2 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 5When 5.00 g of KClO3 is heated it decomposes according to the equation: 2KClO3 2KCl + 3O2a) 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%?
3 Answers 1) Yield: the amount of product Theoretical yield: the amount of product we expect, based on stoichiometric calculationsActual yield: amount of product from a procedure or experiment (this is given in the question)Percent yield:x 100%actual yieldtheoretical yield2)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
4 Sample problemQ - 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 equation2H2 + O2 2H2OStep 2: determine actual and theoretical yield. Actual is given, theoretical is calculated:1 mol H22.02 g H2x2 mol H2O2 mol H2x18.02 g H2O1 mol H2Ox# g H2O=16 g H2143 g=Step 3: Calculate % yieldactualtheoretical138 g H2O143 g H2O=% yield =x 100%x 100%96.7%=
5 Practice problemQ - 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 equationN2 + 3H2 2NH3Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated:2 mol NH33 mol H2x17.04 g NH31 mol NH3x# g NH3=20.0 mol H2227 g=Step 3: Calculate % yieldactualtheoretical40.5 g NH3227 g NH3=% yield =x 100%x 100%17.8%=
6 Answers 4) 2KClO3 2KCl + 3O2 a) b) c) # g O2= (also works if you use mol O2)1 mol KClO3g KClO3x3 mol O22 mol KClO3x32 g O21 mol O2x5.00 g KClO31.958 g=actualtheoretical1.78 g O21.958 g O2=% yield =x 100%x 100%90.9%=actualtheoreticalx g O21.958 g O2=% yield =x 100%x 100%78.5%=78.5% x g O2100%=x g O21.537 g O2=
7 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 first1 mol O232 g O2x2 mol H2O1 mol O2x18.02 g H2O1 mol H2Ox# g H2O=60 g O268 g=1 mol H22.02 g H2x2 mol H2O2 mol H2x18.02 g H2O1 mol H2Ox# g H2O=7.0 g H262.4 g=actualtheoretical58 g H2O62.4 g H2O=% yield =x 100%x 100%92.9%=
8 More Percent Yield Questions Note: try “shortcut” for limiting reagent problemsThe 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 + 3O2What 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
9 More Percent Yield Questions Iron pyrites (FeS2) reacts with oxygen according to the following equation:4FeS2 + 11O2 2Fe2O3 + 8SO2If 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
10 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 O2Next, calculate theoretical yield# g O2=1 mol H2O18.02 g H2Ox1 mol O22 mol H2Ox32 g O21 mol O2x14.0 g H2O12.43 g=Finally, calculate % yieldactualtheoretical12.3 g O212.43 g O2=% yield =x 100%x 100%98.9%=
11 Q2107 g of oxygen is produced by heating 300 grams of potassium chlorate.2KClO3 2KCI + 3O2Actual yield is given: 107 g O2Next, calculate theoretical yield# g O2=1 mol KClO3g KClO3x3 mol O22 mol KClO3x32 g O21 mol O2x300 g KClO3117.5 g=Finally, calculate % yieldactualtheoretical107 g O2117.5 g O2=% yield =x 100%x 100%91.1%=
12 Q3What is % yield of ferrous sulfide if 3 mol Fe produce 220 grams of ferrous sulfide?Fe + S FeSActual yield is given: 220 g FeSNext, calculate theoretical yield1 mol FeS1 mol Fex87.91 g FeS1 mol FeSx# g FeS=3.00 mol Fe263.7 g=Finally, calculate % yieldactualtheoretical220 g O2263.7 g O2=% yield =x 100%x 100%83.4%=
13 First, determine limiting reagent 4FeS2 + 11O2 2Fe2O3 + 8SO2If 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 FeS2g FeS2x2 mol Fe2O34 mol FeS2x159.7 g Fe2O31 mol Fe2O3x300 g FeS2199.7 g Fe2O3=1 mol O232 g O2x2 mol Fe2O311 mol O2x159.7 g Fe2O31 mol Fe2O3x200 g O2g Fe2O3=actualtheoretical143 g Fe2O3g Fe2O3=% yield =x 100%x 100%78.8%=
14 For more lessons, visit www.chalkbored.com 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 MnO286.94 g MnO2x1 mol Cl21 mol MnO2x70.9 g Cl21 mol Cl2x70 g MnO257.08 g Cl2=1 mol Cl24 mol HClx71 g Cl21 mol Cl2x3.5 mol HCl62.13 g Cl2=actualtheoreticalx g Cl257.08 g Cl2=% yield =x 100%x 100%42%=42% x g Cl2100%=x g Cl224 g Cl2=For more lessons, visit
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