1. Stoichiometry Predicting amounts of reagents needed or amounts of products made

2. Stoichiometry Composition stoichiometry –Mass relationships of elements in compounds Reaction stoichiometry –Mass relationships between participants in a chemical reaction Must begin all stoichiometry problems with a BALANCED chemical equation.

3. Conservation of Charge Balanced chemical equation must be balanced for both mass and charge. Total charge on reactant side must equal total charge on product side. So far, most of the equations you have seen are neutral on each side, but this is not always the case.

4. Coefficients in Balanced Equations Give relative number of particles in rxn. Give relative number of moles in rxn. Coefficients give relationship between moles of each component in equation. You use these relationships in MOLE-MOLE problems to convert from moles of 1 substance to moles of another substance.

5. Stoichiometry Problems Come in several flavors. Start with the simplest: mole-mole Given: amount of one substance in moles Unknown: amount of some other substance in moles

6. Mole-mole stoichiometry problem Stoichiometry problems are factor-label problems or conversion problems. Use the balanced chemical equation to generate the conversion factors.

7. 2 C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 O How much CO 2 will be produced if 4 moles of C 2 H 6 are consumed? Identify a stoichiometry problem: 1) You always get a balanced chemical equation. 2) The problem will ask how many moles or how many liters of one species given a certain amount of another species. 3) The equation has a  arrow.

8. 2 C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 O How much CO 2 will be produced if 4 moles of C 2 H 6 are consumed? Identify given: 4 moles of C 2 H 6. Identify conversion factor from equation: 4 moles CO 2 2 moles C 2 H 6

9. Conversion 4 moles C 2 H 6 4 moles CO 2 2 moles C 2 H 6 = 4 X 4 2 moles CO 2 = 8 moles CO 2

10. 2 C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 O How much H 2 O will be produced in the combustion of 5 moles of C 2 H 6 ? Given: 5 mol C 2 H 6 Conversion factor: 6 mol H 2 O 2 mol C 2 H 6

11. Conversion 5 mol C 2 H 6 6 mol H 2 O 2 mol C 2 H 6 = 5 X 6 2 mol H 2 O= 15 mol H 2 O

12. 2 C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 O oxygen C 2 H 6How much oxygen will react with 5 moles of C 2 H 6 ? Given: 5 moles C 2 H 6 Conversion factor: 7 mol O 2 2 mol C 2 H 6

13. Conversion 5 moles C 2 H 6 7 mol O 2 2 mol C 2 H 6 = 5 X 7 mol O 2 2 = 17.5 mol O 2

14. Other flavors of problems Other types of stoichiometry problems include: Volume-Volume (for gas phase only) Mole-mass or mass-mole Mass-Mass

15. Gas-Phase Equations 1 mole of any gas occupies the same volume as any other gas. At STP, 22.4 L. So coefficients in equations represent the ratio of the volumes of gases involved in the reaction. For rxns where all reactants & products are gases,  volume-volume problems. Unit doesn’t matter as long as it’s constant throughout.

16. N 2 (g) + 3 H 2 (g)  2 NH 3 (g) How much hydrogen gas will react with 15 liters of nitrogen gas? Given: 15 L of N 2 (g) Conversion factor: 3 L H 2 1 L N 2

17. Conversion 15 L N 2 (g) 3 L H 2 1 L N 2 = 15 X 3 L H 2 1 = 45 L H 2

18. Other types of stoich problems General strategy: 1.Convert given into moles 2.Perform stoichiometric calculation using mole ratios from balanced equation 3.Convert to desired unit

19. A word about …

20. Reactions in aqueous solution Many reactions, esp. many double replacement reactions, occur in water. What happens when substances dissolve in water? Depends on if they are ionic or covalent.

21. Dissolving Covalent substance – sugar or C 6 H 12 O 6 C 6 H 12 O 6 (s)  C 6 H 12 O 6 (aq) The sugar molecules are spread out among the water molecules.

22. Dissolving Ionic substance – table salt or NaCl NaCl(s)  Na + (aq) + Cl - (aq) The ions are spread out among the water molecules.

23. Double Replacement Reactions Often occur when you mix 2 solutions of ionic compounds. –1 product may be water or –1 product may be a gas or –1 product may be a solid We say these reactions “go to completion.”

24. Reactions producing Solids Precipitation: the opposite of dissolving! What do you see in the following clips: S1043.mov S1045.mov S1046.mov S1050.mov S1057.mov S1058.mov and S1060.movS1058.movS1060.mov

25. Reactions producing Solids 2 NaOH(aq) + CuCl 2 (aq)  2 NaCl(aq) + Cu(OH) 2 (s) Complete Chemical Equation

26. Complete Ionic Equations 2Na + (aq) + 2OH - (aq) + Cu 2+ (aq) + 2Cl - (aq)  2Na + (aq) + 2Cl - (aq) + Cu(OH) 2 (s) Substances that are ions in solution are written as ions in solution.

27. Net Ionic Equations 2Na + (aq) + 2OH - (aq) + Cu 2+ (aq) + 2Cl - (aq)  2Na + (aq) + 2Cl - (aq) + Cu(OH) 2 (s) Notice that some ions do not participate in the reaction. They are spectator ions. Cross out all the spectator ions & you get the net ionic equation. 2OH - (aq) + Cu 2+ (aq)  Cu(OH) 2 (s)

28. Reactions that form Water HBr(aq) + NaOH(aq)  H 2 O(l) + NaBr(aq) H + (aq) + Br - (aq) + Na + (aq) + OH - (aq)  H 2 O(l) + Na + (aq) + Br - (aq) H + (aq) + OH - (aq)  H 2 O(l)

29. Reactions that form Gases HCl(aq) + NaHCO 3 (aq)  H 2 CO 3 (aq) + NaCl(aq) H 2 CO 3 (aq)  H 2 O(l) + CO 2 (g) ______________________________________ HCl(aq) + NaHCO 3 (aq)  H 2 O(l) + CO 2 (g) + NaCl(aq)

30. Reactions that form Gases HCl(aq) + NaHCO 3 (aq)  H 2 O(l) + CO 2 (g) + NaCl(aq) H + (aq) + Cl - (aq) + Na + (aq) + HCO 3 - (aq)  H 2 O(l) + CO 2 (g) + Na + (aq) + Cl - (aq) H + (aq) + HCO 3 - (aq)  H 2 O(l) + CO 2 (g)