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Assignment: Read 5.6 up to sample (238-9) 1. Define the following terms: yield, theoretical yield, actual yield, percentage yield. 2. Based on your reading, give 4 reasons why the actual yield in a chemical reaction often falls short of the theoretical yield. 3. Read the sample problem on the next slide and try the practice problem on slide number 5 4. When 5.00 g of KClO 3 is heated it decomposes according to the equation: 2KClO 3 2KCl + 3O 2 a) Calculate the theoretical yield of oxygen. b) Give the % yield if 1.78 g of O 2 is produced. c) How much O 2 would be produced if the percentage yield was 78.5%?

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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. Mg 3 N 2 ) The reaction does not go to completion

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

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

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4) 2KClO 3 2KCl + 3O 2 a) b) c) Answers 3 mol O 2 2 mol KClO 3 x # g O 2 = (also works if you use mol O 2 ) 5.00 g KClO 3 1.958 g= 32 g O 2 1 mol O 2 x 1 mol KClO 3 122.55 g KClO 3 x 1.78 g O 2 1.958 g O 2 = % yield = x 100%90.9%= actual theoretical x 100% x g O 2 1.958 g O 2 = % yield = x 100%78.5%= actual theoretical x 100% x g O 2 78.5% x 1.958 g O 2 100% = 1.537 g O 2 =

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

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More Percent Yield Questions Note: try shortcut for limiting reagent problems 1. The electrolysis of water forms H 2 and O 2. 2H 2 O 2H 2 + O 2 What is the % yield of O 2 if 12.3 g of O 2 is produced from the decomposition of 14.0 g H 2 O? 2. 107 g of oxygen is produced by heating 300 grams of potassium chlorate. Calculate % yield. 2KClO 3 2KCI + 3O 2 3. 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

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More Percent Yield Questions 4. Iron pyrites (FeS 2 ) reacts with oxygen according to the following equation: 4FeS 2 + 11O 2 2Fe 2 O 3 + 8SO 2 If 300 g of iron pyrites is burned in 200 g of O 2, 143 grams of ferric oxide is produced. What is the percent yield of ferric oxide? 5. 70 grams of manganese dioxide is mixed with 3.5 moles of hydrochloric acid. How many grams of Cl 2 will be produced from this reaction if the % yield for the process is 42%? MnO 2 + 4HCI MnCl 2 + 2H 2 O + Cl 2

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

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Q2 2.107 g of oxygen is produced by heating 300 grams of potassium chlorate. 2KClO 3 2KCI + 3O 2 Actual yield is given: 107 g O 2 Next, calculate theoretical yield 3 mol O 2 2 mol KClO 3 x # g O 2 = 300 g KClO 3 117.5 g= 32 g O 2 1 mol O 2 x 1 mol KClO 3 122.55 g KClO 3 x Finally, calculate % yield 107 g O 2 117.5 g O 2 = % yield = x 100%91.1%= actual theoretical x 100%

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Q3 3.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 # g FeS=3.00 mol Fe 263.7 g= 87.91 g FeS 1 mol FeS x Finally, calculate % yield 220 g O 2 263.7 g O 2 = % yield = x 100%83.4%= actual theoretical x 100%

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4.4FeS 2 + 11O 2 2Fe 2 O 3 + 8SO 2 If 300 g of FeS 2 is burned in 200 g of O 2, 143 g Fe 2 O 3 results. % yield Fe 2 O 3 ? First, determine limiting reagent 2 mol Fe 2 O 3 11 mol O 2 x 200 g O 2 181.48 g Fe 2 O 3 = 159.7 g Fe 2 O 3 1 mol Fe 2 O 3 x 1 mol O 2 32 g O 2 x 2 mol Fe 2 O 3 4 mol FeS 2 x # g Fe 2 O 3 = 300 g FeS 2 199.7 g Fe 2 O 3 = 159.7 g Fe 2 O 3 1 mol Fe 2 O 3 x 1 mol FeS 2 119.97 g FeS 2 x 143 g Fe 2 O 3 181.48 g Fe 2 O 3 = % yield = x 100%78.8%= actual theoretical x 100%

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5.70 g of MnO 2 + 3.5 mol HCl gives a 42% yield. How many g of Cl 2 is produced? MnO 2 + 4HCI MnCl 2 + 2H 2 O + Cl 2 1 mol Cl 2 4 mol HCl x 3.5 mol HCl 62.13 g Cl 2 = 71 g Cl 2 1 mol Cl 2 x 1 mol MnO 2 x # g Cl 2 = 70 g MnO 2 57.08 g Cl 2 = 70.9 g Cl 2 1 mol Cl 2 x 1 mol MnO 2 86.94 g MnO 2 x x g Cl 2 57.08 g Cl 2 = % yield = x 100%42%= actual theoretical x 100% x g Cl 2 42% x 57.08 g Cl 2 100% = 24 g Cl 2 = For more lessons, visit www.chalkbored.com www.chalkbored.com

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