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STOICHIOMETRY

STOICHIOMETRY. 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.

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STOICHIOMETRY

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  1. STOICHIOMETRY

  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.

  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 Units match 3. Convert using factors (ratios). 10 miles = inches 633600 Goal Given Convert Menu

  4. Stoichiometry (more working with ratios) Ratios are found within a chemical equation. 2HCl + Ba(OH)2 2H2O + BaCl2 1 1 coefficients give MOLAR RATIOS 2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O and 1 mole of BaCl2

  5. Information given by chemical equations 2 C6H6(l) +15O2(g)12CO2(g) +6H2O (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. • 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. Since the relationship between the actual number of molecules and the number of moles present is 6.02 x 1023, a common factor between all species involved in the equation, a MOLE RATIO relationship can be discussed.

  6. Information given by chemical equations 2 C6H6(l) +15O2(g)12CO2(g) +6H2O (g) The MOLE RATIO for benzene and oxygen is 2 : 15. It can be written as: 2 moles C6H6or as 15 moles of O2 15 moles O2 2 moles of C6H6 The MOLE RATIO for oxygen and carbon dioxide is 15 : 12. It can be written as: 12 moles CO2or as 15 moles of O2 15 moles O2 12 moles of CO2 • NOTE: The MOLERATIO 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.

  7. 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 H2(g)+ O2(g)2 H2O (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 O2 (2 moles H2 / 1 mole O2) = 4 mole H2

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

  9. STOICHIOMETRY 2 H2(g) + O2(g) 2 H2O (g) Q1. How many moles of hydrogen are necessary to react with 15.0 g of oxygen? A. 15.0g O2(1 mole O2)(2 mole H2)= 0.938 moles H2 32.0 g 1 mole O2 Q2. How many grams of hydrogen are necessary to react with 15.0 g of oxygen? A. 15.0g O2 (1 mole O2) ( 2 mole H2) ( 2.016 g H2) = 1.89 g H2 32.0 g 1 mole O21 mole H2

  10. STOICHIOMETRY 2 H2(g) + O2(g) 2 H2O (g) Q3. How many grams of water are produced from 15.0 g of oxygen? A.15.0g O2 (1 mole O2) ( 2 mole H2O) ( 18.0 g H2O) =16.9 g H2O 32.0 g 1 mole O21 mole H2O Q4. How much hydrogen and oxygen is needed to produce 25.0 grams of water? A. 25.0g H2O (1 mole H2O)(2 mole H2)(2.016 g H2)= 2.80 g H2 18.0 g 2 mole H2O1 mole H2 A. 25.0g H2O (1 mole H2O)(1 mole O2)(32.0 g O2)= 22.2 g O2 18.0 g 2 mole H2O1 mole O2 Notice that the Law of Conservation of Mass still applies.

  11. How many grams of solid are formed when 10.0 g of lead reacts with excess phosphoric acid? • Write the chemical equation: Pb + H3PO4 ? You recognize that this is a single displacement (replacement) reaction. So Pb (a metal) will displace (replace) H (the cation). Pb + H3PO4 Pb3(PO4)2 + H2 2. Balance the equation: 3Pb+2 H3PO4 Pb3(PO4)2 + 3 H2 3. Make a list under the appropriate substance 3Pb+2 H3PO4 Pb3(PO4)2 (s) + 3 H2 (g) 10.0gm=? Start with what is given: 10.0gPb (1 mole Pb)(1 mole Pb3(PO4)2)(811 g Pb3(PO4)2) = 13.1 g Pb3(PO4)2 207 g Pb 3 mole Pb1 mole Pb3(PO4)2

  12. PRACTICE PROBLEM # 18 14.20g 1. How many grams of gas can be produced from 0.8876 moles of HgO? 2 HgO  2 Hg + O2 2. How many moles of fluorine are required to produce 12.0 grams of KrF6? Given the equation: Kr + 3 F2 KrF6 3. How many grams of Na2CO3 will be produced from the thermal decomposition of 250.0 g of NaHCO3? 4. How many grams of CO2 can be produced by the reaction of 75.0 grams of C2H2 with excess oxygen? 5. Cu + 2 AgNO3 Cu(NO3)2 + 2 Ag. How many grams of silver is produced when 125.0 g of copper is reacted with excess silver nitrate solution? 0.182 mol 157.7 g 254 g 424.9 g

  13. GROUP STUDY PROBLEM # 18 ______1. How many grams of liquid product can be produced from 3.55 moles of HgO? 2 HgO 2 Hg + O2 ______2. How many moles of fluorine are required to produce 3.0 grams of KrF6? Given the equation: Kr + 3 F2 KrF6 ______3. How many grams of Na2CO3 will be produced from the decomposition of 20.0g of NaHCO3? ______4. How many grams of O2 are needed to combust 55.0 grams of C2H4? ______5. Cu + 2 AgNO3 Cu(NO3)2 + 2 Ag. How many grams of silver is produced when 50.0 g of copper is reacted with excess silver nitrate solution?

  14. Mole – Mole Conversions When N2O5 is heated, it decomposes: 2N2O5(g) 4NO2(g) + O2(g) a. How many moles of NO2 can be produced from 4.3 moles of N2O5? 2N2O5(g) 4NO2(g) + O2(g) 4.3 mol ? mol Units match 4.3 mol N2O5 = moles NO2 8.6 b. How many moles of O2 can be produced from 4.3 moles of N2O5? 2N2O5(g) 4NO2(g) + O2(g) 4.3 mol ? mol 4.3 mol N2O5 = mole O2 2.2

  15. gram ↔ mole and gram ↔ gram conversions When N2O5 is heated, it decomposes: 2N2O5(g) 4NO2(g) + O2(g) a. How many moles of N2O5 were used if 210g of NO2 were produced? 2N2O5(g) 4NO2(g) + O2(g) ? moles 210g Units match 210 g NO2 = moles N2O5 2.28 b. How many grams of N2O5 are needed to produce 75.0 grams of O2? 2N2O5(g) 4NO2(g) + O2(g) 75.0 g ? grams 75.0 g O2 = grams N2O5 506

  16. Gram to Gram Conversions 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)  AlCl3(aq) + H2(g) 2 6 2 3 First write a balanced equation.

  17. Gram to Gram Conversions 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)  AlCl3(aq) + H2(g) 2 6 2 3 ? grams 3.45 g Now let’s get organized. Write the information below the substances.

  18. gram to gram conversions 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)  AlCl3(aq) + H2(g) 2 6 2 3 ? grams 3.45 g Units match 3.45 g Al = g AlCl3 17.0 Let’s work the problem. We must always convert to moles. Now use the molar ratio. Now use the molar mass to convert to grams.

  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.

  20. Solutions A solution is prepared by dissolving 3.73 grams of AlCl3 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.

  21. Solutions A solution is prepared by dissolving 3.73 grams of AlCl3 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. 1st: 3.73 g = mol L 0.140 200.0 x 10-3 L molar mass of AlCl3 dilution formula M1V1 = M2V2 2nd: (0.140 M)(10.0 mL) = (? M)(100.0 mL) final concentration 0.0140 M = M2

  22. Solution Stoichiometry 50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used? H2SO4(aq) + 2NaHCO3 2H2O(l) + Na2SO4(aq) + 2CO2(g)

  23. Solution Stoichiometry 50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used? H2SO4(aq) + 2NaHCO3 2H2O(l) + Na2SO4(aq) + 2CO2(g) 50.0 mL ? g Our Goal 6.0 M = Look! A conversion factor!

  24. Solution Stoichiometry 50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used? H2SO4(aq) + 2NaHCO3 2H2O(l) + Na2SO4(aq) + 2CO2(g) 50.0 mL ? g Our Goal 6.0 M = NaHCO3 2 mol H2SO4 50.0 mL NaHCO3 84.0 g 50.4 = g NaHCO3 mol NaHCO3 1 mol H2SO4

  25. Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution are needed to neutralize 35.0 mL of 0.125 M H2SO4 solution. 2 1 2 1 ____NaOH + ____H2SO4 ____H2O + ____Na2SO4 First write a balanced Equation.

  26. Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution. 2 1 2 1 ____NaOH + ____H2SO4 ____H2O + ____Na2SO4 0.102 M 35.0 mL ? mL Our Goal Since 1 L = 1000 mL, we can use this to save on the number of conversions Now, let’s get organized. Place numerical Information and accompanying UNITS below each compound.

  27. Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution. 2 1 2 1 ____NaOH + ____H2SO4 ____H2O + ____Na2SO4 0.102 M 35.0 mL ? mL Units Match shortcut H2SO4 35.0 mL H2SO4 0.125 mol 1000 mL H2SO4 NaOH 2 mol 1 mol H2SO4 1000 mL NaOH 0.102 mol NaOH = mL NaOH 85.8 Now let’s get to work converting.

  28. Solution Stoichiometry 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

  29. Solution Stoichiometry 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)  2H2O(l) + BaCl2 47.1 mL 0.75 M 0.40 M ? mL Units match HCl 2 mol Ba(OH)2 47.1 mL HCl 1000 mL 176 = mL HCl 0.40 mol HCl 1 mol Ba(OH)2

  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)? 2 1 2 1 ____HCl(aq) + ____Ba(OH)2(aq)  ____H2O(l) + ____BaCl2(aq) 25.00 mL 23.28 mL ? mol L 0.135 mol L First write a balanced chemical reaction.

  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)? 2 1 2 1 ____HCl(aq) + ____Ba(OH)2(aq)  ____H2O(l) + ____BaCl2(aq) 25.00 mL 23.28 mL Units match on top! ? mol L 0.135 mol L = mol Ba(OH)2 L Ba(OH)2 0.0629 25.00 x 10-3 L Ba(OH)2 Units Already Match on Bottom!

  32. Solution Stochiometry Problem: 48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution. We must first write a balanced equation.

  33. Solution Stochiometry Problem: 48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution. Ca(OH)2(aq) + HNO3(aq)  2 H2O(l) 2 + Ca(NO3)2(aq) 48.0 mL 19.2 mL ? M 0.385 M HNO3 19.2 mL mol(Ca(OH)2) L (Ca(OH)2) = 0.0770 48.0 x 10-3L units match!

  34. Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? b. Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) First copy down the the BALANCED equation! Now place numerical the information below the compounds.

  35. Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? b. Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol ? moles 0.10 mol Hide one Two starting amounts? Where do we start?

  36. Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? b. Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol ? moles 0.10 mol Hide Based on: KO2 0.15 mol KO2 = mol O2 0.1125

  37. Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? b. Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol ? moles Hide 0.10 mol Based on: KO2 0.15 mol KO2 = mol O2 0.1125 0.10 mol H2O Based on: H2O = mol O2 0.150

  38. Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol ? moles 0.10 mol Based on: KO2 0.15 mol KO2 = mol O2 0.1125 It was limited by the amount of KO2. 0.10 mol H2O Based on: H2O = mol O2 0.150 H2O = excess (XS) reactant! What is the theoretical yield? Hint: Which is the smallest amount? The is based upon the limiting reactant?

  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

  40. Limiting/Excess Reactant Problem with % Yield 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced? 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) ? g 120.0 g 47.0 g Hide one 120.0 g KO2 Based on: KO2 = g O2 40.51

  41. Limiting/Excess Reactant Problem with % Yield 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced? 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) ? g 120.0 g 47.0 g Hide 120.0 g KO2 Based on: KO2 = g O2 40.51 47.0 g H2O Based on: H2O = g O2 125.3 Question if only 35.2 g of O2 were recovered, what was the percent yield?

  42. If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced? 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) ? g 120.0 g 47.0 g 120.0 g KO2 Based on: KO2 = g O2 40.51 47.0 g H2O Based on: H2O = g O2 125.3 Determine how many grams of Water were left over. The Difference between the above amounts is directly RELATED to the XS H2O. 125.3 - 40.51 = 84.79 g of O2 that could have been formed from the XS water. 84.79 g O2 31.83 = g XS H2O

  43. Try this problem (then check your answer): Calculate the molarity of a solution prepared by dissolving 25.6 grams of Al(NO3)3 in 455 mL of solution. After you have worked the problem, click here to see setup answer

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