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Stoichiometry Percent Yield

Stoichiometry Percent Yield. Important Terms. Yield : the amount of product Theoretical yield : the maximum amount of product expected, based on stoichiometric calculations Actual yield : amount of product from a procedure or experiment (this is often given in a question)

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Stoichiometry Percent Yield

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  1. StoichiometryPercent Yield

  2. Important Terms • Yield: the amount of product • Theoretical yield: the maximum amount of product expected, based on stoichiometric calculations • Actual yield: amount of product from a procedure or experiment (this is often given in a question) • Percent yield: (actual yield ÷ theoretical yield) × 100

  3. Real reactions usually produce less than the “ideal” or theoretical yield. Why is this so? • •Side Reactions (aka: competing reactions): reactant products • • Reaction does not go to completion: reactant product • • Loss of product (vaporizing, spillage, spattering) • • Reactant impurities due to varying grades of chemicals: • Reagent = The highest quality commercially available for this chemical. • Practical = chemicals of good quality where there are no official standards. • Lab grade = Suitable for histology methods and general laboratory applications. • USP = Chemicals manufactured under current Good Manufacturing Practices and which meet the requirements of the US Pharmacopeia. • Technical = A grade suitable for general industrial use and non critical laboratory tasks. • More details here: Fisher Scientific or SIGMA-ALDRICH

  4. Comes from a measured experimental value or is given in the question or problem. % Yield Comes from the stoichiometry calculation. It is the “ideal” or “perfect” yield.

  5. Finding % Yield 4NH3 + 5O2 6H2O + 4NO Using the equation above, if 39.9 g of water are produced when 26.0 g of ammonia are reacted, what is the % yield of the reaction? Step 1 We find the stoichiometric amount of water that would be produced from 26.0 g of ammonia if we were to obtain a 100% yield.

  6. Finding % Yield 4NH3 + 5O2 6H2O + 4NO Using the equation above, if 39.9 g of water are produced when 26.0 g of ammonia are reacted, what is the % yield of the reaction? Step 2 Knowing that the stoichiometric amount of water produced is 41.3 g H2O, we can now substitute values into the % Yield equation.

  7. 4NH3 + 5O2 6H2O + 4NO Using % Yield to find a product amount 22.0 g of oxygen gas are reacted with excess ammonia as shown above. If the reaction yields 97.0%, what mass of NO will actually be produced? Step 1 We find the stoichiometric amount of NO that would be produced from 22.0 g of oxygen gas if we were to obtain a 100% yield.

  8. 4NH3 + 5O2 6H2O + 4NO Using % Yield to find a product amount 22.0 g of oxygen gas are reacted with excess ammonia as shown above. If the reaction yields 97.0%, what mass of NO will actually be produced? Step 2 With the stoichiometric amount of NO determined, we can now reduce it down by 97.0% to the actual yield.

  9. 4NH3 + 5O2 6H2O + 4NO Using % Yield to find a reactant amount If the reaction above yields 95.0% and 44.0 g of H2O are actually produced, what amount of ammonia gas is required? Step 1 The 44.0 g of H2O are already reduced down to 95.0% of the theoretical yield. In order to go from the mass of H2O back to the mass of NH3, we will need to first find what the 100% amount was.

  10. 4NH3 + 5O2 6H2O + 4NO Using % Yield to find a reactant amount If the reaction above yields 95.0% and 44.0 g of H2O are actually produced, what amount of ammonia gas is required? Step 2 With the 100% yield (stoichiometric amount) of water known, it can now be used to calculate the amount of NH3 reactant needed.

  11. Summary • Finding % Yield ►Given actual yield ►Find stoichiometric yield ►Use % yield formula • Using % Yield to find amount of product ►Given % yield ►Find stoichiometric yield of product ►Multiply % yield and stoichiometric yield • Using % Yield to find amount of reactant ►Given % yield ►Divide actual yield by % yield to get theoretical yield ►Use theoretical yield of product to find reactant amount

  12. Have we learned it yet? Try these on your own: Given: 4NH3 + 5O2 6H2O + 4NO a) What is the % yield of H2O if 7.50 g are actually produced using 5.00 g NH3? b) If 9.50 g of O2 is used to make NO at a 92.5% yield, what mass of NO is actually produced? c) How many grams of NH3 are needed to produce an actual yield of 33.3 g of H2O representing a 94.3% yield?

  13. 1 mol NH3 18.02 g H2O 6 mol H2O 17.03 g NH3 7.50 g H2O Actual g H2O 1 mol H2O % Yield= % Yield= X 100 X 100 4 mol NH3 7.94 g H2O Theoretical g H2O = = = = 94.5% yield 7.94 g H2O 6.59 g NO 22.2 g NH3 1 mol O2 4 mol NO 30.01 g NO 32.00 g O2 5 mol O2 1 mol NO 100 92.5 1 mol H2O 4 mol NH3 100 94.3 18.02 g H2O 6 mol H2O Answers ?g H2O= 5.00 g NH3 4NH3 + 5O2 6H2O + 4NO a) b) c) ?g NO= 9.50 g O2 17.03 g NH3 ?g NH3= 33.3 g H2O 1 mol NH3

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