2. 1 STOICHIOMETRY TUTORIAL Paul Gilletti

3. 2 Instructions: This is a work along tutorial. Each time you click the mouse or touch the space bar on your computer, one step of the problem solving occurs. Pressing the PAGE UP key will backup the steps. Get a pencil and paper, a periodic table and a calculator, and let’s get to work.

4. 3 (1-2-3) General Approach For Problem Solving: 1. Clearly identify the Goal or Goals and the UNITS involved. (What are you trying to do?) 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.

5. 4 Table of Contents: Click on each tab to view problem types. Sample problem 1 Sample problem 2 Converting grams to moles Mole to Mole Conversions View Complete Slide Show Gram-Mole and Gram-Gram Problems Solution Stoichiometry Problems Limiting/Excess/ Reactant and Theoretical Yield Problems :

6. 5 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

7. 6 Sample problem #2 on problem solving. A car is traveling at a speed of 45 miles per hr (45 miles/hr). Determine its speed in kilometers per second using the following conversion factors (ratios). 1 mile = 5280 ft; 1 ft = 12 in; 1 inch = 2.54 cm; k = 1 x 10 3 ; c = 1 x 10 -2 ; 1 hr =60 min; 1 min = 60 s = km s 0.020 Units Match! Goal c cancels c m remains Given This is the same as putting k over k

8. 7 Converting grams to moles. Determine how many moles there are in 5.17 grams of Fe(C 5 H 5 ) 2. Goal = moles Fe(C 5 H 5 ) 2 Given 5.17 g Fe(C 5 H 5 ) 2 Use the molar mass to convert grams to moles. Fe(C 5 H 5 ) 2 2 x 5 x 1.001 = 10.01 2 x 5 x 12.011 = 120.11 1 x 55.85 = 55.85 0.0278 units match

9. 8 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

10. 9 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

11. 10 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

12. 11 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

13. 12 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

14. 13 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

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16. 15 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.

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18. 17 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

19. 18 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

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21. 20 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

22. 21 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

23. 22 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

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25. 24 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

26. 25 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

27. 26 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

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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

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33. 32 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

34. 33 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!

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36. 35 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:

37. 36 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:

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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? 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.

40. 39 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

41. 40 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

42. 41 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 :

43. 42 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!

44. 43 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

45. 44 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

46. 45 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

47. 46 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

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49. 48 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):

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