1 / 95

Chapter 6: Quantities in Chemical Reactions

Chapter 6: Quantities in Chemical Reactions. Questions for Consideration. What do the coefficients in balanced equations represent? How can we use a balanced equation to relate the number of moles of reactants and products in a chemical reaction?

gillians
Télécharger la présentation

Chapter 6: Quantities in Chemical Reactions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 6: Quantities in Chemical Reactions

  2. Questions for Consideration • What do the coefficients in balanced equations represent? • How can we use a balanced equation to relate the number of moles of reactants and products in a chemical reaction? • How can we use a balanced equation to relate the mass of reactants and products in a chemical reaction? • How do we determine which reactant limits the amount of product that can form? • How can we compare the amount of product we actually obtain to the amount we expect to obtain? • How can we describe and measure energy changes? • How are heat changes involved in chemical reactions?

  3. Chapter 6 Topics: • The Meaning of a Balanced Equation • Mole-Mole Conversions • Mass-Mass Conversions • Limiting Reactants • Percent Yield • Energy Changes • Heat Changes in Chemical Reactions

  4. Introduction • How can we predict amounts of reactants and products in a reaction, such as that in an internal combustion engine? • How can we predict the amount of heat generated or absorbed during a reaction? Figure 6.2

  5. Internal Combustion Engine • In an ICE, octane burns with oxygen to produce hot gases that push against a piston to do work. • The amount of oxygen that reacts is dependent upon the amount of octane that burns. • The amount of energy produced also depends on the amount of octane that reacts. Figure 6.3 (A)

  6. Hydrogen Fuel Cell • In a hydrogen fuel cell, hydrogen reacts with oxygen to water and electrical energy. • The amount of oxygen that reacts is dependent upon the amount of hydrogen available. • The amount of energy produced also depends on the amount of hydrogen that reacts. Figure 6.3 (B)

  7. 6.1 The Meaning of a Balanced Equation • What does a balanced equation tell us about the relative amounts of reactants and products? Let’s consider the combustion of propane: Figure 6.4

  8. What do the coefficients in a balanced chemical equation mean? • Balanced, the equation is: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) • The coefficients in the balanced equation tell us about the relative numbers of reactants that combine and products that form. There is an implied coefficient of 1 in front of C3H8(g).

  9. What do the coefficients in a balanced chemical equation mean? C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) • For every 1 molecule of propane that reacts with 5 molecules of oxygen gas, 3 molecules of carbon dioxide and 4 molecules of water are produced. Figure 6.4

  10. What do the coefficients in a balanced chemical equation mean? C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) • The coefficients in the balanced equation also tell us about the relative numbers of moles of reactants and products. • For every 1 mole of C3H8(g) that reacts with 5 moles of O2(g), 3 moles of CO2(g) and 4 moles of H2O(g) are formed.

  11. Summary – The Meaning of the Coefficients

  12. Amounts of Reactants and Products • The process of determining the amount of a reactant or product from another reactant or product in a reaction is called stoichiometry. • We use the mole relationships provided in a balanced equation to calculate amounts of reactants and products in a reactions.

  13. 6.2 Mole-Mole Conversions C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) • A mole ratio is used to relate the number of moles of one reactant or product to another. • Mole ratios are obtained from the coefficients in the balanced equation. • For example, the mole ratio of O2 to C3H8 is 5:1 or:

  14. Mole Ratios C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) • The mole ratio of O2 to C3H8 allows us to calculate the amount of O2 that will react with any amount of C3H8 that reacts. The mole ratio is used as a conversion factor in a dimensional analysis equation. • If 0.40 mol C3H8 reacts:

  15. Activity: Mole-Mole Conversions • Benzene (C6H6) burns in air according to the following equation: 2C6H6(g) + 15O2(g)  12CO2(g) + 6H2O(g) • What is the mole ratio of O2 to C6H6? • How many moles of O2 are required to react with each mole C6H6? • How many moles of O2 are required to react with 0.38 mole of C6H6?

  16. Activity Solutions: Mole-Mole Conversions 2C6H6(g) + 15O2(g)  12CO2(g) + 6H2O(g) • What is the mole ratio of O2 to C6H6? Based on the coefficients in the balanced chemical equation, the mole ratio can be written two ways: 15 moles O2 OR 2 moles C6H6 2 moles C6H6 15 moles O2 • How many moles of O2 are required to react with each mole C6H6? 15/2 or 7.5 moles of O2 are required to react with each mole of C6H6 • How many moles of O2 are required to react with 0.38 mole of C6H6?

  17. Activity: Mole Ratios C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) • What is the mole ratio for determining the moles of CO2 that will be produced when 2.3 mol O2 reacts? • How many moles of CO2 will be produced?

  18. Activity: Mole Ratios 2Al(s) + 3Cl2(g)  2AlCl3(g) • How many moles of Cl2 are required to prepare 0.62 mol AlCl3?

  19. 6.3 Mass-Mass Conversions • 2Na(s) + Cl2(g)  2NaCl(s) • What mass of chlorine gas is required to react with 9.20 grams of sodium? • We don’t measure reactants and products in moles, but we commonly measure their mass. • The balanced equation does not tell us a mass relationship. • How do we convert grams of reactant or product to moles?

  20. Mass-Mass Conversions 2Na(s) + Cl2(g)  2NaCl(s) (9.20 g Na) 1. We convert grams to moles using molar mass: For a review, see Section 4.2 or Math Toolbox 4.1. Figure from p. 220

  21. Mass-Mass Conversions 2Na(s) + Cl2(g)  2NaCl(s) (9.20 g Na) (0.400 mol) 2. Next we relate moles of Na to moles of Cl2 using the mole ratio: Figure from p. 220

  22. Mass-Mass Conversions 2Na(s) + Cl2(g)  2NaCl(s) (9.20 g Na) (0.400 mol Na) (0.200 mol Cl2) 3. The last step is to convert moles of Cl2 to grams using the molar mass of Cl2: Figure from p. 220

  23. Mass-Mass Conversions 2Na(s) + Cl2(g)  2NaCl(s) (9.20 g Na) (14.2 g Cl2) (0.400 mol Na) (0.200 mol Cl2) 3. The last step is to convert moles of Cl2 to grams using the molar mass of Cl2:

  24. Steps for Mass-Mass Conversions From page 221

  25. Mass-Mass Conversions 2Na(s) + Cl2(g)  2NaCl(s) • What mass of NaCl should be produced when 9.20 g Na reacts with 14.2 g Cl2?

  26. Mass-Mass Conversions 2Na(s) + Cl2(g)  2NaCl(s) • What mass of NaCl should be produced when 9.20 g Na reacts with 14.2 g Cl2? Figure 6.6

  27. Activity: Mass-Mass Conversions • Given: 2 H2(g) + O2(g)  2 H2O(g) If 1.8 × 108 g of hydrogen was used in the liftoff during a shuttle launch, then: • What mass of oxygen gas was consumed? • What mass of water vapor was produced? • Does the mass of the water vapor equal the masses of the reactants? If so, then what law describes this observation?

  28. Activity Solutions: Mass-Mass Conversions Mole Ratio Mass H2 (Given in Problem) MM(O2) MM(H2) Moles H2 Moles O2 Mass O2 2 H2(g) + O2(g)  2 H2O(g) • What mass of oxygen gas was consumed? Start with the number that was given in the problem – 1.8×108 g of H2(g). Convert to moles via the MM(H2), then use a mole ratio to relate H2 to O2. Finally, convert from moles of O2 to grams of O2 using the MM(O2).

  29. Activity Solutions: Mass-Mass Conversions Mole Ratio Mass H2 (Given in Problem) MM(H2O) MM(H2) Moles H2 Moles H2O Mass H2O 2 H2(g) + O2(g) → 2 H2O(g) • What mass of water vapor was produced? Start again with the number that was given in the problem: 1.8×108 of H2(g). Convert to moles using the MM(H2), then use a mole ratio to relate H2 to H2O. Finally, convert from moles of H2O to grams of H2O using the MM(H2O).

  30. Activity Solutions: Mass-Mass Conversions • Given: 2 H2(g) + O2(g)  2 H2O(g) 1.8 × 108 g H2 + 1.4 × 109 g O2 = 1.6 × 109 g H2O 3. Does the mass of the water vapor equal the masses of the reactants? Yes or close to it – hydrogen and oxygen added together equals 1.58 × 109 g H2O. If so, then what law describes this observation? The Law of Conservation of Mass

  31. The Law of Conservation of Mass Figure 6.6 • The law of conservation of mass states that the masses of the reactants that are consumed must equal the masses of the products that are formed. Mass consumed reactants = Mass formed products

  32. Activity: Mass-Mass Conversions • When aluminum metal is exposed to oxygen gas, a coating of aluminum oxide forms on the surface of the aluminum. The balanced equation for the reaction of aluminum metal with oxygen gas is: 4Al(s) + 3O2(g)  2Al2O3(s) Suppose a sheet of pure aluminum gains 0.0900 g of mass when exposed to air. Assume that this gain can be attributed to its reaction with oxygen. • What mass of O2 reacted with the Al? • What mass of Al is reacted? • What mass of Al2O3 is formed?

  33. Activity Solutions: Mass-Mass Conversions 4Al(s) + 3O2(g)  2Al2O3(s) Suppose a sheet of pure aluminum gains 0.0900 g of mass when exposed to air. Assume that this gain can be attributed to its reaction with oxygen. • What mass of O2 reacted with the Al? 0.0900 g O2 • What mass of Al is reacted? • What mass of Al2O3 is formed?

  34. 6.4 Limiting Reactants • When two reactants are mixed, one usually does not react completely because there is too much of it. • In this reaction solid magnesium metal reacts with aqueous hydrochloric acid (A). One reactant is in excess in B and the other reactant is in excess in C. • Can you identify the excess reactant in each case? Figure 6.7

  35. Limiting Reactant • The limiting reactant is the reactant that reacts completely and is therefore not present when the reaction is complete. • Since the limiting reactant reacts completely, its amount determines the amount of product that can form.

  36. What is the limiting reactant?Cu(s) + AgNO3(aq)  Figure from Example 6.3

  37. Consider an Analogy:The construction of a model solar car Figure 6.8 (A)

  38. If you have 4 frames, 5 solar cells, 6 motors, and 12 wheels, how many solar cars can you make? Figure 6.8 (A and B)

  39. What is the limiting part? Figure 6.8 (B) • The wheel is the limiting part because it will be completely used up. • The number of cars made depends on the number of wheels available. If there were 4 more wheels, then another car could have been made with the remaining parts.

  40. How many solar cars? Figure 6.8 B( and C)

  41. Limiting Reactants at a Molecular Level • In a chemical reaction, the balanced equation tells us the relative number of molecules (or moles) that combine in the reaction. Figure 6.9

  42. Limiting Reactants at a Molecular Level • If reactants are not present in this ratio, then there will be a limiting reactant and excess of the other reactant. Figure 6.10

  43. Activity: Limiting Reactant2H2(g) + O2(g)  2H2O(g) • Complete the after picture. How many H2O molecules form? • What is the limiting reactant? What is in excess? Figure from p. 251

  44. Activity Solution: Limiting Reactant 2H2(g) + O2(g)  2H2O(g) What is the limiting reactant? Using molecule ratios from the balanced chemical equation: The limiting reactant is O2 because there is only enough of it to create 6 molecules of H2O. Which reactant is left over? H2 is left over at the end of the reaction.

  45. 2H2(g) + O2(g)  2H2O(g) • You can also use “molecule” ratios to determine the limiting reactant. The 8 H2 molecules need 4 O2 molecules to react with them. There are only 3 O2 molecules, so all the H2 cannot react. • H2 is in excess, and O2 is the limiting reactant. Figure from p. 251

  46. Activity Solution: Limiting Reactant • The molecular-level diagram shows a mixture of reactant molecules (three O2 molecules and 8 H2 molecules) for the following reaction: 2H2(g) + O2(g)  2H2O(g) The after picture should have 6 H2O molecules and 2 H2 molecules: Figure from p. 251

  47. Steps for Determining the Limiting Reactant • Calculate the amount of one reactant (B) needed to react with the other reactant (A). • Compare the calculated amount of B (amount needed) to the actual amount of B that is given. • If calculated B = actual B, there is no limiting reactant. Both A and B will react completely. • If calculated B > actual B, B is the limiting reactant. Only B will react completely. • If calculated B < actual B, A is the limiting reactant. Only A will react completely.

  48. Activity: Limiting Reactants (Mole Scale) • The balanced equation for the reaction of phosphorus and oxygen gas to form diphosphorus pentoxide is: P4(s) + 5O2(g)  2P2O5(s) What is the limiting reactant when each of the following sets of quantities of reactants is mixed? • 0.50 mol P4 and 5.0 mol O2 • 0.20 mol P4 and 1.0 mol O2 • 0.25 mol P4 and 0.75 mol O2

  49. Activity Solutions: Limiting Reactants (Mole Scale) P4(s) + 5O2(g) → 2P2O5(s) What is the limiting reactant when each of the following sets of quantities of reactants is mixed? • 0.50 mol P4 and 5.0 mol O2 Each mole of P4 requires 5 moles of O2, so 0.50 mol P4 requires: The amount of O2 present is more than the amount required, so P4 is the limiting reactant and O2 is present in excess.

  50. Activity Solutions: Limiting Reactants (Mole Scale) P4(s) + 5O2(g)  2P2O5(s) What is the limiting reactant when each of the following sets of quantities of reactants is mixed? • 0.20 mol P4 and 1.0 mol O2 Each mole of P4 requires 5 moles of O2, so 0.20 mol P4 requires: Since this is the amount of O2 present, there is no limiting reactant. Both reactants are consumed completely.

More Related