Percentage Yield.

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Percentage Yield

Assignment: Read 5.6 up to sample (238-9)
Define the following terms: yield, theoretical yield, actual yield, percentage yield. Based on your reading, give 4 reasons why the actual yield in a chemical reaction often falls short of the theoretical yield. Read the sample problem on the next slide and try the practice problem on slide number 5 When 5.00 g of KClO3 is heated it decomposes according to the equation: 2KClO3  2KCl + 3O2 a) Calculate the theoretical yield of oxygen. b) Give the % yield if 1.78 g of O2 is produced. c) How much O2 would be produced if the percentage yield was 78.5%?

Answers 1) Yield: the amount of product
Theoretical yield: the amount of product we expect, based on stoichiometric calculations Actual yield: amount of product from a procedure or experiment (this is given in the question) Percent yield: x 100% actual yield theoretical yield 2) Not all product is recovered (e.g. spattering) Reactant impurities (e.g. weigh out 100 g of chemical which has 20 g of junk) A side reaction occurs (e.g. MgO vs. Mg3N2) The reaction does not go to completion

Sample problem Q - What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2? Step 1: write the balanced chemical equation 2H2 + O2  2H2O Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: 1 mol H2 2.02 g H2 x 2 mol H2O 2 mol H2 x 18.02 g H2O 1 mol H2O x # g H2O= 16 g H2 143 g = Step 3: Calculate % yield actual theoretical 138 g H2O 143 g H2O = % yield = x 100% x 100% 96.7% =

Practice problem Q - What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess N2? Step 1: write the balanced chemical equation N2 + 3H2  2NH3 Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: 2 mol NH3 3 mol H2 x 17.04 g NH3 1 mol NH3 x # g NH3= 20.0 mol H2 227 g = Step 3: Calculate % yield actual theoretical 40.5 g NH3 227 g NH3 = % yield = x 100% x 100% 17.8% =

Answers 4) 2KClO3  2KCl + 3O2 a) b) c)
# g O2= (also works if you use mol O2) 1 mol KClO3 g KClO3 x 3 mol O2 2 mol KClO3 x 32 g O2 1 mol O2 x 5.00 g KClO3 1.958 g = actual theoretical 1.78 g O2 1.958 g O2 = % yield = x 100% x 100% 90.9% = actual theoretical x g O2 1.958 g O2 = % yield = x 100% x 100% 78.5% = 78.5% x g O2 100% = x g O2 1.537 g O2 =

Challenging question 2H2 + O2  2H2O
What is the % yield of H2O if 58 g H2O are produced by combining 60 g O2 and 7.0 g H2? Hint: determine limiting reagent first 1 mol O2 32 g O2 x 2 mol H2O 1 mol O2 x 18.02 g H2O 1 mol H2O x # g H2O= 60 g O2 68 g = 1 mol H2 2.02 g H2 x 2 mol H2O 2 mol H2 x 18.02 g H2O 1 mol H2O x # g H2O= 7.0 g H2 62.4 g = actual theoretical 58 g H2O 62.4 g H2O = % yield = x 100% x 100% 92.9% =

More Percent Yield Questions
Note: try “shortcut” for limiting reagent problems The electrolysis of water forms H2 and O H2O  2H2 + O What is the % yield of O2 if 12.3 g of O2 is produced from the decomposition of 14.0 g H2O? 107 g of oxygen is produced by heating 300 grams of potassium chlorate. Calculate % yield. 2KClO3  2KCI + 3O2 What is the % yield of ferrous sulphide if 3.00 moles of Fe reacts with excess sulfur to produce 220 grams of ferrous sulphide? Fe + S  FeS

More Percent Yield Questions
Iron pyrites (FeS2) reacts with oxygen according to the following equation: 4FeS2 + 11O2  2Fe2O3 + 8SO2 If 300 g of iron pyrites is burned in 200 g of O2, 143 grams of ferric oxide is produced. What is the percent yield of ferric oxide? 70 grams of manganese dioxide is mixed with 3.5 moles of hydrochloric acid. How many grams of Cl2 will be produced from this reaction if the % yield for the process is 42%? MnO2 + 4HCI  MnCl2 + 2H2O + Cl2

Q1 The electrolysis of water forms H2 & O2. 2H2O  2H2 + O2
Give the percent yield of O2 if 12.3 g O2 is produced from the decomp. of 14 g H2O? Actual yield is given: 12.3 g O2 Next, calculate theoretical yield # g O2= 1 mol H2O 18.02 g H2O x 1 mol O2 2 mol H2O x 32 g O2 1 mol O2 x 14.0 g H2O 12.43 g = Finally, calculate % yield actual theoretical 12.3 g O2 12.43 g O2 = % yield = x 100% x 100% 98.9% =

Q2 107 g of oxygen is produced by heating 300 grams of potassium chlorate. 2KClO3  2KCI + 3O2 Actual yield is given: 107 g O2 Next, calculate theoretical yield # g O2= 1 mol KClO3 g KClO3 x 3 mol O2 2 mol KClO3 x 32 g O2 1 mol O2 x 300 g KClO3 117.5 g = Finally, calculate % yield actual theoretical 107 g O2 117.5 g O2 = % yield = x 100% x 100% 91.1% =

Q3 What is % yield of ferrous sulfide if 3 mol Fe produce 220 grams of ferrous sulfide? Fe + S  FeS Actual yield is given: 220 g FeS Next, calculate theoretical yield 1 mol FeS 1 mol Fe x 87.91 g FeS 1 mol FeS x # g FeS= 3.00 mol Fe 263.7 g = Finally, calculate % yield actual theoretical 220 g O2 263.7 g O2 = % yield = x 100% x 100% 83.4% =

First, determine limiting reagent
4FeS2 + 11O2  2Fe2O3 + 8SO2 If 300 g of FeS2 is burned in 200 g of O2, 143 g Fe2O3 results. % yield Fe2O3? First, determine limiting reagent # g Fe2O3= 1 mol FeS2 g FeS2 x 2 mol Fe2O3 4 mol FeS2 x 159.7 g Fe2O3 1 mol Fe2O3 x 300 g FeS2 199.7 g Fe2O3 = 1 mol O2 32 g O2 x 2 mol Fe2O3 11 mol O2 x 159.7 g Fe2O3 1 mol Fe2O3 x 200 g O2 g Fe2O3 = actual theoretical 143 g Fe2O3 g Fe2O3 = % yield = x 100% x 100% 78.8% =

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70 g of MnO mol HCl gives a 42% yield. How many g of Cl2 is produced? MnO2 + 4HCI  MnCl2 + 2H2O + Cl2 # g Cl2= 1 mol MnO2 86.94 g MnO2 x 1 mol Cl2 1 mol MnO2 x 70.9 g Cl2 1 mol Cl2 x 70 g MnO2 57.08 g Cl2 = 1 mol Cl2 4 mol HCl x 71 g Cl2 1 mol Cl2 x 3.5 mol HCl 62.13 g Cl2 = actual theoretical x g Cl2 57.08 g Cl2 = % yield = x 100% x 100% 42% = 42% x g Cl2 100% = x g Cl2 24 g Cl2 = For more lessons, visit