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Chapter #8

Chapter #8 . Chemical Quantities. Chapter Outline. 8.2 Mole relationships defined in balanced equations 8.3 Mole conversions- Stoichiometry 8.4 Mass Conversions Stoichiometry 8.5 Excess and Limiting Reactants 8.5 Theoretical Yield 8.6 Theoretical Yield From Initial Reactant Masses

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Chapter #8

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  1. Chapter #8 Chemical Quantities

  2. Chapter Outline • 8.2 Mole relationships defined in balanced equations • 8.3 Mole conversions- Stoichiometry • 8.4 Mass Conversions Stoichiometry • 8.5 Excess and Limiting Reactants • 8.5 Theoretical Yield • 8.6 Theoretical Yield From Initial Reactant Masses • 8.7 Thermochemical Equations

  3. 8.3-8.4 STOICHIOMETRY Stoichiometryis the use of balanced chemical equations in the conversion process.

  4. 8.4 STOICHIOMETRY Stoichiometryis the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2H2O

  5. 8.4 STOICHIOMETRY Stoichiometryis the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2H2O 6.33 g H2

  6. 8.4 STOICHIOMETRY Stoichiometryis the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2H2O 6.33 g H2 mole H2 2.016 g H2

  7. 8.4 STOICHIOMETRY Stoichiometryis the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2H2O 6.33 g H2 Mole H2 2.016 g H2

  8. 8.4 STOICHIOMETRY Stoichiometryis the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2H2O 6.33 g H2 Mole H2 2 mole H2O 2.016 g H2O 2 Mole H2

  9. 8.4 STOICHIOMETRY Stoichiometryis the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2 H2O 6.33 g H2 Mole H2 2 mole H2O 18.02 g H2O 2.016 g H2 2 Mole H2 Mole H2O

  10. 8.4 STOICHIOMETRY Stoichiometryis the use of balanced chemical equations in the conversion process. Examples Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required. 2 H2 + O2 2 H2O 6.33 g H2 Mole H2 2 mole H2O 18.02 g H2O = 28.3 g H2O 2.016 g H2 2 Mole H2 Mole H2O

  11. 8.5 Excess and Limiting Reactants Reactants are substances that can be changed into something else. For example, nails and boards are reactants for carpenters, while thread and fabric are reactants for the seamstress. And for a chemist hydrogen and oxygen are reactants for making water.

  12. 8.5 Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?

  13. 8.5 Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? Yes, only one house!

  14. 8.5 Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we had enough boards?

  15. 8.5 Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we have enough boards?

  16. 8.5 Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we have enough boards? Yes, nails are in excess!

  17. 8.5 Building Houses Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we have enough boards? Yes, nails are in excess! Nine more houses if we have an adequate amount of boards.

  18. 8.5 Excess and Limiting Example If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?

  19. 8.5 Excess and Limiting Example If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.

  20. 8.5 Excess and Limiting Example If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H2 + O2 2 H2O 10.0 g O2

  21. 8.5 Excess and Limiting Example If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H2 + O2 2 H2O mole O2 10.0 g O2 32.0 g O2

  22. 8.5 Excess and Limiting Example If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H2 + O2 2 H2O mole O2 2 mole H2 10.0 g O2 mole O2 32.0 g O2

  23. 8.5 Excess and Limiting Example If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H2 + O2 2 H2O mole O2 2 mole H2 2.02 g H2 10.0 g O2 mole H2 mole O2 32.0 g O2

  24. 8.5 Excess and Limiting Example If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen. 2 H2 + O2 2 H2O mole O2 2 mole H2 2.02 g H2 10.0 g O2 = 1.26 g H2 mole H2 mole O2 32.0 g O2

  25. 8.5 Excess and Limiting Example Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen 10.0 -1.26 = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.

  26. 8.5 Excess and Limiting Example Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen 10.0 -1.26 = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water. mole O2 10.0 g O2 32.0 g O2

  27. 8.5 Excess and Limiting Example Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen 10.0 -1.26 = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water. mole O2 2 mole H2O 10.0 g O2 mole O2 32.0 g O2

  28. 8.5 Excess and Limiting Example Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen 10.0 -1.26 = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water. 18.0 g H2O mole O2 2 mole H2O 10.0 g O2 mole H2O mole O2 32.0 g O2

  29. 8.5 Excess and Limiting Example Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen 10.0 -1.26 = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water. 18.0 g H2O mole O2 2 mole H2O 10.0 g O2 = 11.3 g H2O mole H2O mole O2 32.0 g O2

  30. 8.5 Percentage Yield The percent yield is a comparison of the laboratory answer to the correct answer which is determined by the conversion process. Suppose a student combined 10.0 g of oxygen and 10.0 g of hydrogen in the lab and recovered 8.66 g of water. What would be the percent yield?

  31. 8.5 Percentage Yield The percent yield is a comparison of the laboratory answer to the correct answer which is determined by the conversion process. Suppose a student combined 10.0 g of oxygen and 10.0 g of hydrogen in the lab and recovered 8.66 g of water. What would be the percent yield? Yield (the lab amount) X 100 percent yield = Theoretical Yield (by conversions) 8.66 percent yield = X 100 = 76.6% 11.3

  32. 8.7 Thermochemistry Eact + ΔH - ΔH Exothermic Reaction Endothermic Reaction

  33. 8.7 Thermochemistry When a chemical or physical change takes place energy is either lost of gained. A Thermochemical equation describes this change. Equations gaining energy are called endothermic and equations losing energy are called exothermic.

  34. 8.7 Thermochemical Equations When a chemical or physical change takes place energy is either lost of gained. A Thermochemical equation describes this change. Equations gaining energy are called endothermic and equations losing energy are called exothermic. Examples: C3H6O (l) +4O2 (g) 3CO2(g) + 3 H2O (g) ΔH = -1790 kj Exothermic H2O(l) H2O (g) ΔH = 44.01 kj Endothermic

  35. 8.7 Thermochemical Equations How many kj of heat are released when 709 g of C3H6O are burned?

  36. 8.7 Thermochemical Equations How many kj of heat are released when 709 g of C3H6O are burned? C3H6O (l) + 4O2 (g) 3CO2(g) + 3 H2O (g) ΔH = -1790 kj 709 g C3H6O

  37. 8.7 Thermochemical Equations How many kj of heat are released when 709 g of C3H6O are burned? C3H6O (l) + 4O2 (g) 3CO2(g) + 3 H2O (g) ΔH = -1790 kj 709 g C3H6O mole C3H6O 58.1 g C3H6O

  38. 8.7 Thermochemical Equations How many kj of heat are released when 709 g of C3H6O are burned? C3H6O (l) +4O2 (g) 3CO2(g) + 3 H2O (g) ΔH = -1790 kj 709 g C3H6O mole C3H6O 58.1 g C3H6O

  39. 8.7 Thermochemical Equations How many kj of heat are released when 709 g of C3H6O are burned? C3H6O (l) +4O2 (g) 3CO2(g) + 3 H2O (g) ΔH = -1790 kj 709 g C3H6O mole C3H6O 1790 kj = 21800 kj mole C3H6O 58.1 g C3H6O

  40. The End

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