1. by Steven S. Zumdahl & Donald J. DeCoste University of Illinois Introductory Chemistry: A Foundation, 6 th Ed. Introductory Chemistry, 6 th Ed. Basic Chemistry, 6 th Ed.

2. Chapter 9 Chemical Quantities

3. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 3 Information Given by the Chemical Equation Balanced equations show the relationship between the relative numbers of reacting molecules and product molecules. 2 CO + O 2  2 CO 2 2 CO molecules react with 1 O 2 molecules to produce 2 CO 2 molecules

4. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 4 Since the information given is relative: 2 CO + O 2  2 CO 2 1.200 CO molecules react with 100 O 2 molecules to produce 200 CO 2 molecules 2.2 billion CO molecules react with 1 billion O 2 molecules to produce 20 billion CO 2 molecules 3.2 moles CO molecules react with 1 mole O 2 molecules to produce 2 moles CO 2 molecules 4.12 moles CO molecules react with 6 moles O 2 molecules to produce 12 moles CO 2 molecules Information Given by the Chemical Equation

5. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 5

6. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 6 The balanced breakdown of water molecules.

7. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 7 The coefficients in the balanced chemical equation show the molecules and the mole ratio of the reactants and products. Since moles can be converted to masses, we can determine the mass ratio of the reactants and products as well. Information Given by the Chemical Equation

8. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 8 2 CO + O 2  2 CO 2 2 moles CO = 1 mole O 2 = 2 moles CO 2 Since 1 mole of CO = 28.01 g, 1 mole O 2 = 32.00 g, and 1 mole CO 2 = 44.01 g 2(28.01) g CO = 1(32.00) g O 2 = 2(44.01) g CO 2 Information Given by the Chemical Equation

9. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 9 Stoichiometry Rules Write the balanced equation. Find the necessary molar masses. Convert grams to moles. Determine the mole ratio from the equation. Calculate moles. Calculate grams.

10. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 10 General Format Grams x 1 mole x mole ratio x grams = molar mass 1 mole *Mole ratio comes from coefficients *Grams per mole is molar mass

11. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 11 Example #1 Determine the number of moles of carbon monoxide required to react with 3.2 moles oxygen. Write the balanced equation: 2 CO + O 2  2 CO 2 Use the coefficients to find the mole relationship: 2 moles CO = 1 mol O 2 = 2 moles CO 2

12. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 12 Use dimensional analysis: the coefficients form the mole to mole ratio. Example #1 (cont.)

13. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 13 Determine the moles of carbon dioxide produced.

14. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 14 Example #2 (cont.) Determine the moles of carbon dioxide produced when 48 g of oxygen react. Write the balanced equation: 2 CO + O 2  2 CO 2 Use the coefficients to find the mole relationship: 2 moles CO = 1 mol O 2 = 2 moles CO 2

15. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 15 Determine the molar mass of each: 1 mol CO = 28.01 g 1 mol O 2 = 32.00 g 1 mol CO 2 = 44.01 g

16. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 16 Use the molar mass of the given quantity to convert it to moles. Use the mole relationship to convert the moles of the given quantity to the moles of the desired quantity: Example #2 (cont.)

17. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 17 Determine the number of grams of carbon dioxide this represents. Use the molar mass of the desired quantity to convert the moles to mass: Example #2 (cont.)

18. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 18 Example 3 - Grams to Grams 3Ca(OH) 2 + 2 H 3 PO 4  Ca 3 (PO 4 ) 2 + 6 H 2 O How much calcium hydroxide would be needed to completely react with 10 grams of phosphoric acid ? 10 g H 3 PO 4 x 1 mole H 3 PO 4 x 3 mol Ca(OH) 2 x 74 g = 98 g2 mol H 3 PO 4 1 mol Ca(OH) 2

19. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 19 Limiting and Excess Reactants Limiting reactant: a reactant that is completely consumed when a reaction is run to completion –Determines the amount of product that can be formed. –The limiting reagent is used up. Excess reactant: a reactant that is not completely consumed in a reaction –there is more than enough.

20. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 20 Limiting and Excess Reactants (cont.)

21. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 21 Example #3 Determine the limiting reagent when 3.2 moles oxygen reacts with 4.0 moles of carbon monoxide to produce carbon dioxide.

22. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 22 Example #3 (cont.) Write the balanced equation: 2 CO + O 2  2 CO 2 Use the coefficients to find the mole relationship: 2 moles CO = 1 mol O 2 = 2 moles CO 2

23. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 23 Use dimensional analysis to determine the number of moles of reactant A needed to react with reactant B and B with A: Example #3 (cont.)

24. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 24 Compare the calculated number of moles of reactant A to the number of moles given of reactant A. (Also use reactant B) If the calculated moles is greater, then A is the limiting reactant; if the calculated moles is less, then A is the excess reactant. The calculated moles of CO (6.4 moles) is greater than the given 4.0 moles; therefore CO is the limiting reactant. Example #3 (cont.)

25. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 25 Use the limiting reactant to determine the moles of product: Example #3 (cont.)

26. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 26 Example #4 Determine the mass of carbon dioxide produced when 48.0 g of oxygen reacts with 56.0 g of carbon monoxide. (Hint: start by determining the limiting reagent.)

27. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 27 p. 263 # 45a 45.a. Have 5 g of Na 2 B 4 O 7 5 g of H 2 SO 4 5 g H 2 O Need 2.2 g H 2 O 2.44 g H 2 SO 4 10.26 g Na 2 B 4 O 7 4.5 g H 2 O 5.4 g H 2 SO 4 11.1 g Na 2 B 4 O 7 Na 2 B 4 O 7 limits, 2.56 g H 2 SO 4 excess, 2.76 H 2 O excess

28. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 28 p. 263 # 47a, 51, 55, 57 47. a. O 2 limits, 8.25 g CO 2 and 4.51 g H 2 O produced 51. 1.74 g PbI 2 (KI limits) 55. No, not enough sodium oxalate 57. BaO 2 limits, 0.301 g H 2 O 2 formed

29. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 29 Theoretical Yield Theoretical yield: the maximum amount of a product that can be made by the time the limiting reactant is completely consumed Most reactions do not go to completion. Equations give us the theoretical yield.

30. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 30 Actual Yield Actual yield: the amount of product made in an experiment Experimental data Less due to: –Side reactions –Loss of product –Product solubility –Reaction reaches equilibrium

31. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 31 Percent Yield Percent yield: the percentage of the theoretical yield that is actually made ratio of the actual to the theoretical yield. Percent Yield = Actual Yield Theoretical Yield x 100%

32. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 32 Example #4a If in the last problem 72.0 g of carbon dioxide is actually made, what is the percentage yield ? (We calculated 88 g would be produced.)

33. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 33 Example #4a (cont.) Divide the actual yield by the theoretical yield, then multiply by 100% –The actual yield of CO 2 is 72.0 g –The theoretical yield of CO 2 is 88.0g

34. Copyright © Houghton Mifflin Company. All rights reserved. 9 | 34 Percent Yield Problems P. 265 # 63 to 66