CHAPTER 3-B Stoichiometry

1. CHAPTER 3-B Stoichiometry © 2012 by W. W. Norton & Company

2. Avogadro and the Mole

3. Stoichiometry • Stoichiometry: Relates the moles of products and reactants to each other and to measurable quantities.

4. Stoichiometry Aqueous solutions of NaOCl (household bleach) are prepared by the reaction of NaOH with Cl2: • 2 NaOH(aq) + Cl2(g)  NaOCl(aq) + NaCl(aq) + H2O(l) How many grams of NaOH are needed to react with 25.0 g of Cl2?

5. Stoichiometry 2 NaOH + Cl2→ NaOCl + NaCl + H2O 25.0 g Cl2 reacts with ? g NaOH

6. Calculate the molar mass of the following: Fe2O3 (Rust) C6H8O7 (Citric acid) C16H18N2O4 (Penicillin G) Balance the following, and determine how many moles of CO will react with 0.500 moles of Fe2O3. Fe2O3(s) + CO(g) Fe(s) + CO2(g) Avogadro and the Mole

7. Fe2O3 + CO → Fe + CO2 Balance (not a simple one) Save Fe for last C is balanced, but can’t balance O In the products the ratio C:O is 1:2 and can’t change Make the ratio C:O in reactants 1:2 Fe2O3 + 3CO → 2Fe + 3CO2 Avogadro and the Mole

8. Avogadro and the Mole Fe2O3 + 3CO → 2Fe + 3CO2

9. Stoichiometry

10. Stoichiometry • Aspirin is prepared by reaction of salicylic acid (C7H6O3) with acetic anhydride (C4H6O3) to form aspirin (C9H8O4) and acetic acid (CH3CO2H). Use this information to determine the mass of acetic anhydride required to react with 4.50 g of salicylic acid. How many grams of aspirin will result? How many grams of acetic acid will be produced as a by-product?

11. Stoichiometry Salicylic acid + Acetic anhydride → Aspirin + acetic acid C7H6O3 + C4H6O3 → C9H8O4 + CH3CO2H C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2 Balanced! Equal # moles for all

12. Stoichiometry 4.50 g Salicylic acid (C7H6O3) = ? moles MW C7H6O3 = 7 x 12.01 + 6 x 1.008 + 3 x 16.00 = 138.12 g/mole

13. Stoichiometry Since all compounds have the same S.C., there must be 0.0326 moles of all 4 of them involved in the reaction. g Aspirin (C9H8O4) = 0.0326 moles x MW Aspirin = .0326 x [9x12.01 + 8x1.008 + 4x16.00] =.0326 mole x 180.15 g/mole 5.87 g Aspirin

14. Yields of Chemical Reactions:If the actual amount of product formed in a reaction is less than the theoretical amount, we can calculate apercentage yield. Stoichiometry

15. Stoichiometry • Dichloromethane (CH2Cl2) is prepared by reaction of methane (CH4) with chlorine (Cl2) giving hydrogen chloride as a by-product. How many grams of dichloromethane result from the reaction of 1.85 kg of methane if the yield is 43.1%?

16. Stoichiometry CH4 + Cl2→ CH2Cl2 + HCl Balance CH4 + 2Cl2→ CH2Cl2 + 2HCl 1.85 kg CH4 = ? moles CH4

17. Stoichiometry CH4 + 2Cl2→ CH2Cl2 + 2HCl 1.85 kg CH4 = ? moles CH4 MW CH4 = 1x12.01 + 4x1.008 = 16.04 g/mole

18. Stoichiometry CH4 + 2Cl2→ CH2Cl2 + 2HCl 115 moles CH4 in theory we should produce: 115 moles of CH2Cl2 and 230 moles of HCl And use up 230 moles of Cl2

19. Stoichiometry CH4 + 2Cl2→ CH2Cl2 + 2HCl 115 moles of CH2Cl2 = ? g MW CH2Cl2 = 12.01 + 2x1.008 + 2x35.45 = 84.93 115 moles x (84.03 g/mole) = 9770 g

20. Stoichiometry CH4 + 2Cl2→ CH2Cl2 + 2HCl Expect 9770 g CH2Cl2 but the yield is 43.1% So we produced just 0.431 x 9770 g 4.21 kg CH2Cl2

21. Stoichiometry CH4 + 2Cl2→ CH2Cl2 + 2HCl Suppose the reaction went to completion (100% yield) Is mass conserved?

22. Stoichiometry CH4 + 2Cl2→ CH2Cl2 + 2HCl Start with 115 moles CH4 and 230 moles Cl2 total mass = 115x16.04 + 230x70.90 = 1850 + 16300 = 18150 only 3 sig. figs. → 18.2 kg

23. Stoichiometry CH4 + 2Cl2→ CH2Cl2 + 2HCl End with 115 moles CH2Cl2 and 230 moles HCl total mass = 115x84.93 + 230x36.46 = 9770 + 8390 = 18160 only 3 sig. figs → 18.2 kg

24. Stoichiometry • Limiting Reagents: The extent to which a reaction takes place depends on the reactant that is present in limiting amounts—the limiting reagent.

25. In this figure, ethylene oxide is the limiting reagent. Stoichiometry

26. Limiting Reagent: The reactant that governs the maximum amount of product that can be formed. Ex: Combustion of Octane C8H18 + O2→ CO2 + H2O If 100 g octane and 150 g oxygen are supplied for combustion, which is the limiting reagent? Stoichiometry

27. Limiting Reagent: The reactant that governs the maximum amount of product that can be formed. Ex: Combustion of Octane 2 C8H18 + 25 O2→ 16 CO2 + 18 H2O If 100 g octane and 150 g oxygen are supplied for combustion, which is the limiting reagent? Stoichiometry

28. 2 C8H18 + 25 O2→ 16 CO2 + 18 H2O Moles octane: 100 g x mole/114.22 g = 0.876 moles Moles oxygen: 150 g x mole/32.00 g = 4.69 moles Moles oxygen needed: 0.876 x 25/2 = 11.0 moles Moles octane needed: 4.69 x 2/25 = 0.375 moles Stoichiometry

29. 2 C8H18 + 25 O2→ 16 CO2 + 18 H2O 0.976 moles octane available and 0.375 moles needed 4.69 moles oxygen available and 11.0 moles needed So OXYGEN is the limiting reagent We won’t burn all of the octane Stoichiometry

30. Some Interesting Chemical Reactions 1. Production of Smog N2 + O2 + heat → 2 NO 2 NO + O2 → 2 NO2 (brown gas) 2. The Greenhouse Effect 2 C8H18 + 25 O2 → 18 H2O + 16 CO2 CO2 transmits visible light but absorbs heat

31. Some Interesting Chemical Reactions 3. Reduction of Iron Ore Fe2O3 + 3 CO → 2 Fe + 3 CO2 4. Depletion of Ozone O3 + uv rays → O2 + O CF2Cl2 + O3 → O2 + O

32. Some Interesting Chemical Reactions 5. Photosynthesis 6 CO2 + 6 H2O → C6H12O6 + 6 O2 6. Acid Rain S + O2 + heat → SO2 SO2 + H2O → H2SO3 (acid)