2. 1 STOICHIOMETRY

3. 2 General Approach For Problem Solving 1. Clearly identify the Goal or Goals and the UNITS involved. (starting and ending unit) 2. Determine what is given and the UNITS. 3. Use conversion factors (which are really ratios) and their UNITS to CONVERT what is given into what is desired.

4. 3 Sample problem for general problem solving. Sam has entered into a 10 mile marathon. Use ALL of the following conversions (ratios) to determine how many inches there are in the race. 5280 ft = 1 mile; 12 inches = 1 ft 1. What is the goal and what units are needed? Goal = ______ inches 2. What is given and its units? 10 miles 3. Convert using factors (ratios). 10 miles = inches633600 Units match Given Goal Convert Menu

5. 4 Stoichiometry (more working with ratios) Ratios are found within a chemical equation. 2HCl + Ba(OH) 2  2H 2 O + BaCl 2 1 1 2 moles of HCl react with 1 mole of Ba(OH) 2 to form 2 moles of H 2 O and 1 mole of BaCl 2 coefficients give MOLAR RATIOS

6. Information given by chemical equations 2 C 6 H 6 (l) + 15 O 2 (g)  12 CO 2 (g) + 6 H 2 O (g)  In this equation there are 2 molecules of benzene reacting with 15 molecules of oxygen to produce 12 molecules of carbon dioxide and 6 molecules of water  In this equation there are 2 molecules of benzene reacting with 15 molecules of oxygen to produce 12 molecules of carbon dioxide and 6 molecules of water.  This equation could also be read as 2 moles of benzene reacts with 15 moles of oxygen to produce 12 moles of carbon dioxide and 6 moles of water. MOLE RATIO Since the relationship between the actual number of molecules and the number of moles present is 6.02 x 10 23, a common factor between all species involved in the equation, a MOLE RATIO relationship can be discussed.

7. Information given by chemical equations 2 C 6 H 6 (l) + 15 O 2 (g)  12 CO 2 (g) + 6 H 2 O (g) The MOLE RATIO for benzene and oxygen is 2 : 15. It can be written as: 2 moles C 6 H 6 or as 15 moles of O 2 15 moles O 2 2 moles of C 6 H 6 The MOLE RATIO for oxygen and carbon dioxide is 15 : 12. It can be written as: 12 moles CO 2 or as 15 moles of O 2 15 moles O 2 12 moles of CO 2 NOTE: The MOLE RATIO is used for converting moles of one substance into moles of another substance. Without the balanced equation there is no other relationship between two different compounds.

8. Using the mole ratio to relate the moles of one compound to the moles of another compound is the part of chemistry called STOICHIOMETRY !!!!! 2 H 2 (g) + O 2 (g) 2 H 2 O (g) 2 H 2 (g) + O 2 (g)  2 H 2 O (g) Q. How many mole of hydrogen are necessary to react with 2 moles of oxygen in order to produce exactly 4 moles of water? A. 2 mol O 2 (2 moles H 2 / 1 mole O 2 ) = 4 mole H 2

9. STOICHIOMETRY The Stoichiometry Flow Chart Use Molar mass (A) Use mole ratio from equation Use Molar mass (B)

10. STOICHIOMETRY 2 H 2 (g) + O 2 (g) 2 H 2 O (g) 2 H 2 (g) + O 2 (g)  2 H 2 O (g) Q1. How many moles of hydrogen are necessary to react with 15.0 g of oxygen? A. 15.0g O 2 ( 1 mole O 2 ) ( 2 mole H 2 ) = 0.938 moles H 2 32.0 g 1 mole O 2 Q2. How many grams of hydrogen are necessary to react with 15.0 g of oxygen? A. 15.0g O 2 ( 1 mole O 2 ) ( 2 mole H 2 ) ( 2.016 g H 2 ) = 1.89 g H 2 32.0 g 1 mole O 2 1 mole H 2

11. STOICHIOMETRY 2 H 2 (g) + O 2 (g)  2 H 2 O (g) Q3. How many grams of water are produced from 15.0 g of oxygen? A. 15.0g O 2 ( 1 mole O 2 ) ( 2 mole H 2 O ) ( 18.0 g H 2 O ) =16.9 g H 2 O 32.0 g 1 mole O 2 1 mole H 2 O Q4. How much hydrogen and oxygen is needed to produce 25.0 grams of water? A. 25.0g H 2 O ( 1 mole H 2 O ) ( 2 mole H 2 ) ( 2.016 g H 2 ) = 2.80 g H 2 18.0 g 2 mole H 2 O 1 mole H 2 A. 25.0g H 2 O ( 1 mole H 2 O ) ( 1 mole O 2 ) ( 32.0 g O 2 ) = 22.2 g O 2 18.0 g 2 mole H 2 O 1 mole O 2 Notice that the Law of Conservation of Mass still applies.

12. How many grams of solid are formed when 10.0 g of lead reacts with excess phosphoric acid? 1.Write the chemical equation: Pb + H 3 PO 4  ? You recognize that this is a single displacement (replacement) reaction. So Pb (a metal) will displace (replace) H (the cation). Pb + H 3 PO 4  Pb 3 (PO 4 ) 2 + H 2 2. Balance the equation: 3 Pb+2 H 3 PO 4  Pb 3 (PO 4 ) 2 + 3 H 2 3. Make a list under the appropriate substance 3 Pb+2 H 3 PO 4  Pb 3 (PO 4 ) 2 (s) + 3 H 2 (g) 10.0gm=? Start with what is given: 13.1 g Pb 3 (PO 4 ) 2 10.0gPb ( 1 mole Pb )( 1 mole Pb 3 (PO 4 ) 2 )( 811 g Pb 3 (PO 4 ) 2 ) = 13.1 g Pb 3 (PO 4 ) 2 207 g Pb 3 mole Pb 1 mole Pb 3 (PO 4 ) 2

13. PRACTICE PROBLEM # 18 1. How many grams of gas can be produced from 0.8876 moles of HgO? 2 HgO  2 Hg + O 2 2. How many moles of fluorine are required to produce 12.0 grams of KrF 6 ? Given the equation: Kr + 3 F 2  KrF 6 3. How many grams of Na 2 CO 3 will be produced from the thermal decomposition of 250.0 g of NaHCO 3 ? 4. How many grams of CO 2 can be produced by the reaction of 75.0 grams of C 2 H 2 with excess oxygen? 5. Cu + 2 AgNO 3  Cu(NO 3 ) 2 + 2 Ag. How many grams of silver is produced when 125.0 g of copper is reacted with excess silver nitrate solution? 14.20g 0.182 mol 157.7 g 254 g 424.9 g

14. GROUP STUDY PROBLEM # 18 ______1. How many grams of liquid product can be produced from 3.55 moles of HgO? 2 HgO  2 Hg + O 2 ______2. How many moles of fluorine are required to produce 3.0 grams of KrF 6 ? Given the equation: Kr + 3 F 2  KrF 6 ______3. How many grams of Na 2 CO 3 will be produced from the decomposition of 20.0g of NaHCO 3 ? ______4. How many grams of O 2 are needed to combust 55.0 grams of C 2 H 4 ? ______5. Cu + 2 AgNO 3  Cu(NO 3 ) 2 + 2 Ag. How many grams of silver is produced when 50.0 g of copper is reacted with excess silver nitrate solution?

15. 14 When N 2 O 5 is heated, it decomposes: 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) a. How many moles of NO 2 can be produced from 4.3 moles of N 2 O 5 ? = moles NO 2 4.3 mol N2O5N2O5 8.6 b. How many moles of O 2 can be produced from 4.3 moles of N 2 O 5 ? = mole O 2 4.3 mol N2O5N2O5 2.2 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 4.3 mol? mol 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 4.3 mol ? mol Mole – Mole Conversions Units match

16. 15 When N 2 O 5 is heated, it decomposes: 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) a. How many moles of N 2 O 5 were used if 210g of NO 2 were produced? = moles N 2 O 5 210 g NO 2 2.28 b. How many grams of N 2 O 5 are needed to produce 75.0 grams of O 2 ? = grams N 2 O 5 75.0 g O2O2 506 gram ↔ mole and gram ↔ gram conversions 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 210g? moles 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 75.0 g ? grams Units match

17. 16 Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? First write a balanced equation. Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g) 2 6 2 3 Gram to Gram Conversions

18. 17 Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g) 2 6 2 3 Now let’s get organized. Write the information below the substances. 3.45 g ? grams Gram to Gram Conversions

19. 18 Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g) 2 6 2 3 3.45 g ? grams Let’s work the problem. = g AlCl 3 3.45 g Al We must always convert to moles.Now use the molar ratio.Now use the molar mass to convert to grams. 17.0 Units match gram to gram conversions

20. 19 Molarity Molarity is a term used to express concentration. The units of molarity are moles per liter (It is abbreviated as a capital M) When working problems, it is a good idea to change M into its units.

21. 20 A solution is prepared by dissolving 3.73 grams of AlCl 3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution. What type of problem(s) is this? Molarity followed by dilution. Solutions

22. 21 A solution is prepared by dissolving 3.73 grams of AlCl 3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution. 1 st : = mol L 3.73 g 200.0 x 10 - 3 L 0.140 2 nd : M 1 V 1 = M 2 V 2 (0.140 M)(10.0 mL) = (? M)(100.0 mL) 0.0140 M = M 2 molar mass of AlCl 3 dilution formula final concentration Solutions

23. 22 50.0 mL of 6.0 M H 2 SO 4 (battery acid) were spilled and solid NaHCO 3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO 3 must be used? H 2 SO 4 (aq) + 2NaHCO 3  2H 2 O(l) + Na 2 SO 4 (aq) + 2CO 2 (g) Solution Stoichiometry

24. 23 50.0 mL 6.0 M ? g Look! A conversion factor! 50.0 mL of 6.0 M H 2 SO 4 (battery acid) were spilled and solid NaHCO 3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO 3 must be used? H 2 SO 4 (aq) + 2NaHCO 3  2H 2 O(l) + Na 2 SO 4 (aq) + 2CO 2 (g) Solution Stoichiometry = Our Goal

25. 24 50.0 mL 6.0 M ? g 50.0 mL of 6.0 M H 2 SO 4 (battery acid) were spilled and solid NaHCO 3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO 3 must be used? H 2 SO 4 (aq) + 2NaHCO 3  2H 2 O(l) + Na 2 SO 4 (aq) + 2CO 2 (g) Solution Stoichiometry = Our Goal = g NaHCO 3 H 2 SO 4 50.0 mL 1 mol H 2 SO 4 NaHCO 3 2 mol NaHCO 3 84.0 g mol NaHCO 3 50.4

26. 25 Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution are needed to neutralize 35.0 mL of 0.125 M H 2 SO 4 solution. First write a balanced Equation. ____NaOH + ____H 2 SO 4  ____H 2 O + ____Na 2 SO 4 2 1 2 1

27. 26 Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H 2 SO 4 solution. Now, let’s get organized. Place numerical Information and accompanying UNITS below each compound. ____NaOH + ____H 2 SO 4  ____H 2 O + ____Na 2 SO 4 2 1 2 1 0.102 M ? mL 35.0 mL Since 1 L = 1000 mL, we can use this to save on the number of conversions Our Goal

28. 27 Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H 2 SO 4 solution. Now let’s get to work converting. ____NaOH + ____H 2 SO 4  ____H 2 O + ____Na 2 SO 4 2 1 2 1 0.102 M ? mL 35.0 mL = mL NaOH H 2 SO 4 35.0 mL H 2 SO 4 0.125 mol 1000 mL H 2 SO 4 NaOH 2 mol 1 mol H 2 SO 4 1000 mL NaOH 0.102 mol NaOH 85.8 Units Match Solution Stoichiometry: shortcut

29. 28 What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH) 2 ? 1st write out a balanced chemical equation Solution Stoichiometry

30. 29 What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH) 2 ? 2HCl(aq) + Ba(OH) 2 (aq)  2H 2 O(l) + BaCl 2 0.40 M 47.1 mL 0.75 M ? mL = mL HCl Ba(OH) 2 47.1 mL 1 mol Ba(OH) 2 HCl 2 mol 0.40 mol HCl HCl 1000 mL 176 Units match Solution Stoichiometry

31. 30 Solution Stochiometry Problem: A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)? First write a balanced chemical reaction. ____HCl(aq) + ____Ba(OH) 2 (aq)  ____H 2 O(l) + ____BaCl 2 (aq) 2 1 2 1 23.28 mL 0.135 mol L 25.00 mL ? mol L

32. 31 Solution Stochiometry Problem: A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)? ____HCl(aq) + ____Ba(OH) 2 (aq)  ____H 2 O(l) + ____BaCl 2 (aq) 2 1 23.28 mL 0.135 mol L 25.00 mL ? mol L = mol Ba(OH) 2 L Ba(OH) 2 25.00 x 10 -3 L Ba(OH) 2 Units Already Match on Bottom! 0.0629 Units match on top!

33. 32 48.0 mL of Ca(OH) 2 solution was titrated with 19.2 mL of 0.385 M HNO 3. Determine the molarity of the Ca(OH) 2 solution. We must first write a balanced equation. Solution Stochiometry Problem:

34. 33 48.0 mL of Ca(OH) 2 solution was titrated with 19.2 mL of 0.385 M HNO 3. Determine the molarity of the Ca(OH) 2 solution. Ca(OH) 2 (aq) + HNO 3 (aq)  H 2 O(l) + Ca(NO 3 ) 2 (aq) 2 2 48.0 mL19.2 mL 0.385 M = mol (Ca(OH) 2) L (Ca(OH) 2 ) 19.2 mL HNO 3 48.0 x 10 -3 L ? M units match! 0.0770 Solution Stochiometry Problem:

35. 34 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) First copy down the the BALANCED equation! Now place numerical the information below the compounds.

36. 35 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Two starting amounts? Where do we start? Hide one

37. 36 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Hide Based on: KO 2 = mol O 2 0.15 mol KO 2 0.1125

38. 37 Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Based on: KO 2 = mol O 2 0.15 mol KO 2 0.1125 Hide Based on: H 2 O = mol O 2 0.10 mol H2OH2O 0.150 Limiting/Excess/ Reactant and Theoretical Yield Problems :

39. 38 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Based on: KO 2 = mol O 2 0.15 mol KO 2 0.1125 Based on: H 2 O = mol O 2 0.10 mol H 2 O 0.150 What is the theoretical yield? Hint: Which is the smallest amount? The is based upon the limiting reactant? It was limited by the amount of KO 2. H 2 O = excess (XS) reactant!

40. 39 Theoretical yield vs. Actual yield Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed, but only 12.3 grams of product were recovered. Determine the % yield. Theoretical yield = 19.5 g based on limiting reactant Actual yield = 12.3 g experimentally recovered

41. 40 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) If a reaction vessel contains 120.0 g of KO 2 and 47.0 g of H 2 O, how many grams of O 2 can be produced? 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 120.0 g47.0 g ? g Hide one Based on: KO 2 = g O 2 120.0 g KO 2 40.51 Limiting/Excess Reactant Problem with % Yield

42. 41 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) If a reaction vessel contains 120.0 g of KO 2 and 47.0 g of H 2 O, how many grams of O 2 can be produced? 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 120.0 g47.0 g ? g Based on: KO 2 = g O 2 120.0 g KO 2 40.51 Based on: H 2 O = g O 2 Question if only 35.2 g of O 2 were recovered, what was the percent yield? Hide 47.0 g H 2 O 125.3 Limiting/Excess Reactant Problem with % Yield

43. 42 If a reaction vessel contains 120.0 g of KO 2 and 47.0 g of H 2 O, how many grams of O 2 can be produced? 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 120.0 g47.0 g ? g Based on: KO 2 = g O 2 120.0 g KO 2 40.51 Based on: H 2 O = g O 2 47.0 g H 2 O 125.3 Determine how many grams of Water were left over. The Difference between the above amounts is directly RELATED to the XS H 2 O. 125.3 - 40.51 = 84.79 g of O 2 that could have been formed from the XS water. = g XS H 2 O 84.79 g O 2 31.83

44. 43 Calculate the molarity of a solution prepared by dissolving 25.6 grams of Al(NO 3 ) 3 in 455 mL of solution. After you have worked the problem, click here to see setup answer Try this problem (then check your answer):