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Stoichiometry Percent Yield

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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

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Real reactions usually produce less than the “ideal” or theoretical yield. 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. Fisher Scientific SIGMA-ALDRICH Fisher Scientific SIGMA-ALDRICH More details here: Fisher Scientific or SIGMA-ALDRICHFisher Scientific SIGMA-ALDRICH Why is this so?

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

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Finding % Yield 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? 4NH 3 + 5O 2 6H 2 O + 4NO Step 1We find the stoichiometric amount of water that would be produced from 26.0 g of ammonia if we were to obtain a 100% yield.

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Finding % Yield 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? 4NH 3 + 5O 2 6H 2 O + 4NO Step 2Knowing that the stoichiometric amount of water produced is 41.3 g H 2 O, we can now substitute values into the % Yield equation.

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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? 4NH 3 + 5O 2 6H 2 O + 4NO Step 1We 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.

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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? 4NH 3 + 5O 2 6H 2 O + 4NO Step 2With the stoichiometric amount of NO determined, we can now reduce it down by 97.0% to the actual yield.

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Using % Yield to find a reactant amount If the reaction above yields 95.0% and 44.0 g of H 2 O are actually produced, what amount of ammonia gas is required? 4NH 3 + 5O 2 6H 2 O + 4NO Step 1The 44.0 g of H 2 O are already reduced down to 95.0% of the theoretical yield. In order to go from the mass of H 2 O back to the mass of NH 3, we will need to first find what the 100% amount was.

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Using % Yield to find a reactant amount If the reaction above yields 95.0% and 44.0 g of H 2 O are actually produced, what amount of ammonia gas is required? 4NH 3 + 5O 2 6H 2 O + 4NO Step 2With the 100% yield (stoichiometric amount) of water known, it can now be used to calculate the amount of NH 3 reactant needed.

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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

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Have we learned it yet? Try these on your own: Given: 4NH 3 + 5O 2 6H 2 O + 4NO a) What is the % yield of H 2 O if 7.50 g are actually produced using 5.00 g NH 3 ? b) If 9.50 g of O 2 is used to make NO at a 92.5% yield, what mass of NO is actually produced? c) How many grams of NH 3 are needed to produce an actual yield of 33.3 g of H 2 O representing a 94.3% yield?

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4NH 3 + 5O 2 6H 2 O + 4NO a) b) c) Answers ?g H 2 O=5.00 g NH 3 7.94 g H 2 O = 4 mol NH 3 6 mol H 2 O 17.03 g NH 3 1 mol NH 3 1 mol H 2 O 18.02 g H 2 O % Yield= Theoretical g H 2 O Actual g H 2 O X 100 % Yield= 7.94 g H 2 O 7.50 g H 2 O X 100 94.5% yield = ?g NO=9.50 g O 2 5 mol O 2 4 mol NO 32.00 g O 2 1 mol O 2 1 mol NO 30.01 g NO 100 92.56.59 g NO = ?g NH 3 =33.3 g H 2 O 6 mol H 2 O 4 mol NH 3 18.02 g H 2 O 1 mol H 2 O 1 mol NH 3 17.03 g NH 3 94.3 10022.2 g NH 3 =

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