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Slide 1 / 109 Slide 2 / 109 Stoichiometric Calculations Slide 3 / 109 Table of Contents Click on the topic to go to that section · Stoichiometry Calculations with Moles · Stoichiometry Calculations with Particles and Volume · Stoichiometry Calculations with Mass · · · · Mixed Stoichiometry Problems Limiting Reactants Theoretical, Actual and Percent Yield Calculating Excess Reactants Slide 4 / 109 Stoichiometry Calculations with Moles Return to Table of Contents Stoichiometry Slide 5 / 109 The word stoichiometry is derived from two Greek words. "stoicheion" meaning element and "metron", meaning measure. Stoichiometry in our daily lives Airbags save thousands of lives every year. When a collision happens, the following reaction occurs. 2NaN3(s) --> 2Na(s) + 3N2(g) In order to properly inflate, roughly 50 L of nitrogen gas must be produced. NaN3 capsule Engineers have determined, using stoichiometry, that about 96 grams of NaN3 are needed to react in each airbag capsule to produce enough nitrogen gas. Slide 6 / 109 Stoichiometric Calculations Slide 7 / 109 The coefficients in the balanced equation give the ratio of moles of reactants and products. 2 H2 + O2 --# 2 H2 O can be read as: 2 moles of H plus 1 mole of O yields 2 moles of H O 2 2 2 giving ratios of: 2 mol H2 1 mol O2 2 mol H2 2 mol H2O 1 mol O2 2 mol H2O Stoichiometric Calculations Slide 8 / 109 A balanced chemical equation is needed to perform any stoichiometric calculations. N 2 + 3H ---> 2 1 mol N 3 mol H 3 mol H 2 mol NH 2 2 2NH 3 1 mol N 2 2 2 mol NH 3 3 The above ratios can be used to determine the quantity of any reactant and products. For every 1 mol of N2 , · you would need 3 mol of H2 to completely react with · you would produce 2 mol of NH3 Slide 9 / 109 Stoichiometric Calculations with Moles Using this interpretation it's straightforward to answer questions about the relative number of moles of reactants and products. For instance, use the balanced equation below to determine the maximum number of moles of H O that could be created from reacting 8 moles of H . 2 2 2H +O 2 2 2HO 2 Stoichiometry Calculations with Moles 2H +O 2 Slide 10 / 109 2HO 2 2 1. Use the equation to set up a ratio of the substances of interest. 2 mol H2O 2 mol H2 2. Set that equal to the ratio of the known to unknown quantities of the same substances. 2 mol H2O n mol H2O = 8 mol H2 2 mol H2 3. Solve for the unknown. 8(2) = (2) n mol H2O n = 8 mol H2O Slide 11 / 109 Stoichiometry Calculations with Moles Another Example: Given the equation: 2H +O 2 2 2HO 2 How many moles of oxygen would be needed to react with 12 moles of H2? Stoichiometry Calculations with Moles 2H +O 1. Use the formula to set up a ratio of the substances of interest. 2 2 2. Set that equal to the ratio of the known to unknown quantities of the same substances. 3. Solve for the unknown by cross multiplying. - -# 2 H O 2 1 mol O2 2 mol H2 1 mol O2 n mol O2 = 2 mol H2 12 mol H2 12(1) = (2)n O2 = 6 n O2 Slide 12 / 109 Stoichiometry Calculations 2 H2 Slide 13 / 109 + O2 -# 2 H2O Given 12 moles of H2, how much O2 and H2O would be needed or produced? Using the ratios, you would need..... 1/2 as much O2 as H2 (ratio is 1 mol O2/2 mol H2) 12 mol H2 x 1 mol O2 = 6 mol O2 needed 2 mol H2 Using the ratios, you would produce..... An equal amount of H2O as H2 used (ratio is 2 mol H2O/2 mol H2) 12 mol H2 x 2 mol H2O = 12 mol H2O 2 mol H2 Real World Application 2Al + Fe2O3 Slide 14 / 109 --> 2Fe + Al2O3 + 859 kJ of energy The thermite reaction (above) releases a lot of heat and is used in to weld railroad tracks together. How many moles of Al would be needed to produce 7.8 moles of Al2O3? 7.8 mol Al2O3 move x 2 mol for Al = 15.6 mol Al answer 1 mol Al O 2 1 3 What is the largest number of moles of Al2O3 that could result from reacting 6 moles of O2? 4 Al (s) + 3 O2 (g) - - # 2 Al O 2 3 (s) Slide 15 / 109 Slide 16 / 109 2 How many moles of O2 would be required to create 12 moles of Al2 O3 ? 4 Al (s) + 3 O (g) 2 - # 2 Al O 2 3 (s) Slide 17 / 109 3 How many moles of O2 would be required to completely react with 8 moles of Al? 4 Al (s) + 3 O2 (g) - - # 2 Al O 2 3 (s) Slide 18 / 109 4 When iron rusts in air, iron (III) oxide is produced. How many moles of oxygen react with 2.4 mol of Fe in this reaction? 4 Fe (s) + 3 O2 (g)--> 2 Fe2O3 (s) Slide 19 / 109 5 How many moles of aluminium are needed to react completely with 1.2 mol FeO? 2 Al (s) + 3 FeO (s) --> 3 Fe (s) + Al2O3 (s) 6 How many moles of calcium metal are produced from the decomposition of 8 mol of calcium chloride? Slide 20 / 109 CaCl2 (s) --> Ca (s) + Cl2 (g) 7 How many moles (total) of calcium metal and chlorine gas are produced from the decomposition of 8 mol of calcium chloride? CaCl2 (s) --> Ca (s) + Cl2 (g) Slide 21 / 109 8 How many moles of Ag are needed to react with 40 moles of HNO3? Slide 22 / 109 3 Ag(s) + 4 HNO3(aq) --> 3 AgNO3(aq) + 2 H2O(l) + NO(g) 9 How many moles of AgNO3 could be produced from 40 moles of HNO3? Slide 23 / 109 3Ag(s) + 4HNO3(aq) --> 3AgNO3(aq) + 2 H2O(l) + NO(g) 10 How many moles of water would be produced when 0.4 moles of Ag react with an excess amount of HNO3? 3Ag(s) + 4HNO3(aq) --> 3AgNO3(aq) + 2 H2O(l) + NO(g) Slide 24 / 109 11 How many moles of NO were produced if 16 moles of water were made during the reaction? Slide 25 / 109 3Ag(s) + 4HNO3(aq) --> 3AgNO3(aq) + 2 H2O(l) + NO(g) 12 How many moles of N2H4 are required to produce 57 moles of nitrogen gas? 2 N2H4(l) + N2O4(l) ---> Slide 26 / 109 3 N2(g) + 4 H2O(g) 13 How many moles of dinitrogen tetraoxide would be needed to produce 57 moles of nitrogen gas? 2 N2H4(l) + N2O4(l) ---> 3 N2(g) + 4 H2O(g) Slide 27 / 109 14 How many moles of water are produced if 57 moles of nitrogen gas are produced? 2 N2H4(l) + N2O4(l) ---> 3 N2(g) + 4 H2O(g) 15 How many total moles of gas would be produced when 5 moles of nitrogen tetrahydride reacts with excess N2O4? 2 N2H4(l) + N2O4(l) ---> Slide 28 / 109 Slide 29 / 109 3 N2(g) + 4 H2O(g) Slide 30 / 109 Stoichiometry Calculations with Particles and Volume Return to Table of Contents Stoichiometry Calculations with Particles Slide 31 / 109 The number of particles (atoms, molecules, formula units) is directly to proportional to the number of moles. Therefore... 2 H2 + O2 --> 2 H2 O can be read as: 2 molecules of H2 plus 1 molecule of O2 yields 2 molecules of H2 O. Note...while moles can be expressed as non-whole numbers, particles must be whole numbers. One cannot have 6.1 atoms, molecules, or formula units! 16 What is the largest number of of Li3 N formula units that could result from reacting 6 N2 molecules? Slide 32 / 109 6 Li (s) + N2 (g) --# 2 Li3 N (s) 17 How many N2 molecules would be required to create 4 Li3 N formula units? 6 Li (s) + N2 (g) --# 2 Li3 N (s) Slide 33 / 109 18 How many Li atoms would be required to completely react with 3 N2 molecules? Slide 34 / 109 6 Li (s) + N2 (g) --# 2 Li3 N (s) Slide 35 / 109 Stoichiometry Calculations with Volumes At a given temperature and pressure, the space a sample of a gas takes up (it's volume) is proportional to the number of moles of gas molecules present. Therefore... 2 H2 (g) + O2 (g) --> 2 H2 O(g) can be read as: 2 volumes of H2 plus 1 volume of O2 yields 2 volumes of H2 O. Note: The volume of a material is only proportional to the number of moles when the substance is in the gas phase! Slide 36 / 109 19 The equation below shows the decomposition of lead nitrate. How many liters of oxygen are produced when 12L of NO2 are formed? (STP) 2Pb(NO3 )2 (s) --> 2PbO (s) +4NO2 (g) + O2 (g) Slide 37 / 109 20 What volume of methane is needed to completely react with 500 mL of O2 at STP? (Balance the equation first!!!) __ CH4 + ___ O2 --# ___ CO2 + ___ H2 O 21 How many liters of H2 O (g) will be created from reacting 8.0 L of H2 (g) with a sufficient amount of O2 (g)? Slide 38 / 109 2 H2 (g) + O2 (g) --# 2 H2 O (g) 22 How many liters of NO2 (g) will be created from reacting 36 L of O2 (g) with a sufficient amount of NH3 (g)? 4 NH3 (g) + 7 O2 (g) --# 4 NO2 (g) + 6 H2 O (g) Slide 39 / 109 Slide 40 / 109 Stoichiometry with Particles and Volumes It's common to be asked to report a value in a unit other than the one given. For example: Given the following reaction, how many L of nitrogen gas would be needed to produce 3 moles of ammonia? N2(g) + 3H2(g) --> 2NH3 Slide 41 / 109 Stoichiometry with Particles and Volumes Given the following reaction, how many liters of nitrogen gas would be needed @STP to produce 3 moles of ammonia? N2(g) + 3H2(g) --> 2NH3 mol NH3 --> mol N2 --> L N2 3 mol NH3 x 1 mol N2 2 mol NH3 x 22.4 L = 33.6 L N2 1 mol Slide 42 / 109 Stoichiometry with Particles and Volumes Example 2: How many moles of Cl2 gas would be needed to produce 3 x 1024 formula units of NaCl given the following reaction. 2Na(s) + Cl2(g) --> 2NaCl(s) formula units NaCl --> molecules Cl2 --> mol Cl2 3 x 1024 formula units NaCl x 1 molecule Cl2 x 1 mol Cl2 2 for. units NaCl 6.02 x 1023 molecules = 2.5 mol Cl2 23 How many sodium atoms would be needed to react with 33.6 L of chlorine gas at STP? Slide 43 / 109 2Na(s) + Cl2(g) --> 2NaCl Note: Sodium is a solid and therefore cannot be expressed in L, so first convert the chlorine gas to moles. 24 How many liters of oxygen gas would need to be combusted with excess hydrocarbon to produce 5.5 moles of water @STP? Slide 44 / 109 2 C H O (s) + 163 O (g) --> 114 CO (g) + 110 H O(l) 57 110 6 2 2 2 Note: The water product is a liquid not a gas, so it can't be converted to L. Find moles of oxygen gas first! 25 How many moles of carbon dioxide gas would be produced @STP if 2.24 L of O2 gas react? 2 C H O (s) + 163 O (g) --> 114 CO (g) + 110 H O(l) 57 110 6 2 2 2 Slide 45 / 109 26 How many L of water vapor would be produced @STP when 3.0 x 1018 molecules of hydrogen gas react? Slide 46 / 109 O2(g) + 2H2(g) --> 2H2O(g) 27 If 44 moles of magnesium react, how many molecules of oxygen gas would be needed? 2Mg(s) + O2(g) Slide 47 / 109 --> 2MgO(s) Slide 48 / 109 Stoichiometry Calculations with Mass Return to Table of Contents Mass Relationships in Stoichiometry Slide 49 / 109 Unlike the volume, particles, and moles of a material which are independent of the type of material present, the mass of a material is specific to each substance and therefore different. O2 gas H2 gas 1 mole 1 mole 22.4 L 22.4 L 6.02 x 1023 molecules 6.02 x 1023 molecules 32 grams 2 grams Slide 50 / 109 Stoichiometry Depending on the units, there are many ways to interpret a balanced equation! 2H2 + O2 2H2 O -- # 2 molecules H + 1 molecule O -- # 2 molecules H O 2 mol H -- # 2 mol H O 2 2 2 + 1 mol O 2 2 2 2 volumes H2 + 1 volume O2 --> 2 volumes H2O 4.0 g H 32.0 g O -- # 2 + 2 36.0 g H O 2 Slide 51 / 109 Mass Relationships in Stoichiometry Example: How many grams of hydrogen gas would need to react with 3.4 moles of oxygen gas? 2H2(g) + O2(g) --> 2H2O(g) mol O2 --> mol H2 --> g H2 3.4 mol O2 x 2 mol H2 1 mol O2 x 2 g H2 = 13.6 g H2 1 mol H2 Mass-Mass Calculations Slide 52 / 109 Starting with the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if a product) or used (if a reactant). Given aA Grams of substance A Find bB Grams of substance B Use molar mass of A Use molar mass of B Use coefficients of A and B from balanced equation Moles of substance A Moles of substance B Slide 53 / 109 Mass-Mass Calculations Example: Calculate the mass of ammonia, NH3 , produced by the reaction of 5.4 g hydrogen gas with an excess of nitrogen. N2 + 3H2 ---> Strategize!! g H2 --> mol H2 --> 2NH3 mol NH3 --> g NH3 Move for answer x 2 mol NH x 17 g NH 5.4 g H2 x 1 mol H2 3 2 g H2 3 mol H2 3 1 mol NH3 = 30.6 g NH3 28 What is the mass of sodium produced when 40 grams of sodium azide decompose? 2 NaN (s) --> 2 Na (s) + 3 N (g) 3 2 Slide 54 / 109 29 How many grams of Al2 O3 will be created from reacting 36 g of Al with a sufficient amount of O2 ? Slide 55 / 109 4 Al (s) + 3 O2 (g) --> 2 Al2O3 (s) Slide 56 / 109 30 How many grams of Mg must react in order to to create 84 g of MgO? 2 Mg (s) + O 2 (g) --> 2 MgO (s) Slide 57 / 109 Mixed Stoichiometry Problems Return to Table of Contents Slide 58 / 109 Mixed Stoichiometry Problems Generally speaking, it is easiest to convert to moles first. L of B L of A mol A g of A mol B g of B particles of A particles of B Mixed Stoichiometry Calculations Slide 59 / 109 Every type of stoichiometry calculation may be solved by following this map. (1) From left to right, we convert any "Given" substance to moles. (2) Next, using the mole ratio created with coefficients, one can calculate the moles of the "Wanted" quantity. (3) Finally, if necessary, moles can be converted to either particles, mass or volume. Mixed Stoichiometry Calculations (3) (1) representative x 1 mol G = particles of G 6.02 x 1023 x 6.02 x 1023 representative = particles of W 1 mol W (2) mass 1 mol G x = of G mass G b mol W mol W = mol G x a mol G 1 mol G volume of x 22.4 L G = G at STP x x mass W mass = 1 mol W of W 22.4 L W Volume of = 1 mol W W at STP Slide 60 / 109 31 How many L of water vapor can be produced from the combustion of 1 gram of glucose @STP? Slide 61 / 109 C H O (s) + 6 O (g) --> 6 CO (g) + 6 H O(g) 6 12 6 2 2 2 32 What mass of CaO would be required to completely react with 42 grams of H2O at STP? Slide 62 / 109 CaO (s) + H2O (l) --> Ca(OH)2 (s) 33 How many grams of iron can be extracted from 500 kg of iron ore? (Make sure you balance the equation first and remember to convert your kg --> g) Fe O (s) + 2 3 C (s) --> Fe (s) + CO (g) 2 Slide 63 / 109 34 How many grams of ammonia can be produced by reacting 10 moles of nitrogen gas with excess hydrogen gas @STP? Slide 64 / 109 N2(g) + 3H2(g) --> 2NH3(g) 35 How many L of O2 gas @STP are required to produce 90 grams of aluminum oxide? Slide 65 / 109 4Al(s) + 3O2(g) --> 2Al2O3(s) 36 How many grams of chlorine gas are needed to react with 1 mole of Sb @STP? 2 Sb + 3 Cl --> 2 SbCl 2 3 Slide 66 / 109 37 How many moles of aluminum oxide are produced when 3.580 kg of manganese dioxide are consumed? 3MnO (s) + 4 Al(s) --> 2 Al O (s) + 2 2 3 3Mn(s) 38 How many moles of manganese dioxide will be needed to react with 6 x 1025 atoms of Al? 3MnO (s) + 4 Al(s) --> 2 Al O (s) + 2 2 3 2 2 3 Slide 68 / 109 3Mn(s) 39 If 4.37 moles of Al are consumed, how many formula units of aluminum oxide would be produced? 3MnO (s) + 4 Al(s) --> 2 Al O (s) + Slide 67 / 109 3Mn(s) Slide 69 / 109 Real World Application Slide 70 / 109 The compound tristearin (C57H110O6) is a type of fat which camels store in their hump and is used to make chocolate. (yes, really) tristearin Here 2 C57 H110 O6 (s) + 163 O2 (g) --> 114 CO2 (g) + 110 H2 O(l) At STP, what volume of carbon dioxide is produced when 50 grams of tristearin is burned by the camel? Slide 71 / 109 Limiting Reactants Return to Table of Contents Concept of the Limiting Reactant In a chemical reaction, do all of the reactants turn into products? What happens when one of the reactants gets used up? If the following reaction starts with 10 moles of H2 and 20 moles of Cl2, which reactant will run out of first? H2(g) + Cl2(g) --> 2HCl(g) Slide 72 / 109 Concept of the Limiting Reactant Slide 73 / 109 The 10 moles of H2 are used to produce 20 moles of HCl. When the H2 is all used up, no more HCl can be produced. Were all the moles of Cl2 used up? How many are left over? H2(g) + Cl2(g) --> 2HCl(g) Concept of the Limiting Reactant Slide 74 / 109 No more HCl can be produced once the H2 runs out, therefore, H2 is the Limiting Reactant (limits the amount of product). Since the reaction started with more Cl2 than H2, not all of the Cl2 is used up in the reaction. Cl2 is the Excess Reactant (there's an excess amount). H2(g) + Cl2(g) --> 2HCl(g) Left-over Cl2 not used in the reaction to make HCL Limiting Reactants The limiting reactant, or limiting reagent, is the reactant present in the smallest stoichiometric amount. This is not necessarily the one with the smallest mass. The limiting reactant is the reactant you’ll run out of first, and it is the one that determines the maximum amount of product that can be made. Slide 75 / 109 Limiting Reactants Slide 76 / 109 Limiting reagent problems are worded differently because the quantities of both reactants are given. 10 moles of H2 and 20 moles of Cl2 react to produce HCl. Which quantity is the limiting reagent? It is your job to figure out which reactant is limiting because that will determine the maximum amount of product you can get, also called the maximum yield. There are a variety of methods to determine which reactant is the limiting one. Steps to Determine the Limiting Reactant Slide 77 / 109 A series of steps can be used to determine the limiting reactant in any reaction: Step 1: Convert the given quantities into moles. These are your initial amounts of each reactant. Step 2: Divide each by its stoichiometrical coefficient from the balanced chemical equation. This factors in how much is needed in the reaction. Step 3: Whichever reagent has the smallest quantity must be the limiting reactant! Slide 78 / 109 Determining the Limiting Reactant Example: When 10 grams of hydrogen react with 3.4 moles of nitrogen gas to make ammonia, which substance would be the limiting reactant? N2(g) + 3H2(g) --> 2NH3(g) Step 1: Convert all values to moles. 10 g H2 x 1 mol H2 = 5 mol H2 2 g H2 Initial Amounts = 5 mol H2, 3.4 mol N2 Determining the Limiting Reactant Slide 79 / 109 Step 2: Find the stoichiometrical equivalents of each reactant N2(g) + 3H2(g) --> 2NH3(g) Initial: 3.4 mol 5 mol Divide by coefficient: 3.4/1 5/3 Available Amounts: 3.4 mol 1.66 mol Step 3: Since there is less hydrogen gas, it will be the limiting reactant! Limiting Reactants Slide 80 / 109 Example: If 10 moles of hydrogen gas react with 7 moles of oxygen gas @STP to make water, which is the limiting reactant? 2H2(g) + O2(g) --> 2H2O(g) Step 1: Both amounts are already in moles! Step 2: 10 mol H2/2 = 5 mol H2 available 7 mol O2/1 = 7 mol O2 available Step 3: H2 gas is the limiting reactant. O2 gas will be left over and is the excess reactant. Limiting Reactants Example: Given that 2H2O(g) + O2(g) -->2H2O2(l) If 36 grams of water react with 44.8 L of oxygen gas @STP, which substance is the limiting reactant? Step 1: 36 g H2O x 1 mol = 2 mol H2O 18 g H2O Step 2: 44.8 L O2 x 1 mol = 2 mol O2 24 L O2 2 mol H2O/2 = 1 mol H2O available 2 mol O2/1 = 2 mol O2 available Step 3: Since less water is available, it is the limiting reactant. Oxygen gas is the excess reactant. Slide 81 / 109 40 In this example, the ____ is the limiting reagent. Slide 82 / 109 A Hydrogen B Oxygen C water Before reaction 10H and 7 O 2 After reaction 10 H O and 2O 2 2 2 Slide 83 / 109 41 In this example _______ is the excess reagent. A Hydrogen B Oxygen C Water Before reaction 10H and 7 O 2 2 After reaction 10 H O and 2O 2 2 42 When 33 L of nitrogen gas react with 12 grams of hydrogen gas to make ammonia @STP, the hydrogen gas will be the excess reactant. True False N2(g) + 3H2(g) --> 2NH3(g) Slide 84 / 109 43 When 120 grams of zinc react with 2.1 moles of H+ ion, the zinc will limit the reaction. True Zn(s) + 2H+(aq) --> Zn2+ + H2(g) False 44 When 120 grams of zinc react with 2.1 moles of H+ ion, the zinc will limit the reaction. True False Slide 85 / 109 Slide 86 / 109 Zn(s) + 2H+(aq) --> Zn2+ + H2(g) Real World Application 2C8H18(g) + 25O2(g) --> 16CO2(g) + 18H2O(g) In your car engine, octane is combusted with oxygen to produce carbon dioxide and water (the exhaust). The mix of octane to oxygen must be right on or the mix is too rich (too much octane) or too lean (too little octane). If 0.065 L of oxygen is being mixed with 0.0061 L of octane @ STP, calculate if the mixture is running lean or running rich? 0.065 L x 1 = 0.0029 mol O2 0.0061 L x 1 = 0.00027 mol octane move for answer 0.00027 mol/2 = 0.00014 mol octane av. 0.0029/25 = 0.000116 O2 av. Excess of octane so mixture is too RICH! Slide 87 / 109 Slide 88 / 109 Theoretical, Actual and Percent Yield Return to Table of Contents 3 Types of Yield Slide 89 / 109 Theoretical yield - the amount of product that could form during a reaction; it is calculated from a balanced chemical equation and it represents the maximum amount of product that could be formed from a given amount of reactant. Actual yield - the amount of product that forms when a reaction is carried out in the laboratory. It is measured in the lab. Why is the actual yield different from the percent yield? Percent yield - the ratio of the actual yield to the theoretical yield for a chemical reaction expressed as a percent; it is a measure of the efficiency of a reaction Theoretical Yield and % Yield Theoretical Yield: Maximum amount of product that could be made. Limited by the amount of the limiting reactant. % Yield: The ratio of actual amount produced in the laboratory to the theoretical amount that could have been produced. Expressed as: Actual Yield Theoretical Yield x 100 Slide 90 / 109 Calculating the Theoretical Yield Slide 91 / 109 Example: Find the theoretical yield (in g) of AlCl , if 27g Al and 71g Cl react. 3 2 2Al(s) + 3Cl2(g) --> 2AlCl3(s) Step 1: Determine the limiting reactant 27 g Al x 1 mol = 1 mol Al 71 g Cl2 x 1 mol = 1 mol Cl2 27 g 71 g 1 mol Al/2 = 0.50 mol Al available 1 mol Cl2/3 = 0.33 mol Cl2 available Cl2 Limits Step 2: Use INITIAL amount of Cl2 and stoichiometry to determine yield of desired product. 1 mol Cl2 x 2 mol AlCl3 x 3 mol Cl2 133 g AlCl3 = 87.8 g 1 mol AlCl3 Percent Yield Slide 92 / 109 The efficiency of a reaction can be expressed as a ratio of the actual yield to the theoretical yield. For example, a percent yield of 85% shows that the reaction conditions are more favorable than with a percent yield of only 55%. Percent yield is the ratio comparing the amount actually obtained (actual yield) to the maximum amount that was possible (theoretical yield). Percent Yield = Actual Yield x 100 Theoretical Yield Slide 93 / 109 Calculating Theoretical Yield and % Yield Example: A student burns 24 grams of methane with 30 L of oxygen gas in the laboratory and produces 12.1 L of carbon dioxide gas at STP. What is the % yield? CH4(g) + 2O2(g) --> CO2(g) + 2H2O(g) Calculating Theoretical Yield and % Yield Slide 94 / 109 Step 1: Find the LR 24 g CH4 x 1 mol = 1.5 mol CH4 30 L O2 x 1 mol = 1.33 mol O2 16 g 22.4 L 1.5 mol CH4/1 = 1.5 mol CH4 av. 1.33 mol O2/2 = 0.67 mol O2 av. O2 is LR Step 2: Find the theoretical yield (in L) using INITIAL amount of oxygen gas 1.33 mol O2 x 1 mol CO2 2 mol O2 x 22.4 L = 19.8 L CO2 1 mol Step 3: Calculate the % Yield 12.1 L CO2 Actual yield x 100 = 61.1% yield 19.8 L CO2 Theoretical yield 45 What is the theoretical yield of phosphorus pentachloride if 2 grams of phosphorus trichloride react with 1.5 moles of chlorine gas @STP? Slide 95 / 109 PCl3(g) + Cl2(g) --> PCl5(g) 46 At STP, what volume of laughing gas (dinitrogen monoxide) will be produced from 50 grams of nitrogen gas and 75 grams of oxygen gas? Remember to first write a balanced equation. Slide 96 / 109 47 How many atoms of silver will be produced when 100 grams of copper react with 200 grams of silver nitrate? Slide 97 / 109 Cu(s) + 2 AgNO3(aq) --> Cu(NO3)2(aq) + 2 Ag(s) 48 In the thermite reaction, aluminum reacts with iron (III)oxide to produce aluminum oxide and solid iron. If, when 258 grams of Al react with excess rust to produce 464 grams of pure iron, what is the % yield? (Remember, first write a balanced equation) 49 If 34 grams of ethane react with 84 L of oxygen gas to produce an actual yield in the laboratory of 2.9 moles of water vapor, what is the % yield? 2C2H6(g) + 7O2(g) --> 4CO2(g) + 6H2O(g) Slide 98 / 109 Slide 99 / 109 50 Given the equation below, how many liters of sulfur dioxide would be actually produced if 55 grams of zinc sulfide were reacted with excess oxygen @STP and produced a 75% yield? Slide 100 / 109 2ZnS(s) + 3O2(g) --> 2 SO2(g) + 2ZnO(s) Slide 101 / 109 Calculating Excess Reactants Return to Table of Contents Slide 102 / 109 Calculating Excess Reactant There will always be a certain amount of excess reactant remaining. The following steps are useful in determining how much of the excess reactant is left over. Step 1: Use the limiting reactant to determine how much of the excess reactant was required to react. Step 2: Subtract the amount of excess reactant used from the initial amount. Slide 103 / 109 Calculating Excess Reactant Example: If 6 grams of hydrogen gas react with 160 grams of oxygen gas, how much of the excess reactant remains? 2H2(g) + O2(g) --> 2H2O(g) Calculating Excess Reactant Slide 104 / 109 2H2(g) + O2(g) --> 2H2O(g) Step 1: Find the LR 6 g H2 x 1 mol H2 = 3 mol H2 160 g O2 x 1 mol = 5 mol O2 2 g H2 32 g O2 2 mol H2/2 = 1 mol H2 av. 5 mol/1 = 5 mol O2 av. H2 Limits Step 2: Use LR to find how much oxygen will be required. 3 mol H2 x 1 mol O2 = 1.5 mol O2 required 2 mol H2 Step 3: Subtract required amount from initial amount. 5 mol O2 initial - 1.5 mol O2 required = 3.5 mol Excess 51 How many grams of the excess reactant remain if 400 grams of nitrogen gas are reacted with 800 grams of oxygen gas according to the reaction below? (Don't forget to balance first!) ___N2(g) + ___O2(g) --> ___N2O5(g) Slide 105 / 109 Slide 106 / 109 52 Methanol (CH3OH) can be synthesized from carbon monoxide and hydrogen gas. If 152 kg of carbon monoxide gas is reacted with 1500 L of H2 gas @STP, how many liters of the excess reactant remain? (Remember to first write a balanced reaction!) Slide 107 / 109 Credit to Tom Greenbowe Chemical Education Group at Iowa State University Stoichiometry Practice Problem Calcium hydroxide, Ca(OH)2 , is also known as “slaked lime” and it is produced when water reacts with “quick lime,” CaO. If you start with 2.4 kg of quick lime, add excess water, and produce 2.06 kg of slaked lime, what is the percent yield of the reaction? Is this a limiting reagent problem? Is the 2.06 kg a theoretical yield or actual yield? What quantity must you solve for? Did you write a balanced equation? Slide 108 / 109 Slide 109 / 109