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Stoichiometry: Calculating Mass in Chemical Reactions

Learn how to convert grams of one substance into grams of another in a chemical reaction. Understand the conservation of mass in balanced equations.

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Stoichiometry: Calculating Mass in Chemical Reactions

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  1. Prentice Hall Chemistry(c) 2005 By Daniel R. Barnes Init: 12/17/2008 Section Assessment Answers Chapter 12 WARNING: some images and content in this presentation may have been taken without permission from the world wide web. It is intended for use only by Mr. Barnes and his students. It is not meant to be copied or distributed. Stoichiometry

  2. SWBAT . . . . . . determine the mass of one substance involved in a chemical reaction if given the mass of any one of the other substances involved in the reaction. in other words . . . . . . convert grams of the known into grams of the unknown.

  3. 5. How is a balanced chemical equation similar to a recipe? A recipe says what ingredients you need to make some kind of food. A balanced chemical equation says what reactants you need to make some kind of product. As the teacher’s edition says, “Both a balanced equation and a recipe give quantitative information about the starting and end materials. 12.1 Section Assessment

  4. 6. How do chemists use balanced equations? Chemists use balanced equations as a basis to calculate how much reactant is needed or how much product is formed in a reaction. For instance . . . If you run a factory that makes chemicals, knowing how much of each reactant you’re going to use lets you know how much it’s going to cost you, and knowing how much product will be formed tells you how much money you can expect to make when you sell the product. 12.1 Section Assessment Gross sales–fabrication costs= PROFIT

  5. 7. Chemical reactions can be described in terms of what quantities? Chemical reactions can be described in terms of numbers atoms, molecules, or moles; in terms of mass; or in terms of volume. 12.1 Section Assessment

  6. 8. What quantities are always conserved in chemical reactions? Mass and atoms are always conserved in chemical reactions. For example . . . If you start with 18 kg of reactants, you should end up with . . . . . .18 kg of products. Chemical reactions do not create or destroy matter. If 3 million carbon atoms react with 6 million oxygen atoms to form carbon dioxide, the carbon dioxide will consist of . . . 3 million carbon atoms and 6 million oxygen atoms. 12.1 Section Assessment Volume is NOT conserved. (Example: a small handful of nitrogen triiodide can explode to form several liters of nitrogen gas and iodine vapors.) Molecules are NOT conserved. (Example: two million hydrogen molecules and one million oxygen molecules become two million water molecules. 3 million molecules  2 million molecules.)

  7. Interpret the given equation in terms of relative numbers of representative particles , numbers of moles, and masses of reactants and products. • 2K(s) + 2H2O(l)  2KOH(aq) + H2(g) No coefficient = an invisible coefficient of “1” In terms of “representative particles”: Two atoms of K react with 2 molecules of H2O to form 2 formula units of KOH and one molecule of H2. 12.1 Section Assessment In terms of mass: 78.2 g K + 36.0 g H2O  112.2 g KOH + 2.0 g H2 Two moles of K Two moles of H2O Two moles of KOH One mole of H2

  8. 12.1 Section Assessment 10. Balance this equation: C2H5OH(l) + O2(g)  CO2(g) + H2O(g) Show that the balanced equation obeys the law of conservation of mass. C2H5OH(l) + O2(g)  CO2(g) + H2O(g) 3 2 3 2 1 2 C = C = H = H = O = O = 6 6 2 7 5 7 3 3

  9. 10. Balance this equation: C2H5OH(l) + O2(g)  CO2(g) + H2O(g) Show that the balanced equation obeys the law of conservation of mass. 46 g First, I’d like to show how the unbalanced equation disobeys the law of conservation of mass To do this, I’ll need to calculate the molar mass of each reactant and each product . . . 12.1 Section Assessment C2H5OH: C: 2 x 12 = 24 H: 6 x 1 = 6 O: 1 x 16 = 16 ----------------------- 46 g/mol

  10. 10. Balance this equation: C2H5OH(l) + O2(g)  CO2(g) + H2O(g) Show that the balanced equation obeys the law of conservation of mass. 46 g 32 g First, I’d like to show how the unbalanced equation disobeys the law of conservation of mass To do this, I’ll need to calculate the molar mass of each reactant and each product . . . 12.1 Section Assessment O2: O: 2 x 16 = 32 ----------------------- 32 g/mol

  11. 10. Balance this equation: C2H5OH(l) + O2(g)  CO2(g) + H2O(g) Show that the balanced equation obeys the law of conservation of mass. 46 g 32 g 44 g First, I’d like to show how the unbalanced equation disobeys the law of conservation of mass To do this, I’ll need to calculate the molar mass of each reactant and each product . . . 12.1 Section Assessment CO2: C: 1 x 12 = 12 O: 2 x 16 = 32 ----------------------- 44 g/mol

  12. 10. Balance this equation: C2H5OH(l) + O2(g)  CO2(g) + H2O(g) Show that the balanced equation obeys the law of conservation of mass. 46 g 32 g 44 g 18 g First, I’d like to show how the unbalanced equation disobeys the law of conservation of mass To do this, I’ll need to calculate the molar mass of each reactant and each product . . . 12.1 Section Assessment H2O: H: 2 x 1 = 2 O: 1 x 16 = 16 ----------------------- 18 g/mol

  13. 10. Balance this equation: C2H5OH(l) + O2(g)  CO2(g) + H2O(g) Show that the balanced equation obeys the law of conservation of mass. 46 g 32 g 44 g 18 g First, I’d like to show how the unbalanced equation disobeys the law of conservation of mass To do this, I’ll need to calculate the molar mass of each reactant and each product . . . 12.1 Section Assessment 44 + 18 = 62 grams of products 46 + 32 = 78 grams of reactants You can’t have 16 grams of matter just disappear like that. In its unbalanced state, the equation violates the law of conservation of matter. It needs to be balanced!

  14. 12.1 Section Assessment 10. Balance this equation: C2H5OH(l) + O2(g)  CO2(g) + H2O(g) Show that the balanced equation obeys the law of conservation of mass. C2H5OH(l) + O2(g)  CO2(g) + H2O(g) 3 2 3 32 g 44 g 18 g 46 g x 1 x 3 x 2 x 3 88 g 54 g 46 g 96 g 46 g + 96 g = 142 g 88 g + 54 g = 142 g REACTANTS PRODUCTS = (in terms of mass, anyway)

  15. REALITY CHECK Don’t forget that one “mole” of a material = 6.022 x 1023 molecules* of that material. *If the material is a noble gas or a metal, subsitute the word “atoms” for “molecules”. If the material is ionic, substitute “forumula units” for “molecules”.

  16. 12.2 Practice Problems 13. Acetylene gas (C2H2) is produced by adding water to calcium carbide (CaC2). CaC2(s) + 2H2O(l)  C2H2(g) + Ca(OH)2(aq) How many grams of acetylene are produced by adding water to 5.00 g CaC2? g 2.03 5 g 5 g CaC2 1 mol CaC2 1 mol C2H2 g C2H2 26 x x x 1 64 g CaC2 1 mol CaC2 1 mol C2H2 CaC2: Ca: 1 x 40 = 40 C: 2 x 12 = 24 C2H2: C: 2 x 12 = 24 H: 2 x 1 = 2 64 g/mol 26 g/mol 26 x 5 2.03 ) 130 64 130

  17. 12.2 Practice Problems 14. Using the same equation, determine how many moles of CaC2 are needed to react completely with 49.0 g H2O. CaC2(s) + 2H2O(l)  C2H2(g) + Ca(OH)2(aq) 49 g H2O 1 mol H2O 1 mol CaC2 x x 1 18 g H2O 2 mol H2O H2O: H: 2 x 1 = 2 O: 1 x 16 = 16 ---------------------- 18 g/mol 49 49 mol CaC2 mol CaC2 = = 1.36 mol CaC2 36 18 x 2

  18. 12.2 Practice Problems • How many molecules of oxygen are produced by the decomposition of 6.54 g of potassium chlorate (KClO3)? • 2KClO3  2KCl + 3O2 6 x 1023 molecules O2 6.54 g KClO3 1 mol KClO3 3 mol O2 x x x 2 mol KClO3 122 g KClO3 1 mol O2 1 6.54 x 3 x 6 x 1023 117.72 x 1023 = = 0.482459 x 1023 122 x 2 244 = 4.82459 x 1022 KClO3: K: 1 x 39 = 39 Cl: 1 x 35 = 35 O: 3 x 16 = 48 ---------------------- 122 g/mol = 4.82 x 1022 molecules O2

  19. The last step in the production of nitric acid is the reaction of nitrogen dioxide with water: • 3NO2 + H2O  2HNO3 + NO • How many grams of nitrogen dioxide must react with water to produce 5.00 x 1022 molecules of nitrogen monoxide? 5 x 1022 molecules NO 1 mol NO 3 mol NO2 46 g NO2 1 mol NO2 1 1 mol NO 6 x 1023 molecules NO 5 x 3 x 46 x 1022 690 x 1022 = 6 x 1023 6 x 1023 12.2 Practice Problems NO: N: 1 x 14 = 14 O: 2 x 16 = 32 ---------------------- 46 g/mol = 115 x 10-1 g NO2 = 1.15 x 101 g NO2 = 11.5 g NO2

  20. 12.2 Practice Problems

  21. 12.2 Practice Problems Notice how the 22.4’s just ended up canceling each other out? It almost makes you wonder why I bothered putting them there in the first place . . .

  22. Look back at my work-out for #18 and notice again how the 22.4’s just ended up canceling each other out. If I’d put those two fractions in my work-out for this one, the same thing would have happened, so I just didn’t bother putting them in. I could have also put in two fractions to turn mL to L and then L back to mL again, but the 1000’s would have also have just canceled each other out, so I didn’t bother. 12.2 Practice Problems

  23. 12.2 Practice Problems

  24. 12.2 Section Assessment 21. How are mole ratios used in chemical calculations? Mole ratios are written using the coefficients from a balanced chemical equation. They are used to relate moles of reactants and products in stoichiometric calculations.

  25. 12.2 Section Assessment 22. Outline the sequence of steps needed to solve a typical stoichiometric problem. i. Convert the given quantity to moles using the molar mass of the known chemical. ii. Use the mole ratio (from the appropriate coefficients in the balanced chemical equation for the reaction) to find moles of the desired chemical. iii. Using the molar mass of the desired chemical, convert moles of the desired chemical into grams of the desired chemical.

  26. 12.2 Section Assessment • Write the 12 mole ratios that can be derived from the equation for the combustion of isopropyl alcohol. • 2C3H7OH(l) + 9O2(g)  6CO2(g) + 8H2O(g) Holy crud. You ready? Here they come. 2 mol C3H7OH 2 mol C3H7OH 2 mol C3H7OH 9 mol O2 6 mol CO2 8 mol H2O 9 mol O2 9 mol O2 9 mol O2 2 mol C3H7OH 6 mol CO2 8 mol H2O 6 mol CO2 6 mol CO2 6 mol CO2 2 mol C3H7OH 9 mol O2 8 mol H2O 8 mol H2O 8 mol H2O 8 mol H2O 2 mol C3H7OH 9 mol O2 6 mol CO2

  27. 12.2 Section Assessment 24. The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(g) How many grams of CO2 and H2O are produced when 52.0 g C2H2 burns in oxygen? 176 g 52.0 g 1 mol C2H2 52.0 g C2H2 4 mol CO2 44 g CO2 x x x 1 26 g C2H2 2 mol C2H2 1 mol CO2 C2H2: C: 2 x 12 = 24 H: 2 x 1 = 2 ---------------------- 26 g/mol CO2: C: 1 x 12 = 12 O: 2 x 16 = 32 ---------------------- 44 g/mol = 176 g CO2

  28. 12.2 Section Assessment 24. The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(g) How many grams of CO2 and H2O are produced when 52.0 g C2H2 burns in oxygen? 176 g 36 g 1 mol C2H2 52.0 g C2H2 2 mol H2O 18 g H2O x x x 1 26 g C2H2 2 mol C2H2 1 mol H2O C2H2: C: 2 x 12 = 24 H: 2 x 1 = 2 ---------------------- 26 g/mol H2O: H: 2 x 1 = 2 O: 1 x 16 = 16 ---------------------- 18 g/mol = 36 g H2O

  29. 12.3 Practice Problems The math shows that 2.7 moles of C2H4 can create 5.4 moles of CO2, but 6.3 moles of O2 can only create 4.2 moles CO2. O2 is the weakest link, so it’s the limiting reactant. In this reaction, all the O2 would be used up, with leftover C2H4.

  30. 12.3 Practice Problems

  31. 12.3 Practice Problems 2.6 mol C2H4 can produce less H2O than 6.3 mol O2 can, so C2H4 is the limiting reactant. (I do this a little different from the book. I see how much product each amount of reactant can make. Whoever makes the least product is the limiting reactant.)

  32. 12.3 Practice Problems As with #27, I calculated how much product each amount of reactant could make. the 2.4 mol C2H2 produced the lesser amount of H2O, so C2H2 is the limiting reactant.

  33. 12.3 Practice Problems This is just an ordinary, g-this  g-that, stoichiometry problem. The only difference is that now we’re referring to the results of such calculations as “theoretical yield”.

  34. 12.3 Practice Problems #30 is extra-evil because it doesn’t tell you what the formulas are of the chemicals in the reaction, and, since you have to make the equation from scratch, you have to balance it, too. You may have to go back to chapter 9 if you forget how to figure out the formula of a chemical when given only its name.

  35. 12.3 Practice Problems . . .

  36. 12.3 Practice Problems #32 = evil (no formulas or equation given AND 2 stoich calcs)

  37. 12.3 Section Assessment • 33. “In a chemical equation, an insufficient quantity of any of the reactants will limit the amount of product that forms.” • 34. “The efficiency of a reaction carried out in a laboratory can be measured by calculating the percent yield.”

  38. 12.3 Section Assessment 35. What is the percent yield if 4.65 g of copper is produced when 1.87 g of aluminum reacts with an excess of copper (II) sulfate? 2Al + 3CuSO4 Al2(SO4)3 + 3Cu 1.87 g Al 1 mol Al 3 mol Cu 63.55 g Cu = 6.6 g Cu 27 g Al 2 mol Al 1 mol Cu actual yield 4.65 g Cu % yield = = = 0.705 = 70.5% theoretical yield 6.6 g Cu

  39. Chapter 12 Assessment • The reaction of fluorine with ammonia produces dinitrogen tetrafluoride and hydrogen fluoride. • 5F2(g) + 2NH3(g)  N2F4(g) + 6HF(g) • a. If you have 66.6 g of NH3, how many grams of F2 are required for complete reaction?

  40. Chapter 12 Assessment • The reaction of fluorine with ammonia produces dinitrogen tetrafluoride and hydrogen fluoride. • 5F2(g) + 2NH3(g)  N2F4(g) + 6HF(g) • a. If you have 66.6 g of NH3, how many grams of F2 are required for complete reaction? 66.6 g g 372 5F2 + 2NH3 N2F4 + 6HF 66.6 g NH3 1 mol NH3 5 mol F2 38 g F2 x x x = 372 g F2 1 17 g NH3 2 mol NH3 1 mol F2 F2: F: 2 x 19 = 38 38 g/mol NH3: N: 1 x 14 = 14 H: 3 x 1 = 3 17 g/mol

  41. Chapter 12 Assessment • The reaction of fluorine with ammonia produces dinitrogen tetrafluoride and hydrogen fluoride. • 5F2(g) + 2NH3(g)  N2F4(g) + 6HF(g) • b. How many grams of NH3 are required to produce 4.65 g HF? g 4.65 g 5F2 + 2NH3 N2F4 + 6HF

  42. Chapter 12 Assessment • The reaction of fluorine with ammonia produces dinitrogen tetrafluoride and hydrogen fluoride. • 5F2(g) + 2NH3(g)  N2F4(g) + 6HF(g) • b. How many grams of NH3 are required to produce 4.65 g HF? g 1.32 4.65 g 5F2 + 2NH3 N2F4 + 6HF 4.65 g HF 1 mol HF 2 mol NH3 17 g NH3 x x x = 1.32 g NH3 1 20 g HF 6 mol HF 1 mol NH3 HF: H: 1 x 1 = 1 F: 1 x 19 = 19 20 g/mol NH3: N: 1 x 14 = 14 H: 3 x 1 = 3 17 g/mol

  43. Chapter 12 Assessment • The reaction of fluorine with ammonia produces dinitrogen tetrafluoride and hydrogen fluoride. • 5F2(g) + 2NH3(g)  N2F4(g) + 6HF(g) • c. How many grams of N2F4 can be produced from 225 g F2? g ? 225 g 5F2 + 2NH3 N2F4 + 6HF

  44. Chapter 12 Assessment • The reaction of fluorine with ammonia produces dinitrogen tetrafluoride and hydrogen fluoride. • 5F2(g) + 2NH3(g)  N2F4(g) + 6HF(g) • c. How many grams of N2F4 can be produced from 225 g F2? g 123 225 g 5F2 + 2NH3 N2F4 + 6HF 104 g N2F4 225 g F2 1 mol F2 1 mol N2F4 x x x = 123 g N2F4 1 38 g F2 5 mol HF2 1 mol N2F4 F2: F: 2 x 19 = 38 38 g/mol N2F4: N: 2 x 14 = 28 F: 4 x 19 = 76 104 g/mol

  45. Chapter 12 Assessment 44. Lithium nitride reacts with water to form ammonia and aqueous lithium hydroxide. Li3N + 3H2O  NH3 + 3LiOH a. What mass of water is needed to react with 32.9 g Li3N? g ? 32.9 g Li3N + 3H2O  NH3 + 3LiOH

  46. Chapter 12 Assessment 44. Lithium nitride reacts with water to form ammonia and aqueous lithium hydroxide. Li3N + 3H2O  NH3 + 3LiOH a. What mass of water is needed to react with 32.9 g Li3N? g 51 32.9 g Li3N + 3H2O  NH3 + 3LiOH 18 g H2O 32.9 g Li3N 1 mol Li3N 3 mol H2O x x x = 51 g H2O 1 35 g Li3N 1 mol Li3N 1 mol H2O Li3N: Li: 3 x 7 = 21 N: 1 x 14 = 14 35 g/mol H2O: H: 2 x 1 = 2 O: 1 x 16 = 16 18 g/mol

  47. This Power Point may soon again be . . . UNDER CONSTRUCTION I’m working on the honors stuff, mostly . . . 1,000,000,000,000,000,000,000,000,000,000,000,000,000,000 10Q + “E”

  48. hyperTABLE of CONTENTS Title Page 12.3 Section Assessment SWBATS Ch 12 Assessment 12.1 Section Assessment 12.2 Practice Problems 12.2 Section Assessment 12.3 Practice Problems

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