Stoichiometric Calculations - PDF Free Download

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 Slide 6 / 109 Airbags save thousands of lives every year. When a collision happens, the following reaction occurs. 2NaN 3(s) --> 2Na(s) + 3N 2(g) NaN3 capsule In order to properly inflate, roughly 50 L of nitrogen gas must be produced. Engineers have determined, using stoichiometry, that about 96 grams of NaN 3 are needed to react in each airbag capsule to produce enough nitrogen gas.

Stoichiometric Calculations Slide 7 / 109 The coefficients in the balanced equation give the ratio of moles of reactants and products. 2 H 2 + O 2 --# 2 H 2 O can be read as: 2 moles of H2 plus 1 mole of O2 yields 2 moles of H2O giving ratios of: 2 mol H 2 2 mol H 2 1 mol O 2 1 mol O 2 2 mol H 2O 2 mol H 2O Stoichiometric Calculations Slide 8 / 109 A balanced chemical equation is needed to perform any stoichiometric calculations. N2 + 3H2 ---> 2NH3 1 mol N2 3 mol H2 1 mol N2 3 mol H2 2 mol NH3 2 mol NH3 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 H2O that could be created from reacting 8 moles of H2. 2 H2 + O2 2 H2O

Stoichiometry Calculations with Moles Slide 10 / 109 1. Use the equation to set up a ratio of the substances of interest. 2 H2 + O2 2 H2O 2 mol H 2O 2 mol H 2 2. Set that equal to the ratio of the known to unknown quantities of the same substances. n mol H 2O 2 mol H = 2O 8 mol H 2 2 mol H 2 3. Solve for the unknown. 8(2) = (2) n mol H 2O n = 8 mol H 2O Slide 11 / 109 Stoichiometry Calculations with Moles Another Example: Given the equation: 2 H2 + O2 2 H2O How many moles of oxygen would be needed to react with 12 moles of H 2? Stoichiometry Calculations with Moles Slide 12 / 109 2 H2 + O2 - -# 2 H2O 1. Use the formula to set up a ratio of the substances of interest. 1 mol O 2 2 mol H 2 2. Set that equal to the ratio of the known to unknown quantities of the same substances. n mol O 2 1 mol O 2 = 12 mol H 2 2 mol H 2 3. Solve for the unknown by cross multiplying. 12(1) = (2)n O 2 = 6 n O 2

Stoichiometry Calculations Slide 13 / 109 2 H 2 + O 2 -# 2 H 2 O Given 12 moles of H 2, how much O 2 and H 2O would be needed or produced? Using the ratios, you would need... 1/2 as much O 2 as H 2 (ratio is 1 mol O 2/2 mol H 2) 12 mol H2 x 1 mol O 2 = 6 mol O 2 needed 2 mol H 2 Using the ratios, you would produce... An equal amount of H 2O as H 2 used (ratio is 2 mol H 2O/2 mol H 2) 12 mol H 2 x 2 mol H 2O = 12 mol H 2O 2 mol H 2 Real World Application Slide 14 / 109 2Al + Fe 2O 3 --> 2Fe + Al 2O 3 + 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 Al 2O 3? move for answer 7.8 mol Al 2O 3 x 2 mol Al = 15.6 mol Al 1 mol Al 2O 3 1 What is the largest number of moles of Al 2O 3 that could result from reacting 6 moles of O 2? Slide 15 / 109 4 Al (s) + 3 O2 (g) - - # 2 Al2O3 (s)

Slide 16 / 109 2 How many moles of O 2 would be required to create 12 moles of Al 2 O 3? 4 Al (s) + 3 O 2 (g) - - # 2 Al 2O 3 (s) 3 How many moles of O 2 would be required to completely react with 8 moles of Al? 4 Al (s) + 3 O 2 (g) - - # 2 Al 2O 3 (s) Slide 17 / 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? Slide 18 / 109 4 Fe (s) + 3 O 2 (g)--> 2 Fe 2O 3 (s)

5 How many moles of aluminium are needed to react completely with 1.2 mol FeO? Slide 19 / 109 2 Al (s) + 3 FeO (s) --> 3 Fe (s) + Al 2O 3 (s) 6 How many moles of calcium metal are produced from the decomposition of 8 mol of calcium chloride? CaCl 2 (s) --> Ca (s) + Cl 2 (g) Slide 20 / 109 7 How many moles (total) of calcium metal and chlorine gas are produced from the decomposition of 8 mol of calcium chloride? Slide 21 / 109 CaCl 2 (s) --> Ca (s) + Cl 2 (g)

8 How many moles of Ag are needed to react with 40 moles of HNO 3? 3 Ag(s) + 4 HNO 3(aq) --> 3 AgNO 3(aq) + 2 H 2O(l) + NO(g) Slide 22 / 109 9 How many moles of AgNO 3 could be produced from 40 moles of HNO 3? Slide 23 / 109 3Ag(s) + 4HNO 3(aq) --> 3AgNO 3(aq) + 2 H 2O(l) + NO(g) 10 How many moles of water would be produced when 0.4 moles of Ag react with an excess amount of HNO 3? 3Ag(s) + 4HNO 3(aq) --> 3AgNO 3(aq) + 2 H 2O(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) + 4HNO 3(aq) --> 3AgNO 3(aq) + 2 H 2O(l) + NO(g) 12 How many moles of N 2H 4 are required to produce 57 moles of nitrogen gas? Slide 26 / 109 2 N 2H 4(l) + N 2O 4(l) ---> 3 N 2(g) + 4 H 2O(g) 13 How many moles of dinitrogen tetraoxide would be needed to produce 57 moles of nitrogen gas? Slide 27 / 109 2 N 2H 4(l) + N 2O 4(l) ---> 3 N 2(g) + 4 H 2O(g)

14 How many moles of water are produced if 57 moles of nitrogen gas are produced? Slide 28 / 109 2 N 2H 4(l) + N 2O 4(l) ---> 3 N 2(g) + 4 H 2O(g) 15 How many total moles of gas would be produced when 5 moles of nitrogen tetrahydride reacts with excess N 2O 4? Slide 29 / 109 2 N 2H 4(l) + N 2O 4(l) ---> 3 N 2(g) + 4 H 2O(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 H 2 + O 2 --> 2 H 2 O can be read as: 2 molecules of H 2 plus 1 molecule of O 2 yields 2 molecules of H 2 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 N 2 molecules? Slide 32 / 109 6 Li (s) + N 2 (g) --# 2 Li 3 N (s) 17 How many N2 molecules would be required to create 4 Li 3 N formula units? Slide 33 / 109 6 Li (s) + N 2 (g) --# 2 Li 3 N (s)

18 How many Li atoms would be required to completely react with 3 N 2 molecules? Slide 34 / 109 6 Li (s) + N 2 (g) --# 2 Li 3 N (s) Stoichiometry Calculations with Volumes Slide 35 / 109 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 H 2 (g) + O 2 (g) --> 2 H 2 O(g) can be read as: 2 volumes of H 2 plus 1 volume of O 2 yields 2 volumes of H 2 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 NO 2 are formed? (STP) 2Pb(NO 3 ) 2 (s) --> 2PbO (s) +4NO 2 (g) + O 2 (g)

20 What volume of methane is needed to completely react with 500 ml of O 2 at STP? (Balance the equation first!!!) Slide 37 / 109 CH 4 + O 2 --# CO 2 + H 2 O 21 How many liters of H2 O (g) will be created from reacting 8.0 L of H 2 (g) with a sufficient amount of O 2 (g)? Slide 38 / 109 2 H 2 (g) + O 2 (g) --# 2 H 2 O (g) 22 How many liters of NO2 (g) will be created from reacting 36 L of O 2 (g) with a sufficient amount of NH 3 (g)? Slide 39 / 109 4 NH 3 (g) + 7 O 2 (g) --# 4 NO 2 (g) + 6 H 2 O (g)

Stoichiometry with Particles and Volumes Slide 40 / 109 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? N 2(g) + 3H 2(g) --> 2NH 3 Stoichiometry with Particles and Volumes Slide 41 / 109 Given the following reaction, how many liters of nitrogen gas would be needed @STP to produce 3 moles of ammonia? N 2(g) + 3H 2(g) --> 2NH 3 mol NH 3 --> mol N 2 --> L N 2 3 mol NH 3 x 1 mol N 2 x 22.4 L = 33.6 L N 2 2 mol NH 3 1 mol Stoichiometry with Particles and Volumes Slide 42 / 109 Example 2: How many moles of Cl 2 gas would be needed to produce 3 x 10 24 formula units of NaCl given the following reaction. 2Na(s) + Cl 2(g) --> 2NaCl(s) formula units NaCl --> molecules Cl 2 --> mol Cl 2 3 x 10 24 formula units NaCl x 1 molecule Cl 2 x 1 mol Cl 2 2 for. units NaCl 6.02 x 10 23 molecules = 2.5 mol Cl 2

23 How many sodium atoms would be needed to react with 33.6 L of chlorine gas at STP? 2Na(s) + Cl 2(g) --> 2NaCl Slide 43 / 109 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 C57H110O6(s) + 163 O2(g) --> 114 CO2(g) + 110 H2O(l) 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 O 2 gas react? Slide 45 / 109 2 C57H110O6(s) + 163 O2(g) --> 114 CO2(g) + 110 H2O(l)

26 How many L of water vapor would be produced @STP when 3.0 x 10 18 molecules of hydrogen gas react? O 2(g) + 2H 2(g) --> 2H 2O(g) Slide 46 / 109 27 If 44 moles of magnesium react, how many molecules of oxygen gas would be needed? Slide 47 / 109 2Mg(s) + O 2(g) --> 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. O 2 gas H 2 gas 1 mole 1 mole 22.4 L 22.4 L 6.02 x 10 23 molecules 6.02 x 10 23 molecules 32 grams 2 grams Stoichiometry Slide 50 / 109 Depending on the units, there are many ways to interpret a balanced equation! 2H 2 + O 2 -- # 2H 2 O 2 molecules H2 + 1 molecule O2 -- # 2 molecules H2O 2 mol H2 + 1 mol O2 -- # 2 mol H2O 2 volumes H 2 + 1 volume O 2 --> 2 volumes H 2O 4.0 g H2 + 32.0 g O2 -- # 36.0 g H2O Mass Relationships in Stoichiometry Slide 51 / 109 Example: How many grams of hydrogen gas would need to react with 3.4 moles of oxygen gas? 2H 2(g) + O 2(g) --> 2H 2O(g) mol O 2 --> mol H 2 --> g H 2 3.4 mol O 2 x 2 mol H 2 x 2 g H 2 = 13.6 g H 2 1 mol O 2 1 mol H 2

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 Grams of substance A aa bb Find Grams of substance B Use molar mass of A Use molar mass of B Moles of substance A Use coefficients of A and B from balanced equation Moles of substance B Mass-Mass Calculations Slide 53 / 109 Example: Calculate the mass of ammonia, NH 3, produced by the reaction of 5.4 g hydrogen gas with an excess of nitrogen. N 2 + 3H 2 ---> 2NH 3 Strategize!! g H 2 --> mol H 2 --> mol NH 3 --> g NH 3 Move for answer 5.4 g H 2 x 1 mol H 2 x 2 mol NH 3 x 17 g NH 3 2 g H 2 3 mol H 2 1 mol NH 3 = 30.6 g NH 3 28 What is the mass of sodium produced when 40 grams of sodium azide decompose? Slide 54 / 109 2 NaN3 (s) --> 2 Na (s) + 3 N2 (g)

29 How many grams of Al2 O 3 will be created from reacting 36 g of Al with a sufficient amount of O 2? 4 Al (s) + 3 O 2 (g) --> 2 Al 2O 3 (s) Slide 55 / 109 30 How many grams of Mg must react in order to to create 84 g of MgO? Slide 56 / 109 2 Mg (s) + O 2 (g) --> 2 MgO (s) Slide 57 / 109 Mixed Stoichiometry Problems Return to Table of Contents

Mixed Stoichiometry Problems Slide 58 / 109 Generally speaking, it is easiest to convert to moles first. L of A L of B g of A mol 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 Slide 60 / 109 (1) representative particles of G x 1 mol G 6.02 x 10 23 = x 6.02 x 10 23 1 mol W (3) representative = particles of W mass of G x 1 mol G mass G = (2) mol G x b mol W = a mol G mol W x mass W 1 mol W = mass of W volume of G at STP x 1 mol G 22.4 L G = x 22.4 L W 1 mol W = Volume of W at STP

31 How many L of water vapor can be produced from the combustion of 1 gram of glucose @STP? C6H12O6(s) + 6 O2(g) --> 6 CO2(g) + 6 H2O(g) Slide 61 / 109 32 What mass of CaO would be required to completely react with 42 grams of H 2O at STP? Slide 62 / 109 CaO (s) + H 2O (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) Slide 63 / 109 Fe2O3 (s) + C (s) --> Fe (s) + CO2 (g)

34 How many grams of ammonia can be produced by reacting 10 moles of nitrogen gas with excess hydrogen gas @STP? N 2 (g) + 3H 2 (g) --> 2NH 3 (g) Slide 64 / 109 35 How many L of O 2 gas @STP are required to produce 90 grams of aluminum oxide? 4Al(s) + 3O 2(g) --> 2Al 2O 3(s) Slide 65 / 109 36 How many grams of chlorine gas are needed to react with 1 mole of Sb @STP? Slide 66 / 109 2 Sb + 3 Cl2 --> 2 SbCl3

37 How many moles of aluminum oxide are produced when 3.580 kg of manganese dioxide are consumed? Slide 67 / 109 3MnO2(s) + 4 Al(s) --> 2 Al2 O3(s) + 3Mn(s) 38 How many moles of manganese dioxide will be needed to react with 6 x 10 25 atoms of Al? Slide 68 / 109 3MnO2(s) + 4 Al(s) --> 2 Al2 O3(s) + 3Mn(s) 39 If 4.37 moles of Al are consumed, how many formula units of aluminum oxide would be produced? Slide 69 / 109 3MnO2(s) + 4 Al(s) --> 2 Al2 O3(s) + 3Mn(s)

Real World Application Slide 70 / 109 The compound tristearin (C 57H 110O 6) is a type of fat which camels store in their hump and is used to make chocolate. (yes, really) Here tristearin 2 C 57 H 110 O 6 (s) + 163 O 2 (g) --> 114 CO 2 (g) + 110 H 2 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 Slide 72 / 109 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 H 2 and 20 moles of Cl 2, which reactant will run out of first? H 2(g) + Cl 2(g) --> 2HCl(g)

Concept of the Limiting Reactant Slide 73 / 109 The 10 moles of H 2 are used to produce 20 moles of HCl. When the H 2 is all used up, no more HCl can be produced. Were all the moles of Cl 2 used up? How many are left over? H 2(g) + Cl 2(g) --> 2HCl(g) Concept of the Limiting Reactant No more HCl can be produced once the H 2 runs out, therefore, H 2 is the Limiting Reactant (limits the amount of product). Since the reaction started with more Cl 2 than H 2, not all of the Cl 2 is used up in the reaction. Cl 2 is the Excess Reactant (there's an excess amount). H 2(g) + Cl 2(g) --> 2HCl(g) Slide 74 / 109 Left-over Cl 2 not used in the reaction to make HCL Limiting Reactants Slide 75 / 109 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.

Limiting Reactants Slide 76 / 109 Limiting reagent problems are worded differently because the quantities of both reactants are given. 10 moles of H 2 and 20 moles of Cl 2 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! Determining the Limiting Reactant Slide 78 / 109 Example: When 10 grams of hydrogen react with 3.4 moles of nitrogen gas to make ammonia, which substance would be the limiting reactant? N 2(g) + 3H 2(g) --> 2NH 3(g) Step 1: Convert all values to moles. 10 g H 2 x 1 mol H 2 = 5 mol H 2 2 g H 2 Initial Amounts = 5 mol H 2, 3.4 mol N 2

Determining the Limiting Reactant Slide 79 / 109 Step 2: Find the stoichiometrical equivalents of each reactant N 2(g) + 3H 2(g) --> 2NH 3(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? Step 1: 2H 2(g) + O 2(g) --> 2H 2O(g) Both amounts are already in moles! Step 2: 10 mol H 2/2 = 5 mol H 2 available 7 mol O 2/1 = 7 mol O 2 available Step 3: H 2 gas is the limiting reactant. O 2 gas will be left over and is the excess reactant. Limiting Reactants Slide 81 / 109 Example: Given that 2H 2O(g) + O 2(g) -->2H 2O 2(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 H 2O x 1 mol = 2 mol H 2O 44.8 L O 2 x 1 mol = 2 mol O 2 24 L O 2 Step 2: 18 g H 2O 2 mol H 2O/2 = 1 mol H 2O available 2 mol O 2/1 = 2 mol O 2 available Step 3: Since less water is available, it is the limiting reactant. Oxygen gas is the excess reactant.

40 In this example, the is the limiting reagent. Slide 82 / 109 A B C Hydrogen Oxygen water Before reaction 10H2 and 7 O2 After reaction 10 H2O and 2O2 41 In this example is the excess reagent. Slide 83 / 109 A B Hydrogen Oxygen C Water Before reaction 10H2 and 7 O2 After reaction 10 H2O and 2O2 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. Slide 84 / 109 True False N 2(g) + 3H 2(g) --> 2NH 3(g)

43 When 120 grams of zinc react with 2.1 moles of H+ ion, the zinc will limit the reaction. True False Zn(s) + 2H+(aq) --> Zn 2+ + H 2(g) Slide 85 / 109 44 When 120 grams of zinc react with 2.1 moles of H+ ion, the zinc will limit the reaction. True False Zn(s) + 2H+(aq) --> Zn 2+ + H 2(g) Slide 86 / 109 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). Real World Application 2C 8H 18(g) + 25O 2(g) --> 16CO 2(g) + 18H 2O(g) Slide 87 / 109 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 O 2 0.0061 L x 1 = 0.00027 mol octane move for answer 0.0029/25 = 0.000116 O 2 av. 0.00027 mol/2 = 0.00014 mol octane av. Excess of octane so mixture is too RICH!

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 Slide 90 / 109 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 x 100 Theoretical Yield

Calculating the Theoretical Yield Slide 91 / 109 Example: Find the theoretical yield (in g) of AlCl3, if 27g Al and 71g Cl2 react. 2Al(s) + 3Cl 2(g) --> 2AlCl 3(s) Step 1: Determine the limiting reactant 27 g Al x 1 mol = 1 mol Al 71 g Cl2 x 1 mol = 1 mol Cl 2 27 g 71 g 1 mol Al/2 = 0.50 mol Al available 1 mol Cl 2/3 = 0.33 mol Cl 2 available Cl 2 Limits Step 2: Use INITIAL amount of Cl 2 and stoichiometry to determine yield of desired product. 1 mol Cl 2 x 2 mol AlCl 3 x 133 g AlCl 3 = 87.8 g 3 mol Cl 2 1 mol AlCl 3 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 Calculating Theoretical Yield and % Yield Slide 93 / 109 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? CH 4(g) + 2O 2(g) --> CO 2(g) + 2H 2O(g)

Calculating Theoretical Yield and % Yield Slide 94 / 109 Step 1: Find the LR 24 g CH 4 x 1 mol = 1.5 mol CH 4 30 L O2 x 1 mol = 1.33 mol O 2 16 g 22.4 L 1.5 mol CH 4/1 = 1.5 mol CH 4 av. 1.33 mol O 2/2 = 0.67 mol O 2 av. O 2 is LR Step 2: Find the theoretical yield (in L) using INITIAL amount of oxygen gas 1.33 mol O 2 x 1 mol CO 2 x 22.4 L = 19.8 L CO 2 2 mol O 2 1 mol Step 3: Calculate the % Yield 12.1 L CO 2 Actual yield x 100 = 61.1% yield 19.8 L CO 2 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? PCl 3 (g) + Cl 2 (g) --> PCl 5 (g) Slide 95 / 109 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 AgNO 3(aq) --> Cu(NO 3) 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) Slide 98 / 109 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? 2C 2H 6(g) + 7O 2(g) --> 4CO 2(g) + 6H 2O(g) 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? 2ZnS(s) + 3O 2(g) --> 2 SO 2(g) + 2ZnO(s) Slide 100 / 109 Slide 101 / 109 Calculating Excess Reactants Return to Table of Contents Calculating Excess Reactant Slide 102 / 109 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.

Calculating Excess Reactant Slide 103 / 109 Example: If 6 grams of hydrogen gas react with 160 grams of oxygen gas, how much of the excess reactant remains? 2H 2(g) + O 2(g) --> 2H 2O(g) Calculating Excess Reactant 2H 2(g) + O 2(g) --> 2H 2O(g) Slide 104 / 109 Step 1: Find the LR 6 g H 2 x 1 mol H 2 = 3 mol H 2 160 g O 2 x 1 mol = 5 mol O 2 2 g H 2 32 g O 2 2 mol H 2/2 = 1 mol H 2 av. 5 mol/1 = 5 mol O 2 av. H 2 Limits Step 2: Use LR to find how much oxygen will be required. 3 mol H 2 x 1 mol O 2 = 1.5 mol O 2 required 2 mol H 2 Step 3: Subtract required amount from initial amount. 5 mol O 2 initial - 1.5 mol O 2 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!) Slide 105 / 109 N2(g) + O2(g) --> N2O5(g)

52 Methanol (CH 3OH) 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 106 / 109 Slide 107 / 109 Credit to Tom Greenbowe Chemical Education Group at Iowa State University Stoichiometry Practice Problem Slide 108 / 109 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 109 / 109