Stoichiometric Calculations

Transcription

1 Slide 1 / 109 Slide 2 / 109 Stoichiometric Calculations Slide 3 / 109 Slide 4 / 109 Table of Contents Stoichiometry Calculations with Moles Click on the topic to go to that section Stoichiometry Calculations with Particles and Volume Stoichiometry Calculations with Mass Mixed Stoichiometry Problems Limiting Reactants Theoretical, Actual and Percent Yield Calculating Excess Reactants Stoichiometry Calculations with Moles Return to Table of Contents Slide 5 / 109 Stoichiometry Slide 6 / 109 Stoichiometry in our daily lives The word stoichiometry is derived from two Greek words. "stoicheion" meaning element and "metron", meaning measure. 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.

2 Slide 7 / 109 Stoichiometric Calculations 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 Slide 8 / 109 Stoichiometric Calculations 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. 1. Use the equation to set up a ratio of the substances of interest. 2. Set that equal to the ratio of the known to unknown quantities of the same substances. Slide 10 / 109 Stoichiometry Calculations with Moles 2 H2 + O2 2 H2O 2 mol H 2O 2 mol H 2 n mol H 2O 2 mol H = 2O 8 mol H 2 2 mol H 2 2 H2 + O2 2 H2O 3. Solve for the unknown. 8(2) = (2) n mol H 2O n = 8 mol H 2O Slide 11 / 109 Stoichiometry Calculations with Moles Slide 12 / 109 Stoichiometry Calculations with Moles Another Example: Given the equation: 2 H2 + O2 2 H2O 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 How many moles of oxygen would be needed to react with 12 moles of 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

3 Slide 13 / 109 Stoichiometry Calculations 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 Slide 15 / 109 2Al + Fe 2O 3 Slide 14 / 109 Real World Application --> 2Fe + Al 2O 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 Slide 15 () / What is the largest number of moles of Al 2O 3 that could result from reacting 6 moles of O 2? 4 Al (s) + 3 O2 (g) - - # 2 Al2O3 (s) 1 What is the largest number of moles of Al 2O 3 that could result from reacting 6 moles of O 2? 4 Al (s) + 3 O2 (g) - - # 2 Al2O3 (s) 4 Slide 16 / 109 Slide 16 () / 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) 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) 18

4 Slide 17 / 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 () / 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) 6 Slide 18 / 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 O 2 (g)--> 2 Fe 2O 3 (s) Slide 18 () / 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 O 2 (g)--> 2 Fe 2O 3 (s) 1.8 Slide 19 / 109 Slide 19 () / How many moles of aluminium are needed to react completely with 1.2 mol FeO? 2 Al (s) + 3 FeO (s) --> 3 Fe (s) + Al 2O 3 (s) 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) + Al 2O 3 (s) 0.8

5 Slide 20 / 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 () / 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) 8 Slide 21 / How many moles (total) of calcium metal and chlorine gas are produced from the decomposition of 8 mol of calcium chloride? CaCl 2 (s) --> Ca (s) + Cl 2 (g) Slide 21 () / How many moles (total) of calcium metal and chlorine gas are produced from the decomposition of 8 mol of calcium chloride? CaCl 2 (s) --> Ca (s) + Cl 2 (g) 16 Slide 22 / 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 () / 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) 30

6 Slide 23 / How many moles of AgNO 3 could be produced from 40 moles of HNO 3? Slide 23 () / How many moles of AgNO 3 could be produced from 40 moles of HNO 3? 3Ag(s) + 4HNO 3(aq) --> 3AgNO 3(aq) + 2 H 2O(l) + NO(g) 3Ag(s) + 4HNO 3(aq) --> 3AgNO 3(aq) + 2 H 2O(l) + NO(g) 30 Slide 24 / 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 () / 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) 0.26 Slide 25 / How many moles of NO were produced if 16 moles of water were made during the reaction? 3Ag(s) + 4HNO 3(aq) --> 3AgNO 3(aq) + 2 H 2O(l) + NO(g) Slide 25 () / How many moles of NO were produced if 16 moles of water were made during the reaction? 3Ag(s) + 4HNO 3(aq) --> 3AgNO 3(aq) + 2 H 2O(l) + NO(g) 8

7 Slide 26 / How many moles of N 2H 4 are required to produce 57 moles of nitrogen gas? Slide 26 () / How many moles of N 2H 4 are required to produce 57 moles of nitrogen gas? 2 N 2H 4(l) + N 2O 4(l) ---> 3 N 2(g) + 4 H 2O(g) 2 N 2H 4(l) + N 2O 4(l) ---> 3 N 2(g) + 4 H 2O(g) 38 Slide 27 / How many moles of dinitrogen tetraoxide would be needed to produce 57 moles of nitrogen gas? Slide 27 () / How many moles of dinitrogen tetraoxide would be needed to produce 57 moles of nitrogen gas? 2 N 2H 4(l) + N 2O 4(l) ---> 3 N 2(g) + 4 H 2O(g) 2 N 2H 4(l) + N 2O 4(l) ---> 3 N 2(g) + 4 H 2O(g) 19 Slide 28 / How many moles of water are produced if 57 moles of nitrogen gas are produced? Slide 28 () / How many moles of water are produced if 57 moles of nitrogen gas are produced? 2 N 2H 4(l) + N 2O 4(l) ---> 3 N 2(g) + 4 H 2O(g) 2 N 2H 4(l) + N 2O 4(l) ---> 3 N 2(g) + 4 H 2O(g) 76

8 Slide 29 / How many total moles of gas would be produced when 5 moles of nitrogen tetrahydride reacts with excess N 2O 4? 2 N 2H 4(l) + N 2O 4(l) ---> 3 N 2(g) + 4 H 2O(g) Slide 29 () / How many total moles of gas would be produced when 5 moles of nitrogen tetrahydride reacts with excess N 2O 4? 2 N 2H 4(l) + N 2O 4(l) ---> 3 N 2(g) + 4 H 2O(g) 17.5 Slide 30 / 109 Slide 31 / 109 Stoichiometry Calculations with Particles Stoichiometry Calculations with Particles and Volume 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. Return to Table of Contents 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! Slide 32 / What is the largest number of of Li3 N formula units that could result from reacting 6 N 2 molecules? 6 Li (s) + N 2 (g) --# 2 Li 3 N (s) Slide 32 () / What is the largest number of of Li3 N formula units that could result from reacting 6 N 2 molecules? 6 Li (s) + N 2 (g) --# 2 Li 3 N (s) 6 N2 molecules x 2 Li3N formula units = 12 Formula units Li3N 1 molecule N2

9 Slide 33 / How many N2 molecules would be required to create 4 Li 3 N formula units? 6 Li (s) + N 2 (g) --# 2 Li 3 N (s) Slide 33 () / How many N2 molecules would be required to create 4 Li 3 N formula units? 6 Li (s) + N 2 (g) --# 2 Li 3 N (s) 2 Slide 34 / How many Li atoms would be required to completely react with 3 N 2 molecules? 6 Li (s) + N 2 (g) --# 2 Li 3 N (s) Slide 34 () / How many Li atoms would be required to completely react with 3 N 2 molecules? 6 Li (s) + N 2 (g) --# 2 Li 3 N (s) 18 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 H 2 (g) + O 2 (g) --> 2 H 2 O(g) Slide 36 / 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) 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!

10 Slide 36 () / 109 Slide 37 / 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!!!) CH 4 + O 2 --# CO 2 + H 2 O 12 L NO2 x 1 L O2 = 3 L of O2 4 L NO2 Slide 37 () / What volume of methane is needed to completely react with 500 ml of O 2 at STP? (Balance the equation first!!!) CH 4 + O 2 --# CO 2 + H 2 O Slide 38 / 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)? 2 H 2 (g) + O 2 (g) --# 2 H 2 O (g) 250 Slide 38 () / 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)? 2 H 2 (g) + O 2 (g) --# 2 H 2 O (g) Slide 39 / 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)? 4 NH 3 (g) + 7 O 2 (g) --# 4 NO 2 (g) + 6 H 2 O (g) 8

11 Slide 39 () / 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)? 4 NH 3 (g) + 7 O 2 (g) --# 4 NO 2 (g) + 6 H 2 O (g) 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? N 2(g) + 3H 2(g) --> 2NH 3 Slide 41 / 109 Stoichiometry with Particles and Volumes Given the following reaction, how many liters of nitrogen gas would be to produce 3 moles of ammonia? N 2(g) + 3H 2(g) --> 2NH 3 Slide 42 / 109 Stoichiometry with Particles and Volumes Example 2: How many moles of Cl 2 gas would be needed to produce 3 x formula units of NaCl given the following reaction. 2Na(s) + Cl 2(g) --> 2NaCl(s) formula units NaCl --> molecules Cl 2 --> mol Cl 2 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 3 x formula units NaCl x 1 molecule Cl 2 x 1 mol Cl 2 2 for. units NaCl 6.02 x molecules = 2.5 mol Cl 2 Slide 43 / 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 () / How many sodium atoms would be needed to react with 33.6 L of chlorine gas at STP? 2Na(s) + Cl 2(g) --> 2NaCl Note: Sodium is a solid and therefore cannot be expressed in L, so first convert the chlorine gas to moles. Note: Sodium is a solid and therefore cannot be expressed Strategize!! L Cl2 --> in n L, Cl2 --> so n Na first --> atoms convert Na the chlorine gas to moles L Cl2 x 1 mol Cl2 x 2 mol Na x 6.02 x atoms Na 22.4 L 1 mol Cl2 mol Na = 1.8 x atoms Na

12 Slide 44 / How many liters of oxygen gas would need to be combusted with excess hydrocarbon to produce 5.5 moles of 2 C57H110O6(s) O2(g) --> 114 CO2(g) 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! Slide 44 () / How many liters of oxygen gas would need to be combusted with excess hydrocarbon to produce 5.5 moles of 2 C57H110O6(s) O2(g) --> 114 CO2(g) 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! Strategize!! mol H2O --> mol O2 --> L O2 5.5 mol H2O x 163 mol O2 x 22.4 L = 183 L O2 110 mol H2O 1 mol Slide 45 / How many moles of carbon dioxide gas would be if 2.24 L of O 2 gas react? 2 C57H110O6(s) O2(g) --> 114 CO2(g) H2O(l) Slide 45 () / How many moles of carbon dioxide gas would be if 2.24 L of O 2 gas react? 2 C57H110O6(s) O2(g) --> 114 CO2(g) H2O(l) Slide 46 / How many L of water vapor would be when 3.0 x molecules of hydrogen gas react? O 2(g) + 2H 2(g) --> 2H 2O(g) Slide 46 () / How many L of water vapor would be when 3.0 x molecules of hydrogen gas react? O 2(g) + 2H 2(g) --> 2H 2O(g) 1.12 x 10 4

13 Slide 47 / If 44 moles of magnesium react, how many molecules of oxygen gas would be needed? Slide 47 () / If 44 moles of magnesium react, how many molecules of oxygen gas would be needed? 2Mg(s) + O 2(g) --> 2MgO(s) 2Mg(s) + O 2(g) --> 2MgO(s) 1.35 x Slide 48 / 109 Slide 49 / 109 Mass Relationships in Stoichiometry Stoichiometry Calculations with Mass 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 molecules 6.02 x molecules 32 grams 2 grams Return to Table of Contents Slide 50 / 109 Stoichiometry 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 volume O 2 --> 2 volumes H 2O 4.0 g H g O2 -- # 36.0 g H2O 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? 2H 2(g) + O 2(g) --> 2H 2O(g) mol O 2 --> mol H 2 --> g H 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

14 Slide 52 / 109 Mass-Mass Calculations 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 Slide 53 / 109 Mass-Mass Calculations 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 Use molar mass of A Use molar mass of B Move for answer 5.4 g H 2 x 1 mol H 2 x 2 mol NH 3 x 17 g NH 3 Moles of substance A Use coefficients of A and B from balanced equation Moles of substance B 2 g H 2 3 mol H 2 1 mol NH 3 = 30.6 g NH 3 Slide 54 / 109 Slide 54 () / What is the mass of sodium produced when 40 grams of sodium azide decompose? 28 What is the mass of sodium produced when 40 grams of sodium azide decompose? 2 NaN3 (s) --> 2 Na (s) + 3 N2 (g) 2 NaN3 (s) --> 2 Na (s) + 3 N2 (g) 14.1 Slide 55 / 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 () / 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) 68

15 30 Slide 56 / 109 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) 30 Slide 56 () / 109 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) 50.4 Slide 57 / 109 Slide 58 / 109 Mixed Stoichiometry Problems Mixed Stoichiometry Problems 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 Return to Table of Contents particles of A particles of B Slide 59 / 109 Mixed Stoichiometry Calculations Slide 60 / 109 Mixed Stoichiometry Calculations Every type of stoichiometry calculation may be solved by following this map. (1) From left to right, we convert any "Given" substance to moles. (1) representative particles of G x 1 mol G 6.02 x = x 6.02 x mol W (3) representative = particles of W (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. 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

16 Slide 61 / How many L of water vapor can be produced from the combustion of 1 gram of C6H12O6(s) + 6 O2(g) --> 6 CO2(g) + 6 H2O(g) Slide 61 () / How many L of water vapor can be produced from the combustion of 1 gram of C6H12O6(s) + 6 O2(g) --> 6 CO2(g) + 6 H2O(g) 0.75 Slide 62 / What mass of CaO would be required to completely react with 42 grams of H 2O at STP? CaO (s) + H 2O (l) --> Ca(OH) 2 (s) Slide 62 () / What mass of CaO would be required to completely react with 42 grams of H 2O at STP? CaO (s) + H 2O (l) --> Ca(OH) 2 (s) 131 Slide 63 / 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) Fe2O3 (s) + C (s) --> Fe (s) + CO2 (g) Slide 63 () / 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) Fe2O3 (s) + C (s) --> Fe (s) + CO2 (g) 350,000

17 Slide 64 / How many grams of ammonia can be produced by reacting 10 moles of nitrogen gas with excess hydrogen N 2 (g) + 3H 2 (g) --> 2NH 3 (g) Slide 64 () / How many grams of ammonia can be produced by reacting 10 moles of nitrogen gas with excess hydrogen N 2 (g) + 3H 2 (g) --> 2NH 3 (g) 3.40 Slide 65 / How many L of O 2 are required to produce 90 grams of aluminum oxide? 4Al(s) + 3O 2(g) --> 2Al 2O 3(s) Slide 65 () / How many L of O 2 are required to produce 90 grams of aluminum oxide? 4Al(s) + 3O 2(g) --> 2Al 2O 3(s) 29.6 Slide 66 / How many grams of chlorine gas are needed to react with 1 mole of 2 Sb + 3 Cl2 --> 2 SbCl3 Slide 66 () / How many grams of chlorine gas are needed to react with 1 mole of 2 Sb + 3 Cl2 --> 2 SbCl

18 Slide 67 / How many moles of aluminum oxide are produced when kg of manganese dioxide are consumed? Slide 67 () / How many moles of aluminum oxide are produced when kg of manganese dioxide are consumed? 3MnO2(s) + 4 Al(s) --> 2 Al2 O3(s) + 3Mn(s) 3MnO2(s) + 4 Al(s) --> 2 Al2 O3(s) + 3Mn(s) 27.4 Slide 68 / How many moles of manganese dioxide will be needed to react with 6 x atoms of Al? Slide 68 () / How many moles of manganese dioxide will be needed to react with 6 x atoms of Al? 3MnO2(s) + 4 Al(s) --> 2 Al2 O3(s) + 3Mn(s) 3MnO2(s) + 4 Al(s) --> 2 Al2 O3(s) + 3Mn(s) 75 Slide 69 / If 4.37 moles of Al are consumed, how many formula units of aluminum oxide would be produced? Slide 69 () / If 4.37 moles of Al are consumed, how many formula units of aluminum oxide would be produced? 3MnO2(s) + 4 Al(s) --> 2 Al2 O3(s) + 3Mn(s) 3MnO2(s) + 4 Al(s) --> 2 Al2 O3(s) + 3Mn(s) 1.3 x 10 24

19 Slide 70 / 109 Slide 71 / 109 Real World Application 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) Limiting Reactants Here tristearin 2 C 57 H 110 O 6 (s) O 2 (g) --> 114 CO 2 (g) H 2 O(l) At STP, what volume of carbon dioxide is produced when 50 grams of tristearin is burned by the camel? Return to Table of Contents Slide 72 / 109 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 H 2 and 20 moles of Cl 2, which reactant will run out of first? H 2(g) + Cl 2(g) --> 2HCl(g) Slide 73 / 109 Concept of the Limiting Reactant 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) Slide 74 / 109 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 75 / 109 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. Left-over Cl 2 not used in the reaction to make HCL

20 Slide 76 / 109 Limiting Reactants Slide 77 / 109 Steps to Determine the Limiting Reactant 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. 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 Slide 79 / 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? Step 1: Convert all values to moles. N 2(g) + 3H 2(g) --> 2NH 3(g) 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 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! Slide 80 / 109 Limiting Reactants Example: If 10 moles of hydrogen gas react with 7 moles of oxygen 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. Example: Given that Slide 81 / 109 Limiting Reactants 2H 2O(g) + O 2(g) -->2H 2O 2(l) If 36 grams of water react with 44.8 L of oxygen 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.

21 Slide 82 / In this example, the is the limiting reagent. A Hydrogen B Oxygen C water Before reaction After reaction 10H2 and 7 O2 10 H2O and 2O2 Slide 82 () / In this example, the is the limiting reagent. A Hydrogen B Oxygen C water Before reaction 10H2 and 7 O2 A After reaction 10 H2O and 2O2 Slide 83 / 109 Slide 83 () / In this example is the excess reagent. 41 In this example is the excess reagent. A Hydrogen A Hydrogen B Oxygen B Oxygen C Water Before reaction 10H2 and 7 O2 After reaction 10 H2O and 2O2 C Water Before reaction 10H2 and 7 O2 After reaction B 10 H2O and 2O2 Slide 84 / When 33 L of nitrogen gas react with 12 grams of hydrogen gas to make the hydrogen gas will be the excess reactant. Slide 84 () / When 33 L of nitrogen gas react with 12 grams of hydrogen gas to make the hydrogen gas will be the excess reactant. True N 2(g) + 3H 2(g) --> 2NH 3(g) True N 2(g) + 3H 2(g) --> 2NH 3(g) False False True

22 Slide 85 / 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 () / 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) False Slide 86 / 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 () / 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) False Slide 87 / 109 Slide 88 / 109 Real World Application 2C 8H 18(g) + 25O 2(g) --> 16CO 2(g) + 18H 2O(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 L of oxygen is being mixed with L of STP, calculate if the mixture is running lean or running rich? Theoretical, Actual and Percent Yield L x 1 = mol O L x 1 = mol octane move for answer /25 = O 2 av mol/2 = mol octane av. Excess of octane so mixture is too RICH! Return to Table of Contents

23 Slide 89 / Types of Yield 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: % Yield: Slide 90 / 109 Theoretical Yield and % Yield Maximum amount of product that could be made. Limited by the amount of the limiting reactant. 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 Example: Slide 91 / 109 Calculating the Theoretical Yield 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 Slide 92 / 109 Percent Yield 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 3 mol Cl 2 1 mol AlCl 3 Slide 93 / 109 Calculating Theoretical Yield and % Yield Slide 94 / 109 Calculating Theoretical Yield and % Yield Step 1: Find the LR 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) 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 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

24 Slide 95 / What is the theoretical yield of phosphorus pentachloride if 2 grams of phosphorus trichloride react with 1.5 moles of chlorine PCl 3 (g) + Cl 2 (g) --> PCl 5 (g) Slide 95 () / What is the theoretical yield of phosphorus pentachloride if 2 grams of phosphorus trichloride react with 1.5 moles of chlorine PCl 3 (g) + Cl 2 (g) --> PCl 5 (g) 3.04 Slide 96 / 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 () / 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 97 / How many atoms of silver will be produced when 100 grams of copper react with 200 grams of silver nitrate? Cu(s) + 2 AgNO 3(aq) --> Cu(NO 3) 2(aq) + 2 Ag(s) Slide 97 () / How many atoms of silver will be produced when 100 grams of copper react with 200 grams of silver nitrate? Cu(s) + 2 AgNO 3(aq) --> Cu(NO 3) 2(aq) + 2 Ag(s) 7.08 x 10 23

25 Slide 98 / 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 () / 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) 87% Slide 99 / 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 () / 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) 91% Slide 100 / Given the equation below, how many liters of sulfur dioxide would be actually produced if 55 grams of zinc sulfide were reacted with excess and produced a 75% yield? 2ZnS(s) + 3O 2(g) --> 2 SO 2(g) + 2ZnO(s) Slide 100 () / Given the equation below, how many liters of sulfur dioxide would be actually produced if 55 grams of zinc sulfide were reacted with excess and produced a 75% yield? 2ZnS(s) + 3O 2(g) --> 2 SO 2(g) + 2ZnO(s) 9.5

26 Slide 101 / 109 Slide 102 / 109 Calculating Excess Reactant Calculating Excess Reactants 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. Return to Table of Contents 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? 2H 2(g) + O 2(g) --> 2H 2O(g) Slide 104 / 109 Calculating Excess Reactant 2H 2(g) + O 2(g) --> 2H 2O(g) Step 1: Find the LR 6 g H 2 x 1 mol H 2 = 3 mol H 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 mol O 2 required = 3.5 mol Excess Slide 105 / 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 () / 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) 120.4

27 Slide 106 / 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 how many liters of the excess reactant remain? (Remember to first write a balanced reaction!) Slide 106 () / 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 how many liters of the excess reactant remain? (Remember to first write a balanced reaction!) 120,850 Slide 107 / 109 Slide 108 / 109 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? Credit to Tom Greenbowe Chemical Education Group at Iowa State University Slide 109 / 109