1 / 12

Stoichiometry Percent Yield

Stoichiometry Percent Yield. Important Terms. Yield : the amount of product Theoretical yield : Actual yield : Percent yield : (actual yield ÷ theoretical yield) × 100. Real reactions usually produce less than the “ideal” or theoretical yield. . Why is this so?.

viola
Download Presentation

Stoichiometry Percent Yield

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. StoichiometryPercent Yield

  2. Important Terms • Yield: the amount of product • Theoretical yield: • Actual yield: • 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 • • • • 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. % 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 stoichiometricyield ►_____________________________________________ • Using % Yield to find amount of product ►Given % yield ►Find stoichiometric yield of product ►Multiply _______________________________________ • Using % Yield to find amount of reactant ►Given % yield ►Divide ______________________ 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?

More Related