2. 1 STOICHIOMETRY

3. 2 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

4. 3 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

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. 5 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

7. 6 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.

8. 7 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. Solutions

9. 8 1 st : = mol or M L 3.73 g 200.0 x 10 - 3 L Or 0.2 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 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.

10. 9 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

11. 10 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

12. 11 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

13. 12 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

14. 13 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

15. 14 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

16. 15 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

17. 16 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

18. 17 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

19. 18 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!

20. 19 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:

21. 20 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:

22. 21 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.

23. 22 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

24. 23 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

25. 24 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 :

26. 25 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!

27. 26 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

28. 27 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

29. 28 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

30. 29 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

31. 30 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):