 (s) 12.0 mol O 2 x 4 mol Fe = 3 mol O 2 Learning Check 16.0 mol of Fe

10. Conservation of Mass The Law of Conservation of Mass indicates No change in _____ _____ occurs in a reaction. Mass of products is ______ to mass of reactants.

11. Mass Calculations for Reactions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

12. Stoichiometry Chemical Stoichiometry: using mass and quantity relationships among reactants and products in a chemical reaction to make predictions about how much product will be made.

13. In an ordinary chemical reaction, Matter cannot be created nor destroyed. The number of atoms of each element are equal. The mass of reactants equals the mass of products. H 2 (g) + Cl 2 (g) 2HCl(g) 2 mol H, 2 mol Cl = 2 mol HCl atoms atoms molecules 2(1.008) + 2(35.45) = 2(36.46) 72.92 g = 72.92 g Law of Conservation of Mass

14. Conservation of Mass 2 Ag + S → Ag 2 S 2 mol Ag + 1 mol S = 1 mol Ag 2 S 2 (107.9 g) + 1(32.07 g) = 1 (247.9 g) 247.9 g reactants = 247.9 g product Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

15. Stoichiometric Mantra Grams A to Moles A to Moles B to Grams B 1 2 3

16. The reaction between H 2 and O 2 produces 13.1 g H 2 O. How many grams of O 2 reacted? 2H 2 (g) + O 2 (g) 2H 2 O(g) STEP 1 Given 13.1 g H 2 O Need grams O 2 STEP 2 Plan: g H 2 O mol H 2 O mol O 2 g O 2 STEP 3 1 mol O 2 = 2 mol H 2 O 1 mol H 2 O = 18.02 g 1 mol O 2 = 32.00 g O 2 STEP 4 13.1 g H 2 O x 1 mol H 2 O x 1 mol O 2 x 32.00 g O 2 18.02 g H 2 O 2 mol H 2 O 1 mol O 2 = Calculating the Mass of a Reactant 11.6 g O 2

17. How many grams of O 2 are needed to produce 0.400 mol Fe 2 O 3 ? 4Fe(s) + 3O 2 (g) 2 Fe 2 O 3 (s) mole factor molar mass 0.400 mol Fe 2 O 3 x 3 mol O 2 x 32.00 g O 2 2 mol Fe 2 O 3 1 mol O 2 = 19.2 g O 2 Learning Check

18. Acetylene gas C 2 H 2 burns in the oxyacetylene torch for welding. How many grams of C 2 H 2 are burned if the reaction produces 75.0 g CO 2 ? 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O(g) C 12.01 g/mol H 1.008 g/mol O 16.00 g/mol

19. Limiting Reactants Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

20. Limiting Reactant A limiting reactant in a chemical reaction is the substance that Is used up first. Stops the reaction. Limits the amount of product that can form.

21. Reacting Amounts In a table setting, there is 1 plate, 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? Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

22. Reacting Amounts Four table settings can be made. Initially Use Left over plates 5 4 1 forks 6 4 2 spoons 4 4 0 knives 7 4 3 The limiting item is the spoon.

23. Example of Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices bread and 1 jar of peanut butter? With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item. Copyright © 2008 by Pearson Education, Inc. publishing as Benjamin Cummings

24. Example of 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. Copyright © 2008 by Pearson Education, Inc. publishing as Benjamin Cummings

25. Limiting Reactants When 4.00 mol H 2 is mixed with 2.00 mol Cl 2,how many moles of HCl can form? H 2 (g) + Cl 2 (g)  2HCl (g) 4.00 mol 2.00 mol ??? mol Calculate the moles of product from each reactant, H 2 and Cl 2. The limiting reactant is the one that produces the smaller amount of product.

26. Limiting Reactants Using Moles H 2 (g) + Cl 2 (g)  2HCl (g) 4.00 mol 2.00 mol ??? mol HCl from H 2 4.00 mol H 2 x 2 mol HCl = 8.00 mol HCl 1 mol H 2 HCl from Cl 2 2.00 mol Cl 2 x 2 mol HCl = 4.00 mol HCl 1 mol Cl 2 The limiting reactant is Cl 2 because it produces the smaller number of moles of HCl and is used up first. (not possible) (smaller amount)

27. Checking Calculations InitiallyH 2 4.00 mol Cl 2 2.00 mol 2HCl 0 mol Reacted/ Formed -2.00 mol +4.00 mol Left after reaction 2.00 mol Excess 0 mol Limiting 4.00 mol

28. Limiting Reactants Using Mass If 4.80 mol Ca mixed with 2.00 mol N 2, which is the limiting reactant? 3Ca(s) + N 2 (g)  Ca 3 N 2 (s) Moles of Ca 3 N from Ca 4.80 mol Ca x 1 mol Ca 3 N 2 = 1.60 mol Ca 3 N 2 3 mol Ca Moles of Ca 3 N 2 from N 2 2.00 mol N 2 x 1 mol Ca 3 N 2 = 2.00 mol Ca 3 N 2 1 mol N 2 All Ca is used up when 1.60 mol Ca 3 N 2 forms. Thus, Ca is the limiting reactant. N 2 is in excess. (smaller amount) (not possible)

29. Limiting Reactants Using Mass Calculate the mass of water produced when 8.00 g H 2 and 24.0 g O 2 react? 2H 2 (g) + O 2 (g) 2H 2 O(l) What do we need to do? Determine amount of water is produced for each of the reactants.

30. Limiting Reactants Using Mass Calculate the mass of water produced when 8.00 g H 2 and 24.0 g O 2 react? 2H 2 (g) + O 2 (g) 2H 2 O(l) Calculate the grams of H 2 O produced for each reactant. H 2 : 8.00 g H 2 x 1 mol H 2 x 2 mol H 2 O x 18.02 g H 2 O 2.016 g H 2 2 mol H 2 1 mol H 2 O = 71.5 g H 2 O O 2 : 24.0 g O 2 x 1 mol O 2 x 2 mol H 2 O x 18.02 g H 2 O 32.00 g O 2 1 mol O 2 1 mol H 2 O = 27.0 g H 2 O O 2 is the limiting reactant. (not possible) (smaller amount)

31. Limiting Reactants Using Mass Calculate the masses of reactants when (1) 8.00 g H 2 and (2) 24.0 g O 2 react? 2H 2 (g) + O 2 (g) 2H 2 O(l) Calculate the grams of other reactant needed for each reactant. H2:H2: 8.00 g H 2 x 1 mol H 2 x 1 mol O 2 x 32.00 g O 2 2.016 g H 2 2 mol H 2 1 mol O 2 = 63.5 g O 2 O2:O2: 24.0 g O 2 x 1 mol O 2 x 2 mol H 2 x 2.016 g H 2 32.00 g O 2 1 mol O 2 1 mol H 2 = 3.02 g H 2 O 2 is the limiting reactant. Some H 2 remains in excess. (not possible) (is possible)

32. Percent Yield Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

33. Theoretical, Actual, and Percent Yield Theoretical yield The ___________ amount of product calculated using the balanced equation. Actual yield The ___________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)

34. To calculate the percent yield, the actual yield and theoretical yield are needed. You prepared cookie dough to make 5 dozen cookies. The phone rings and you answer. While talking, a sheet of 12 cookies burn and you have to throw them out. The rest of the cookies 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

35. Learning Check Without proper ventilation and limited 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 CO are produced when 30.0 g O 2 are used? theoretical yield of CO 30.0 g O 2 x 1 mol O 2 x 2 mol CO x 28.01 g CO 32.00 g O 2 1 mol O 2 1 mol CO = 52.5 g CO (theoretical) percent yield 40.0 g CO (actual) x 100 = 76.2 % yield 52.5 g CO (theoretical)

36. Learning Check When N 2 and 5.00 g H 2 are mixed, the reaction produces 16.0 g NH 3. What is the percent yield for the reaction? N 2 (g) + 3H 2 (g) 2NH 3 (g) 5.00 g H 2 x 1 mol H 2 x 2 mol NH 3 x 17.03 g NH 3 2.016 g H 2 3 mol H 2 1 mol NH 3 = 28.2 g NH 3 (theoretical) Percent yield = 16.0 g NH 3 x 100 = 56.7 % 28.2 g NH 3

37. Limiting Reactant What is the limiting reactant when 2.00g of Na and 2.00g of Cl 2 combine as follows: 2Na + Cl 2 2NaCl Na 22.99 g/mol Cl 35.45 g/mol

38. Problem continued…. How many grams of the remaining reactant would be left over once the reaction has run to completion? 2Na + Cl 2 2NaCl

39. Problem continued…. If the actual yield of NaCl is 2.29g, what is the percent yield?

40. Limiting Reactant Ethylene burns in air according to the following equation: C 2 H 4 (g) + 3 O 2 (g) 2 CO 2 (g) + 2 H 2 O(l) a. How many grams of CO 2 will be formed when a mixture of 2.93g of C 2 H 4 and 4.29g of O 2 combine? C 12.01 g/mol H 1.008 g/mol O 16.00 g/mol

41. Problem continued…. b. How many grams of the reactant in excess would be left over once the reaction has gone to completion?

42. Problem continued…. c. If the percent yield of CO 2 is 72.1%, what was the actual yield?

43. Limiting Reactant Sulfur trioxide is prepared from SO 2 according to the following equation: 2SO 2 + O 2 2SO 3 In this reaction, not all SO 2 is converted to SO 3 even with excess O 2. In a given experiment, 21.2g of SO 3 is actually produced from 24.0g of SO 2. a) What is the theoretical yield of SO 3 ? b) What is the percent yield? S 32.07 g/mol O 16.00 g/mol