Chapter 9

1. Chapter 9 Stoichiometry

2. Section 9.1 The Arithmetic of Equations • Objectives: • Calculate the amount of reactants required or product formed in a nonchemical process • Interpret balanced chemical equations in terms of interacting moles, representative particles, masses, and gas volume at STP.

3. Stoichiometry • Stoichiometry is derived from the words stoicheion , meaning “element,” and metron, meaning “measure.” • The calculation of quantities in chemical reactions in a subject of chemistry called stoichiometry.

7. Interpreting Chemical Equations • Particles – atoms, molecules, formula units & ions • 1 mole = 6.02 x 1023 particles • Moles – tells the amount of a substance • Mass – mass is related to the number of atoms in a chemical equation • 1 mole = molar mass of substance (use periodic table) • Volume – All gases have the same value at STP (Standard Temperature, 0 °C, and Standard Pressure, 1 atm) • 1 mole = 22.4 L at STP

8. Mole Ratio • Mole ratio is determined by the coefficients in front of the element or compound. Ex. 2H2 + O2 2H2O 2 moles of H2:1 mole of O2:2 moles H2O 2 mol H2 ,2 mol H2, 1 mol O2 1 mol O2 2 mol H2O 2 mol H2O 1 mol O2 ,2 mol H2O, 2 mol H2O 2 mol H2 2 mol H2 1 mol O2

10. Ex #1: Interpret the number of moles of each substance. C2H5OH(l) + 3O2(g)→ 2CO2(g) + 3H2O(g) 1 mole : 3 moles : 2 moles : 3 moles

11. Ex #2: Interpret the following equation in terms of representative particles. 2K(s) + 2H2O(l)→ 2KOH(aq) + H2(g) 2 atoms : 2 molecules : 2 formula : 1 molecule K H2O units KOH H2

12. Section 9.1 The Arithmetic of Equations • Did We Meet Our Objectives? • Calculate the amount of reactants required or product formed in a nonchemical process • Interpret balanced chemical equations in terms of interacting moles, representative particles, masses, and gas volume at STP.

13. Section 9.2 Chemical Calculations • Objectives: • Construct mole ratios from balanced chemical equations and apply these ratios in mole-mole stoichiometric calculations • Calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, representative particles, and volumes of gases at STP

14. Mole to Mole CalculationsEx #3: How many moles of aluminum are needed to form 3.7 mol Al2O3? 4Al + 3O2→ 2Al2O3 3.7 mol Al2O3 x 4 mol Al = 7.4 mol Al 2 mol Al2O3

15. Mole to Mole CalculationsEx #4: How many moles of oxygen are needed to form 3.7 mol Al2O3? 4Al + 3O2→ 2Al2O3 3.7 mol Al2O3 x 3 mol O2= 5.6 mol O2 2 mol Al2O3

16. Mass to Mole CalculationsEx #5: How many moles of oxygen are needed to form 3.7 g Al2O3? 4Al + 3O2→ 2Al2O3 3.7 g Al2O3 x 1 mol Al2O3 x 3 mol O2 = 101.96 g Al2O3 2 mol Al2O3 0.054 mol O2

17. Mole to Mass CalculationsEx #6: How many grams of oxygen are needed to form 3.7 moles Al2O3? 4Al + 3O2→ 2Al2O3 3.7 mol Al2O3 x 3 mol O2 x 32 g O2 = 2 mol Al2O3 1 mol O2 177.6 g O2 or 180 g O2

18. Mass to Mass CalculationsEx #7: How many grams of oxygen are needed to form 3.7 grams Al2O3? 4Al + 3O2→ 2Al2O3 3.7 g Al2O3 x 1 mol Al2O3 x 3 mol O2 x 32 g O2 = 101.96 g Al2O3 2 mol Al2O3 1 mol O2 1.7 g O2

19. Other Stoichiometric CalculationsEx #8: Assuming STP, how many liters of oxygen are needed to produce 19.8 L SO3? 2SO2(g) + O2(g)→ 2SO3(g) 19.8 L SO3 x 1 mol SO3 x 1 mol O2 x 22.4 L O2 = 22.4 L SO3 2 mol SO3 1 mol O2 9.9 L O2

20. Section 9.2 Chemical Calculations • Did We Meet Our Objectives? • Construct mole ratios from balanced chemical equations and apply these ratios in mole-mole stoichiometric calculations • Calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, representative particles, and volumes of gases at STP

21. Section 9.3 Limiting Reagent and Percent Yield • Objectives: • Identify and use the limiting reagent in a reaction to calculate the maximum amount of product(s) produced and the amount of excess reagent • Calculate theoretical yield, actual yield, or percent yield given appropriate information

22. Limiting Reagent • Have you ever started to make a pizza, but ran out of ingredients halfway through? • You can only make as much as the ingredient which runs out first • Limiting reagent – determines the amount of product that can be formed in a reaction, and is the reactant that runs out first • Excess reagent – reactant that is not completely used up in the reaction (left over)

23. Limiting Reagents • 50.0g of N2 reacts with 10g of H2 . What is the limiting reagent? There are 2 ways to solve: • Calculate amount of H2 actually needed to react completely with 50g N2 . If the amount of H2 needed is bigger than 10g, then N2 is limiting. If H2 needed is smaller than 10g, then it is limiting. • Or, calculate the amount of NH3 produced when 50g N2 reacts completely, and do the same for when 10g H2 reacts completely. The one that produces a smaller amount of product is the limiting reagent.

24. Ex #9: 50.0 g of N2 reacts with 10 g of H2. What is the limiting reagent? 1 mol N2 50.0 g N2 2 mol NH3 X X = 28.02 g N2 1 mol N2 10 g H2 1 mol H2 2 mol NH3 X X = 2.02 g H2 3 mol H2

25. Percent Yield • Percent yield – ratio of actual to theoretical yield • Theoretical yield – maximum amount of products that can be formed from reactants • Actual yield – amount of products actually produced, often less than theoretical

26. Ex #10: When 84.8 g of iron (III) oxide reacts with an excess of carbon monoxide, 54.3 g of iron is produced. What is the percent yield of this reaction? Fe2O3 + 2CO → 2Fe + 3CO2 84.8 g Fe2O3 1 mol Fe2O3 2 mol Fe 55.85 g Fe X X X 1 mol Fe2O3 159.70 g Fe2O3 1 mol Fe

27. Section 9.3 Limiting Reagent and Percent Yield • Did We Meet Our Objectives? • Identify and use the limiting reagent in a reaction to calculate the maximum amount of product(s) produced and the amount of excess reagent • Calculate theoretical yield, actual yield, or percent yield given appropriate information