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Chapter 9: Stoichiometry

Chapter 9: Stoichiometry. 9.1 Mole to Mole. Objective: To perform mole to mole conversion problems. Stoichiometry. Stoichiometry is the branch of chemistry that deals with quantities of substances in chemical reactions. Mole to Mole Problems: The Steps. 1. Write chemical equation

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Chapter 9: Stoichiometry

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  1. Chapter 9: Stoichiometry

  2. 9.1 Mole to Mole Objective: To perform mole to mole conversion problems.

  3. Stoichiometry • Stoichiometry is the branch of chemistry that deals with quantities of substances in chemical reactions.

  4. Mole to Mole Problems: The Steps 1. Write chemical equation 2. Balance chemical equation using coefficients 3. Use the following setup to perform calculation: A is the known quantity B is the unknown quantity 4. Don’t forget units on your final answer!! Mole Ratio

  5. 9.2 Mole to Mass and Mass to Mole • Objective: To perform mole to mass and mass to mole conversion problems.

  6. Mole to Mass Molar Mass Unknown Moles Given Unknown Moles (Eqn.) Grams Unknown Given Moles (Eqn.) 1 mole Unknown Mole Ratio

  7. Mass to Mole Unknown Moles (Eqn.) Grams Given 1 mol Given Moles Unknown Molar Mass Given Given Moles (Eqn.) Mole Ratio

  8. 9.3 Mass to Mass • Objective: To perform mass to mass conversion problems.

  9. Stoichiometry Roadmap Unknown Moles (Eqn.) Molar Mass Unknown Grams Given 1 mol Given Molar Mass Given Given Moles (Eqn.) 1 mole Unknown Grams Unknown Mole Ratio

  10. 9.4 Limiting Reactant • Objectives: • To calculate the theoretical yield of a chemical reactions. • To determine the limiting reactant and excess reactant in a chemical reaction.

  11. Limiting Reactant • Any reactant that is used up first in a chemical reaction. • It determines the amount of product that can be formed in the reaction.

  12. Excess Reactant • The reactant that is not completely used up in a reaction.

  13. Limiting Reactant Problems • Use the mass to mass conversions

  14. Example • Copper reacts with sulfur to form copper(I) sulfide according to the following balanced equation: 2Cu + S  Cu2S What is the limiting reactant when 80.0 grams of Cu reacts with 25.0 grams of S?

  15. Example 2Cu + S  Cu2S The general equation is: Start with Copper: = 100.19 g Cu2S Now use Sulfur: = 124.08 g Cu2S

  16. Example • The limiting reactant is copper. • The excess reactant is sulfur. • The amount of Cu2S that is produced is 100.19 g Cu2S.

  17. 9.5 Percent Yield • Objective: To calculate percent yield.

  18. Introduction to Percent Yield… • If you get 15 out of 20 questions correct on a test, what percentage did you receive on the test? How did you figure this out? 15/20 x 100 = 75% • If Sammy Sosa gets 25 hits in the month of May and has 113 bats, what is his batting average for the month of May? Explain how you arrived at your answer. 25/113 = 0.221 As a percentage, this is written, 25/113 x 100 = 22.1%

  19. Percent Yield ·Percent yield is the ratio of the actual yield to the theoretical yield for a chemical reaction expressed as a percentage. · It is a measure of efficiency of a reaction. Percent Yield = actual yield x 100% theoretical yield

  20. Actual Yield • The amount of product that forms when a reaction is carried out in the laboratory.

  21. Theoretical Yield • The amount of product that could form during a reaction calculated from a balanced chemical equation. • It represents the maximum amount of product that could be formed from a given amount of reactant.

  22. Example #1 The equation for the complete combustion of ethene (C2H2) is 2C2H2 + 5O2 4CO2 + 2H2O • If 0.10 g of C2H2 is reacted with 201.60 g of O2, identify the limiting reactant. • What is the theoretical yield of H2O? • If the actual yield of H2O is 0.05 g, calculate the percent yield.

  23. Example #1 Start with C2H2: 2C2H2 + 5O2 4CO2 + 2H2O = 0.07 g H2O Theoretical Yield Limiting Reactant = C2H2 Now start with O2: = 45.36 g H2O

  24. Example #1 2C2H2 + 5O2 4CO2 + 2H2O Actual Yield = 0.05 g H2O Theoretical Yield = 0.07 g H2O Percent Yield = actual yield x 100% theoretical yield Percent Yield = 0.05 g H2O x 100% = 71.43 % 0.07 g H2O

  25. Example #2 • Determine the percent yield for the reaction between 3.74 g of Na and excess O2 if 5.34 g of Na2O2 is recovered? • First, write the chemical equation: Na + O2 Na2O2 • Second, balance the chemical equation: 2Na + O2 Na2O2

  26. Example #22Na + O2 Na2O2 • Solve the mass-mass problem, starting with Na: Actual Yield = 5.34 g Na2O2 Theoretical Yield = 6.34 g Na2O2 Percent Yield = actual yield x 100% theoretical yield Percent Yield = 5.34 g x 100% = 84.23% 6.34 g = 6.34 g Na2O2

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