2. Stoichiometry

3. Needs a balanced equation Use the balanced equation to predict ending and / or starting amounts Coefficients are now mole ratios

4. In terms of Moles The coefficients tell us how many moles of each kind. Mole ratio - conversion ratio that relates the amounts in moles of any two substances in a chemical reaction. Molar mass - mass, in grams, of one mole of a substance.

5. Review: Molar Mass A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound. CO 2 = 44.01 grams per mole H 2 O = 18.02 grams per mole Ca(OH) 2 = 74.10 grams per mole

6. Mole Ratios 2 Al 2 O 3 (l)  4 Al(s) + 3 O 2 (g) Mole RatiosMole Ratio (Fraction) 2 mol Al 2 O 3 : 4 mol Al 2 mol Al 2 O 3 : 3 mol O 2 4 mol Al : 3 mol O 2 2 mol Al 2 O 3 4 mol Al 2 mol Al 2 O 3 3 mol O 2 4 mol Al 3 mol O 2

7. III. Stoichiometric “road map” (Use the balanced chemical equation) aA + bB cC + dD Mass A Atoms Molecules Formula Units A Mol AMol B Mass B Atoms Molecules Formula Units B Mol Ratio Using the Coefficients from the balanced chemical equation Molar mass 6.022 x 10 23 Molar mass 6.022 x 10 23

8. 3 A + 4 B  2 D + 1 F How many moles of F are produced from 1.00 mol of A? 1 mol A 3 mol A 1 mol F Molecules 6.022 x 10 23 Mass AMass B Atoms Molecules A Atoms B MolA MolB Molar mass 6.022 x 10 23 Mol Ratio Molecules = 0.33 mol F How many moles of D are produced from 5.00 mol of B? 5 mol B 4 mol B 2 mol D = 2.50 mol D

9. How many moles of lithium carbonate are produced when 5.3 mol CO 2 are reacted? CO 2 (g) + 2LiOH(s)  Li 2 CO 3 (s) + H 2 O(l) 1. What is your starting point? 2. What is your ending point? 5.3 mol of CO 2 mol of Li 2 CO 3 6.022 x 10 23 Mass AMass B Atoms Molecules A Atoms B MolA MolB Molar mass 6.022 x 10 23 Mol Ratio Molecules 5.3 mol CO 2 1 mol CO 2 1 mol Li 2 CO 3 = 5.3 mol Li 2 CO 3

10. 3 A + 4 B  2 D + 1 F How many grams of F are produced from 1.00 mol of A? If MM of F is 10.0g/mol. 1 mol A 3 mol A 1 mol F = 3.33g F How many grams of D are produced from 5.00 mol of B? MM of D is 20.0g/mol 5 mol B 4 mol B 2 mol D = 50.0g D 1 mol F 10 g F 1 mol D 20 g D 6.022 x 10 23 Mass AMass B Atoms Molecules A Atoms B MolA MolB Molar mass 6.022 x 10 23 Mol Ratio Molecules

11. What is the mass of glucose (C 6 H 12 O 6 ) produced from 3.00 mol of water (H 2 O)? 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (s) + 6O 2 (g) 6.022 x 10 23 Mass A Mass B Atoms Molecules A Atoms B MolA MolB Molar mass 6.022 x 10 23 Mol Ratio Molecules 3 mol H 2 O 1. What is your starting point? 2. What is your ending point? 3.00 mol of H 2 O g of C 6 H 12 O 6 6 mol H 2 O 1 mol C 6 H 12 O 6 =90.1 g C 6 H 12 O 6 180.81g C 6 H 12 O 6

12. 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (s) + 6O 2 (g) What is the mass of oxygen (O 2 ) produced from 2.50 mol of water (H 2 O)? 6.022 x 10 23 Mass AMass B Atoms Molecules A Atoms B MolA MolB Molar mass 6.022 x 10 23 Mol Ratio Molecules 1. What is your starting point? 2. What is your ending point? 2.50 mol of H 2 O g of O 2 2.5 mol H 2 O 6 mol H 2 O 6 mol O 2 1 mol O 2 32.0 g O 2 =80.0 g O 2

13. 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) How many moles of NO are formed from 824 g of NH 3 ? 1. What is your starting point? 2. What is your ending point? 824 g of NH 3 mol of NO 6.022 x 10 23 Mass AMass B Atoms Molecules A Atoms B MolA MolB Molar mass 6.022 x 10 23 Mol Ratio Molecules 824 g NH 3 17.03g 1 mol NH 3 4 mol NH 3 4 mol NO = 48.4 mol NO

14. 3 A + 4 B  2 D + 1 F How many grams of F are produced from 5.00g of A? If MM of F is 10.0g/mol and MM of A is 25.0g/mol 1 mole A 3 mole A 1 mole F = 0.67g F How many grams of D are produced from 5.00g of B? MM of D is 20.0g and MM of B is 10.0g/mol 1 mole B 4 mole B 2 mole D =5.00g D 1 mole F 10 g F 1 mole D 20 g D 6.022 x 10 23 Mass A Mass B Atoms Molecules A Atoms B MolA MolB Molar mass 6.022 x 10 23 Mol Ratio Molecules 5 g A 25 g A 5 g B 10 g B

15. Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) How many grams of SnF 2 are produced from the reaction of 30.00 g HF? 1. What is your starting point? 2. What is your ending point? 30.00 g of HF g SnF 2 1 mol HF 2 mol HF 1 mol SnF 2 = 117.5g SnF 2 1 mol SnF 2 156.71 g SnF 2 30.00g HF 20.01g HF 6.022 x 10 23 Mass AMass B Atoms Molecules A Atoms B MolA MolB Molar mass 6.022 x 10 23 Mol Ratio Molecules

16. Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 1. Identify reactants and products and write the balanced equation. Al+O2O2 Al 2 O 3 b. What are the reactants? a. Every reaction needs a yield sign! c. What are the products? d. What are the balanced coefficients? 43 2

17. = 6.50 g Al ? g Al 2 O 3 1 mol Al 26.98 g Al 4 mol Al 2 mol Al 2 O 3 1 mol Al 2 O 3 101.96 g Al 2 O 3 6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 = 12.3 g Al 2 O 3 1. What is your starting point? 6.50 g of Aluminum 2. What is your ending point?g of aluminum oxide Al+O2O2 Al 2 O 3 43 2 6.022 x 10 23 Mass AMass B Atoms Molecules A Atoms B MolA MolB Molar mass 6.022 x 10 23 Mol Ratio Molecules

18. Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) How many grams of HF are produced from the reaction of 150.5 g H 2 ? 6.022 x 10 23 Mass AMass B Atoms Molecules A Atoms B MolA MolB Molar mass 6.022 x 10 23 Mol Ratio Molecules 1 mol H 2 2 mol HF 1 mol HF 20.01g HF150.5 g H 2 2.02g H 2 = 2982 g HF

19. II. Limiting Reagent A. Stoichiometric amounts: The proportions indicated in the _________ rxn. B. Most reactions do not have stoichiometric amounts. Generally, one reactant will be _________ before the other. The reactant that is depleted first is known as the ___________________. The reactant that is left at the end of the reaction is called the ___________________. C. Analogy: How to make a cheese sandwich. 2 slices of bread + 1 slice of cheese → 1 cheese sandwich If you have 8 slices of bread and 6 slices of cheese, how many sandwiches can you make? __ (theoretical yield) What is the limiting reagent? ______ What is the excess reagent? _______ How much of the excess reagent is left at the end of the rxn? ______________ balanced depleted limiting reagent (LR) excess reagent (ER) 4 bread cheese 2 slices cheese

20. D. Theoretical yield: The amount of product in grams that forms if all of the ________________ has reacted. (This number is CALCULATED on paper! Units are in grams only!) E. Actual yield: The amount of product (in grams) that is actually made (This number is from the EXPERIMENT). F. Percent yield: The comparison of the actual yield to the theoretical yield. limiting reagent

21. Zn + 2 HCl  ZnCl 2 + H 2 If you have 1 mol of Zn, how much H 2 would you make? If you have 1 mol of HCl, how much H 2 would you make? 1 mol Zn 1 mol H 2 = 1 mol H 2 1 mol HCl 2 mol HCl 1 mol H 2 = 0.5 mol H 2 What is the limiting reagent? HCl How much H 2 is produced? 0.5 mol – theoretical yield

22. Zn + 2 HCl  ZnCl 2 + H 2 If you have 0.25 mol of Zn, how much H 2 would you make? If you have 1.00 mol of HCl, how much H 2 would you make? 0.25 mol Zn 1 mol Zn 1 mol H 2 = 0.25 mol H 2 1.00 mol HCl 2 mol HCl 1 mol H 2 = 0.5 mol H 2 What is the limiting reagent? Zn How much H 2 is produced? 0.25 mol - theoretical yield

23. PCl 3 + 3 H 2 O  H 3 PO 3 + 3 HCl 3.00 mol PCl 3 and 3.00 mol H 2 O react. Determine the limiting reactant and theoretical yield of HCl. 1. Determine the limiting reactant 3.00 mol H 2 O 3 mol H 2 O 3 mol HCl = 3.00 mol HCl 3.00 mol PCl 3 1 mol PCl 3 3 mol HCl = 9.00 mol HCl  EXCESS  LIMITING 2. Determine the theoretical yield of HCl 3.00 mol

24. PCl 3 + 3 H 2 O  H 3 PO 3 + 3 HCl 2. Determine the theoretical yield of HCl 3.00 mol 3. Determine the theoretical yield of HCl in grams 3.00 mol HCl 1 mol HCl 36.46g HCl = 109 g HCl

25. PCl 3 + 3 H 2 O  H 3 PO 3 + 3 HCl Determine the limiting reactant and theoretical yield (g) of H 3 PO 3 if 225 g of PCl 3 and 123 g of H 2 O are reacted. 1. Determine the limiting reactant 1 mol PCl 3 1 mol H 3 PO 3 82.00g H 3 PO 3 225 g PCl 3 137.32g PCl 3 = 134g H 3 PO 3 1 mol H 2 O 3 mol H 2 O 1 mol H 3 PO 3 82.00g H 3 PO 3 123 g H 2 O 18.02g H 2 O = 187g H 3 PO 3 EXCESS LIMITING

26. PCl 3 + 3 H 2 O  H 3 PO 3 + 3 HCl The theoretical yield of this reaction is 134g H 3 PO 3. However, the actual yield from the experiment is 120g. Calculate the percent yield. 120 g H 3 PO 4 134 g H 3 PO 4 X 100%= 89.6 %

27. N 2 + 3 H 2  2 NH 3 Determine the limiting reagent, the theoretical yield and the percentage yield if 14.0g N 2 are mixed with 9.0g H 2 and the 16.1g NH 3 actually formed. 1 mol N 2 2 mol NH 3 1 mol NH 3 17.04g NH 3 14 g N 2 = 17.0g NH 3 LIMITING 28.01g N 2 1 mol N 2 1 mol H 2 2 mol NH 3 1 mol NH 3 17.04g NH 3 9 g H 2 = 50.6g NH 3 2.02g H 2 3 mol H 2 EXCESS 16.1 g NH 3 X 100%= 94.2 % 17.0 g NH 3

28. 16.1g of bromine are mixed with 8.42g of chlorine to give an actual yield of 21.1g of bromine monochloride. Determine the limiting reactant and the percentage yield. 1 mol Br 2 2 mol BrCl 1 mol BrCl 115.35g BrCl16.1 g Br 2 = 23.2g BrCl LIMITING 159.8g Br 2 1 mol Br 2 EXCESS 21.1 g BrCl X 100%= 90.9 % 23.2 g BrCl Br 2 + Cl 2  2 BrCl 1 mol Cl 2 2 mol BrCl 1 mol BrCl 115.35g BrCl8.42 g Cl 2 = 27.4g BrCl 70.9g Cl 2 1 mol Cl 2