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Slide 1 / 139 Stoichiometric Calculations Stoichiometry Slide 2 / 139 Stoichiometry is the quantitative determination of reactants or products using a balanced chemical equation. We can interpret a balanced equation in several ways. The most common way is to interpret it as indicating the number of moles of each substance. For instance, this equation, 3 H2 + N2 ® 2 NH3 can be read as: 3 moles of hydrogen gas reacts with 1 mole of nitrogen gas to yield 2 moles of ammonia, or 3 mol H2 + 1 mol N2 to yield 2 mol NH3 The coefficients in the balanced equation give the ratio of moles of reactants and products. Therefore, a balanced chemical equation is needed to perform any stoichiometric calculation. Stoichiometric Calculations with Moles Since the coefficients in the balanced equation represent moles of substances, then we can write "mole ratios" which show relative amounts of reactants and products. Here is an example showing how to write "mole ratios" : Consider the reaction that produces ammonia, NH3. N2 + 3 H2 ® 2 NH3 Here are three different mole ratios based on this equation. 1 mol N2 3 mol H2 3 mol H2 2 mol NH3 1 mol N2 2 mol NH3 Slide 3 / 139 Slide 4 / 139 Stoichiometric Calculations with Moles 1 mol N2 3 mol H2 2 mol NH3 1 mol N2 3 mol H2 2 mol NH3 The above ratios can be used to determine the quantity of any reactant or product. N2 + 3 H2 ® 2 NH3 For every 1 mol of N2, · you would need 3 mol of H2 to completely react with it, and · you would produce 2 mol of NH3 Stoichiometric Calculations with Moles N2 + 3 H2 Slide 5 / 139 ® 2 NH3 In fact, there are three more mole ratios that can be written simply by taking the reciprocal of the first three. 1 mol N2 3 mol H2 1 mol N2 3 mol H2 2 mol NH3 2 mol NH3 3 mol H2 2 mol NH3 2 mol NH3 1 mol N2 3 mol H2 1 mol N2 It is not important to write down all of the possible mole ratios for a given reaction. But you do need to know how to create a mole ratio for a specific problem. Slide 6 / 139 Stoichiometric Calculations with Moles N2 + 3 H2 ® 2 NH3 Sample Problem #1 How many moles of NH3 can be made by reacting 4 moles of N2? Stoichiometric Calculations with Moles Slide 7 / 139 Sample Problem #1 How many moles of NH3 can be made by reacting 4 moles of N2? Sample Problem #1 - Solution · In the problem, circle the "given" quantity and underline the "wanted" quantity. · Write the "given" quantity. 4 moles of N2 · Select the mole ratio that has the "given" unit in the denominator. 3 mol H2 2 mol NH3 1 mol N2 1 mol N2 Choose the mole ratio that has the "wanted" unit in the numerator. 3 mol H2 2 mol NH3 1 mol N2 1 mol N2 Stoichiometric Calculations with Moles Slide 8 / 139 Sample Problem #1 How many moles of NH3 can be made by reacting 4 moles of N2? Sample Problem #1 - Solution (con't) We are now ready to set up our problem: 4 moles of N2 1 x 2 mol NH3 1 mol N2 Just as you multiply fractions, multiply the numbers across the top and divide by any numbers in the denominators. 4 moles of N2 1 x 2 mol NH3 1 mol N2 = 8 mol NH3 Slide 9 / 139 Stoichiometry Calculations with Moles The same approach can tell you how many moles of one reactant is needed to completely react with another reactant. N2 + 3 H2 ® 2 NH3 Sample Problem #2 How many moles of H2 will be needed to react with 15 mol N2? 1 mol N2 3 mol H2 3 mol H2 1 mol N2 2 mol NH3 2 mol NH3 3 mol H2 2 mol NH3 2 mol NH3 1 mol N2 3 mol H2 1 mol N2 Stoichiometric Calculations with Moles Slide 10 / 139 Sample Problem #2- Solution How many moles of H2 will be needed to react with 15 mol N2? 15 moles of N2 x 1 3 mol H2 = 45 mol H2 1 mol N2 We choose the mole ratio that has mol N2 in the denominator and mol H2 in the numerator. 1 mol N2 3 mol H2 1 1 mol N2 3 mol H2 2 mol NH3 2 mol NH3 3 mol H2 2 mol NH3 2 mol NH3 1 mol N2 3 mol H2 1 mol N2 What is the largest number of moles of Al2O3 that could result from reacting 6 moles of O2? Slide 11 / 139 4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s) A 2 mol B 3 mol C 4 mol D 18 mol E 24 mol = 6 mol O2 x 2Answer molAl2O3 3 mol O2 = 4 mol Al2O3 2 How many moles of O2 would be required to create 12 moles of Al2O3? 4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s) A 3 mol B 4 mol C 8 mol D 18 mol = 12 mol Al2Answer O3 x 3 mol O2 2 mol Al2O3 = 18 mol O2 Slide 12 / 139 3 Slide 13 / 139 How many moles of O2 would be required to completely react with 8 moles of Al? 4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s) A 2 mol B 3 mol C 6 mol D 12 mol Answer 8mol Al x 3 mol O2 ----------4 mol Al = 6 mol O2 Slide 14 / 139 4 When iron rusts in air, iron (III) oxide is produced. How many moles of oxygen react with 2 .4 mol Fe? 4 Fe (s) + 3 O2 (g) ® 2 Fe2O3 (s) A 1 .2 mol B 1 .8 mol C 2 .4 mol D 3 .2 mol E 4 .8 mol Answer 2.4 mol Fe x 3 mol O2 ----------4 mol Fe = 1.8 mol O2 Slide 15 / 139 5 How many moles of Al are needed to react completely with 1 .2 mol FeO? 2 Al (s) + 3 FeO (g) ® 3 Fe (s) A 0 .8 mol B 1 .2 mol C 1 .6 mol D 2 .4 mol E 4 .8 mol Answer 1.2 mol FeO x 2 mol Al ----------3 mol FeO = 0.8 mol Al + Al2O3 (s) Slide 16 / 139 6 How many moles of calcium metal are produced from the decomposition of 8 mol of calcium chloride? CaCl2 (s) ® Ca (s) + Cl2 (g) Answer 8 mol CaCl2 x 1 mol Ca ---------------1 mol CaCl2 = 8 mol Ca Slide 17 / 139 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) Answer 8 mol Ca 8 mol Cl2 Total of 16 moles Stoichiometric Calculations with Moles Silver and nitric acid react according to the following balanced equation: 3 Ag(s) + 4 HNO3(aq) à 3 AgNO3(aq) + 2 H2O(l) + NO(g) A. How many moles of silver are needed to react with 40 moles of nitric acid, HNO3? B. Using 40 mol HNO3, how many moles of silver nitrate will be produced? C. Using 40 mol HNO3, how many moles of water will be produced? D. Using 40 mol HNO3, how many moles of nitrogen monoxide will be made? Slide 18 / 139 Slide 19 / 139 Answers: A) 30 mol Ag B) 30 mol AgNO3 C) 20 mol H2O D) 10 mol NO Slide 20 / 139 Stoichiometric Calculations with Moles 2 N2H4(l) + N2O4(l) à 3 N2(g) + 4 H2O(g) A. How many moles of dinitrogen tetrahydride are required to produce 57 moles of nitrogen? B. How many moles of dinitrogen tetroxide are required to produce 57 moles of nitrogen? C. How many moles of water are produced when 57 moles of nitrogen are made? Slide 21 / 139 Answers: A) 38 mol N2H4 B) 19 mol N2O4 C) 76 mol H2O Slide 22 / 139 Stoichiometry So far, we have interpreted the coefficients in a balanced equation as indicating the number of moles of each substance. However, we said before that there are many different ways to interpret a balanced equation. The coefficients in a balanced equation can represent · moles · representative particles (atoms, molecules, formula units) · liters of gas (at constant temperature & pressure) The chart on the next slide shows these relationships. Interpreting a Balanced Equation Slide 23 / 139 The coefficients in a balanced equation can represent · moles · representative particles (atoms, molecules, formula units) · liters of gas * 3H2 + N2 à 2NH3 3 H2 N2 2 NH3 3 moles of H2 1 mole of N2 2 moles of NH3 3 molecules of hydrogen (H2) 1 molecule of nitrogen (N2) 2 molecules of ammonia (NH3) 3 L of H2 * 1 L of N2 * 2 L of NH3 * * at same Temperature & Pressure Interpreting a Balanced Equation 3H2 + N2 à 2NH3 · In terms of moles, the equation above shows a total of 4 moles reacting to yield 2 moles of product. · In terms of particles, a total of 4 molecules react to form 2 molecues. · In terms of liters, a total of 4 liters of gas react to form 2 liters of gas (at STP, 89.2 L of gas react to yield 44.8 L). The number of moles, particles and liters before the arrow is not the same as after the arrow. Therefore, these quantities are not conserved in a balanced equation. Slide 24 / 139 Interpreting a Balanced Equation 3H2 + N2 à Slide 25 / 139 2NH3 We have seen various ways to interpret a balanced equation. Recall that a balanced equation is one in which both sides have the same number of each kind of atom. As a result, the Law of Conservation of Mass is obeyed, since no atoms can be created nor destroyed. Therefore, the quantities that are conserved in a balanced equation are mass and atoms. Conserved means that the amount that appears before the arrow is the same as the amount after the arrow. Only mass and the number of atoms are conserved in a balanced equation. Interpreting a Balanced Equation Slide 26 / 139 Only mass and the number of atoms are conserved in a balanced equation. 3H2 + N2 à 2NH3 3 H2 N2 2 NH3 6 atoms of hydrogen (H) 2 atoms of nitrogen (N) 6 atoms of hydrogen (H) and 2 atoms of nitrogen (N) 3 mol x 2 g/mol 1 mol x 28 g/mol 2 mol x 17 g/mol = 6 g of H2 = 28 g of N2 = 34 g of NH3 Slide 27 / 139 8 In a chemical reaction that is represented by a balanced equation, the quantities that are conserved are A mass and molecules B mass and atoms C moles and liters D moles and molecules E mass and liters Slide 28 / 139 9 In a chemical reaction that is represented by a balanced equation, the quantities that are NOT conserved are A mass and moles B mass and atoms C moles and particles D mass and atoms Stoichiometry Calculations with Particles Slide 29 / 139 We will now consider a second interpretation of a balanced equation in terms of particles (which could be molecules, atoms, or formula units). Multiplying the number of moles by 6.02 x 1023 yields the numbers of particles. So a formula that's balanced for moles must also be balanced for particles. can be read as: 3 H2 + N2 ® 2 NH3 3 molecules of H2 plus 1 molecule of N2 yields 2 molecules of NH3 Slide 30 / 139 Stoichiometry Calculations with Particles Solving stoichiometry problems with particles is similar to those with moles. One big difference is that your answers for particles cannot be fractions or decimal numbers; they must be whole numbers. Stoichiometry Calculations with Particles Slide 31 / 139 So here are the six ratios that you could use for solving problems involving particles: 3 H2 + N2 ® 2 NH3 3 molecules H2 3 molecules H2 1 molecule N2 2 molecules NH3 1 molecule N2 2 molecules NH3 and their reciprocals 1 molecule N2 2 molecules NH3 2 molecules NH3 3 molecules H2 3 molecules H2 1 molecule N2 Note that there is no abbreviation for "molecule." Slide 32 / 139 Stoichiometry Calculations with Particles Sample Problem #3 How many molecules of N2 are needed to completely react with 24 molecules of H2? Sample Problem #3 - Solution · In the problem, circle the "given" quantity and underline the "wanted" quantity. How many molecules of N2 are needed to completely react with 24 molecules of H2? · Write the "given" quantity. 24 molecules of H2 1 Stoichiometry Calculations with Particles Sample Problem #3 How many molecules of N2 are needed to completely react with 24 moles of H2? Sample Problem #3 - Solution (con't) · Select the ratio that has the "given" unit in the denominator and the "wanted" unit in the numerator. 24 molecules of H2 1 3 molecules H2 3 molecules H2 1 molecule N2 2 molecules NH3 1 molecule N2 2 molecules NH3 1 molecule N2 2 molecules NH3 2 molecules NH3 3 molecules H2 3 molecules H2 1 molecule N2 Slide 33 / 139 Slide 34 / 139 Stoichiometry Calculations with Particles Sample Problem #3 How many molecules of N2 are needed to completely react with 24 moles of H2? Sample Problem #3 - Solution (con't) · Select the ratio that has the "given" unit in the denominator and the "wanted" unit in the numerator. 24 molecules of H2 1 10 x 1 molecule N2 3 molecules H2 = 8 molecules N2 What is the largest number of of Li3N formula units that could result from reacting 6 N2 molecules? Slide 35 / 139 6 Li (s) + N2 (g) ® 2 Li3N (s) Answer 6 molecules N2 x 2 formula units Li3N ---------------------1 molecule N2 = 12 formula units Li3N 11 How many N2 molecules would be required to create 4 Li3N formula units? 6 Li (s) + N2 (g) ® 2 Li3N (s) Answer 4 formula units Li3N x 1 molecule N2 ---------------------2 formula units Li3N = 2 molecules N2 Slide 36 / 139 12 How many Li atoms would be required to completely react with 3 N2 molecules? Slide 37 / 139 6 Li (s) + N2 (g) ® 2 Li3N (s) Answer 3 molecules N2 x 6 atoms Li ---------------------1 molecule N2 = 18 atoms Li Stoichiometry Calculations with Volume Slide 38 / 139 So far, we have seen that coefficients in a balanced equation can represent moles or representative particles. A third interpretation applies only to gases involved. If all the gases are at the same temperature and pressure, then the coefficients give the relative volumes of gas in liters. CS2 (g) + 3 O2 (g)® CO2 (g) + 2 SO2 (g) This formula can be read as:* 1 L of CS2 + 3 L of O2 yields 1 L of CO2 + 2 L of SO2 Stoichiometry Calculations with Volume Using coefficients to represent volume only works when comparing two gases in a formula. It cannot be used to compare a gas to a liquid or solid. CS2 (g) + 3 O2 (g)® CO2 (g) + 2 SO2 (g) At STP conditions: 1 x 22.4 L + 3 x 22.4 L ® 1 x 22.4 L + 2 x 22.4 L Slide 39 / 139 Stoichiometry Calculations with Gases Slide 40 / 139 Using this interpretation we can directly answer questions about the relative volumes of gas reactants and products. For the next problem, we will not write out all of the ratios ahead of time. Sample Problem #4 How many liters of SO2 will be produced when 6 liters of O2 are reacted as shown below? CS2 (g) + 3 O2 (g)® CO2 (g) + 2 SO2 (g) Stoichiometry Calculations with Gases Slide 41 / 139 Sample Problem #4 - Solution How many liters of SO2 will be produced when 6 liters of O2 are reacted as shown below? CS2 (g) + 3 O2 (g)® CO2 (g) + 2 SO2 (g) · Write the "given" quantity. · Create a ratio that has the "given" unit in the denominator and the "wanted" unit in the numerator. 6 liters of O2 x 2 L of SO2 3 L of O2 = 4 L of SO2 Slide 42 / 139 13 The coefficients in a balanced equation can represent A mass, in grams B volume, in liters (gases only) C representative particles D Both A and B E Both B and C 14 Slide 43 / 139 How many liters of H2O (g) will be created from reacting 8.0 L of H2 (g) with a sufficient amount of O2 (g)? 2 H2 (g) + O2 (g) ® 2 H2O (g) Answer = 8 L of H2O 15 The equation below shows the decomposition of solid lead (II) nitrate, Pb(NO3)2. How many liters of O2 will be produced when 12 L of NO2 are formed? 2 Pb(NO3)2 (s) ® 2 PbO (s) + Slide 44 / 139 4 NO2 (g) + O2 (g) A 1.0 L B 2.0 L C 3.0 L D 4.0 L E 12 L = 12 L NO2 Answer x 1 L O2 4 L NO2 = 3 L of O2 Slide 45 / 139 16 What volume of methane is needed to completely react with 500 L of O2 at STP? You must first balance the equation. __ CH4 + ___ O2 - - > ___ CO2 + ___ H2O A 125 L B 250 L C 500 L D 1000 L Answer = 500L O2 x 1 L CH4 2 L O2 = 250 L of O2 17 Slide 46 / 139 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 H2O (g) = 36 L O2 Answer x 4 L NO2 7 L O2 = 21 L of NO2 Stoichiometric Calculations with Mass Slide 47 / 139 So far, we have seen stoichiometry problems that can be solved in one step, using only coefficients. These are: · mole-mole problems, · particle-particle problems and · volume-volume problems. Coefficients can represent moles, particles, or liters. When mass is involved, however, we must now take into consideration the fact that all particles have a different molar mass. Coefficients do not represent mass in grams. Stoichiometric Calculations with Mass 3 H2 + N2 ® 2 NH3 Using this example once again, it is clear that 3 grams of H2 + 1 gram of N2 will not equal 2 grams of NH3. Coefficients do not represent mass in grams. 3 H2 N2 2 NH3 3 moles of H2 1 mole of N2 2 moles of NH3 3 mol x 2 g/mol = 6g of H2 1 mol x 28 g/mol 2 mol x 17 g/mol = 28g of N2 = 34g of NH3 6 g of H2 + 28 g of N2 will equal 34 g of NH3 Slide 48 / 139 Stoichiometric Calculations with Mass Slide 49 / 139 Consider this question asked earlier: 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 Al2O3 (s) Now, instead of asking for moles of Al2O3 , consider this question: Sample Problem #5 What is the mass, in grams, of Al2O3 that could result from reacting 6 moles of O2? Stoichiometric Calculations with Mass Slide 50 / 139 4 Al (s) + 3 O2 (g) --> 2 Al2O3 (s) Sample Problem #5 - Solution What is the mass, in grams, of Al2O3 that could result from reacting 6 moles of O2? We proceed similarly as before, changing from moles of the "given" quantity to the moles of the "wanted" quantity. 6 moles of O2 1 x 2 mol Al2O3 3 mol O2 = 4 mol Al2O3 This answer becomes the new "given" which we simply convert to grams, using the molar mass of Al2O3. Stoichiometric Calculations with Mass 4 Al (s) + 3 O2 (g) --> 2 Al2O3 (s) Sample Problem #5 - Solution (con't) What is the mass, in grams, of Al2O3 that could result from reacting 6 moles of O2? 4 mol Al2O3 1 x 102 g Al2O3 1 mol Al2O3 = 408 g Al2O3 Remember: Do not use coefficients when converting between moles and mass. The molar mass is for only ONE mole! Slide 51 / 139 Slide 52 / 139 Stoichiometric Calculations with Mass An alternative way to solve the previous problem is to set-up both ratios, then perform the calculation without stopping after the first step. 6 moles of O2 1 x 2 mol Al2O3 3 mol O2 x 102 g Al2O3 1 mol Al2O3 = 408 g Al2O3 The key is to make sure that the unit in any numerator appears in the next denominator so that units will cancel out. 18 What mass, in grams, of O2 would be required to create 12 moles of Al2O3? Slide 53 / 139 4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s) = 12 mol Al2O3 x 3 mol O2 Answer 2 mol Al2O3 = 18 mol O2 x 32 g O2 1 mol O2 = 576 g O2 19 How many grams of O2 would be required to completely react with 8 moles of Al? 4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s) Answer 8mol Al x 3 mol O2 ----------4 mol Al = 6 mol O2 x = 192 g O2 32 g O2 1 mol O2 Slide 54 / 139 Slide 55 / 139 20 When iron rusts in air, iron (III) oxide is produced. How many grams of oxygen react with 2 .4 mol Fe? 4 Fe (s) + 3 O2 (g) ® 2 Fe2O3 (s) 2.4 mol Fe x 3 Answer mol O2 ----------4 mol Fe = 1.8 mol O2 x 32 g O2 1 mol O2 = 57.6 g O2 Slide 56 / 139 21 How many grams of Al are needed to react completely with 1 .2 mol FeO? 2 Al (s) + 3 FeO (g) ® 3 Fe (s) 1.2 mol FeO x + Al2O3 (s) 2Answer mol Al ----------3 mol FeO = 0.8 mol Al x 27 g Al 1 mol Al = 21.6 g Al Slide 57 / 139 Mass-Mass Calculations We started this unit by using coefficients in a balanced equation to solve mole-mole problems. Given Find Slide 58 / 139 Mass-Mass Calculations Next, we extended these problems by calculating the mass of the "wanted" quantity. Find Given Slide 59 / 139 Mass-Mass Calculations Now, we will solve problems when both the "given" quantity and the "wanted" quantity are in grams. Recall that you cannot simply use the coefficients to relate mass in grams. Given Find Grams of substance A 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 Slide 60 / 139 Mass-Mass Calculations A typical "mass-mass" problem is has three steps as outlined below. Remember that coefficients are used ONLY between moles. Given Find Grams of substance A Use molar mass of A Grams of substance B Step 1 Moles of substance A Step 3 Step 2 Use coefficients of A and B from balanced equation Use molar mass of B Moles of substance B Slide 61 / 139 Mass-Mass Calculations Sample Problem #6 Calculate the mass of ammonia, NH3, produced by the reaction of 5.4 g hydrogen gas with an excess of nitrogen. N2 + 3H2 ---> 2NH3 Sample Problem #6 - Solution Calculate the mass of ammonia, NH3, produced by the reaction of 5.4 g hydrogen gas with an excess of nitrogen. We will proceed using these steps: a) convert 5.4 g H2 to moles H2 b) convert moles H2 to moles NH3 c) convert moles NH3 to grams NH3 Slide 62 / 139 Mass-Mass Calculations Sample Problem #6 - Solution (con't) gH2 mol H2 - - > --> 1 mol H2 5.40 g H2 X 2.0 g H2 a) convert 5.4 g H2 to moles H2 mol NH3 2 mol NH3 X 3 mol H X 2 g NH3 --> 17.0 g NH3 1 mol NH3 = 31 g NH3 c) convert moles NH3 to grams NH3 b) convert moles H2 to moles NH3 Mass-Mass Calculations Sample Problem #7 How many grams of water can be produced from the combustion of 1.00 g of glucose? C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O Sample Problem #7 - Solution Our general strategy is Step (a): convert mass of glucose to moles of glucose · use the molar mass of glucose Step (b): convert moles glucose to moles water · use the ratio of coefficients in the balanced equation Step (c): convert moles of water to grams of water · use the molar mass of water Slide 63 / 139 Slide 64 / 139 Mass-Mass Calculations C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O Sample Problem #7 - Solution (con't) How many grams of water can be produced from the combustion of 1.00 g of glucose? no direct calculation 1.00 g glucose 0.600 g water (a) X (c) 1mol glucose X 180.0 g glucose 5.56 x 10-3 mol glucose X 6 mol water 18.0 g water 1 mol water 3.33 x 10-2 mol water 1 mol glucose (b) for every 1 mol glucose you get 6 mols of water Slide 65 / 139 Mass-Mass Calculations 1.00 g glucose 1 x 1mol glucose 180.0 g glucose x 6 mol water 1 mol glucose x 18.0 g water 1 mol water = 0.600 g water Again, the key to this method is making sure that the unit in any numerator appears in the next denominator so that units will cancel out. Given Find Grams of substance A Grams of substance B Use molar mass of A Moles of substance A Use molar mass of B Use coefficients of A and B from balanced equation Moles of substance B Slide 66 / 139 22 What mass of Na is produced by the decomposition of 6.5 g NaN3? 2 NaN3 (s) ® 2 Na (s) + 3 N2 (g) A 2.3 g B 4.6 g C 6.5 g D 23 g E 46 g How many grams of Al2O3 will be created from reacting 54 g of Al with a sufficient amount of O2? 23 Slide 67 / 139 4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s) A 27 g B 54 g C 102 g D 204 g Slide 68 / 139 24 How many grams of Mg must react in order to to create 160 g of MgO? 2 Mg (s) + O2 (g) ® 2 MgO (s) A 48 g B 80 g C 96 g D 160 g Slide 69 / 139 25 Approximately how many grams of oxygen are needed to react with 240 g of Mg? 2 Mg (s) + O2 (g) ® 2 MgO (s) A 80 g B 120 g C 160 g D 240 g Mixed Stoichiometry Calculations Slide 70 / 139 Thus far, our problem-solving has focused on one of four main types. Most practical applications of chemistry are not this narrowly defined. Many problems in advanced chemistry give a quantity in one unit (e.g. moles, grams, or liters) and ask for the proportional quantity in a different unit. For example, this problem gives a quantity in moles, but asks for mass, in grams. How many grams of ammonia can be produced by using 10 mol of nitrogen gas and an unlimited or excess amount of hydrogen gas? Mixed Stoichiometry Calculations Slide 71 / 139 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. (3) (1) represenative X particles of G 1 mol G x 6.02 x10-23 6.02 x 1023 1 mol W = representative particles of W (2) 1 mol G mass X of G mass G mol G X b mol W a mol G mol W mass W x 1 mol W x 1 mol G volume of G X at STP 22.4 L G 22.4 L W 1 mol W mass = of W Volume of = W at STP Slide 72 / 139 Mixed Stoichiometry Calculations Sample Problem #8 How many grams of ammonia can be produced by using 10 mol of nitrogen gas and an unlimited or excess amount of hydrogen gas? N2 + 3H2 ---> 2NH3 Sample Problem #8 - Solution Since the "Given" quantity is already in moles, we can skip to step (2). Overview: mol N2 ---> mol NH3 ---> grams NH3 Slide 73 / 139 Mixed Stoichiometry Calculations N2 + 3H2 ---> 2NH3 Sample Problem #8 - Solution (con't) 10 mol N2 x 2 mol NH3 x 1 mol N2 17 g NH3 = 340 g NH3 1 mol NH3 Remember: Do not use coefficients when using molar mass. Mixed Stoichiometry Calculations Slide 74 / 139 Sample Problem #9 The airbag in a car generates a relatively large amount of gas quickly through the following reaction. 2 NaN3 (s) ® 2 Na (s) + 3 N2 (g) How many grams of NaN3 will be required to generate enough gas to fill an airbag with a volume of 36 L N2 (assume STP conditions)? Sample Problem #9 - Solution How many grams of NaN3 will be required to generate enough gas to fill an airbag with a volume of 36 L N2 (assume STP conditions)? Mixed Stoichiometry Calculations 2 NaN3 (s) ® 2 Na (s) + 3 N2 (g) Sample Problem #9 - Solution Overview Step (1) - Liters of N2 --> Moles of N2 Step (2) - Moles of N2 --> Moles of NaN3 Step (3) - Moles of NaN3 --> Grams of NaN3 represenative particles of G X 1 mol G 1 mol G mass of G X mass G (1) x 6.02 x10-23 1 mol G volume of G X at STP 22.4 L G (2) mol G X b mol W a mol G mol W (3) 6.02 x 1023 1 mol W mass W x 1 mol W x 22.4 L W 1 mol W = representative particles of W mass = of W Volume of = W at STP Remember that coefficients are used ONLY between moles. Slide 75 / 139 Slide 76 / 139 Mixed Stoichiometry Calculations 2 NaN3 (s) ® 2 Na (s) + 3 N2 (g) Sample Problem #9 - Solution Step (1) - Liters of N2 --> Moles of N2 36 L N2 1 mol N2 x 1 22.4 L N2 = 1.6 mol N2 Step (2) - Moles of N2 --> Moles of NaN3 1.6 mol N2 2 mol NaN3 x 1 3 mol N2 = 1.1 mol NaN3 Step (3) - Moles of NaN3 --> Grams of NaN3 1.1 mol NaN3 1 65 g NaN3 x 1 mol NaN3 = 72 g NaN3 Remember that coefficients are used ONLY between moles. 26 Slide 77 / 139 How many liters of O2 (g) at STP are required to create 102 g of Al2O3 (s)? 4 Al (s) + 3O2 (g) ® 2 Al2O3 (s) A 11.2 L B 22.4 L C 33.6 L D 44.8 L E 67.2 L Answer 22.4 L O2 3 mol O2 102 g Al2O3 x 1 mol Al2O3 x x 102 g Al2O3 2 mol Al2O3 1 mol O2 1 = 33.6 L O2 Slide 78 / 139 27 How many grams of Cl2 are needed to react with 1 mol of antimony, Sb? 2 Sb + 3 Cl2 à 2 SbCl3 A 71 g B 107 g C 142 g D 213 g 1 mol Sb 1 x Answer 3 mol Cl2 2 mol Sb x 71 g Cl2 1 mol Cl2 = 107 g Cl2 Mixed Stoichiometry Practice 3MnO2(s) + 4 Al(s) à 2 Al2O3(s) + * Slide 79 / 139 3Mn(s) A. How many manganese atoms are produced if 55 moles of MnO2 react with excess aluminum. B. How many moles of aluminum oxide are made if 3580 g of manganese oxide are consumed? Mixed Stoichiometry Practice 3MnO2(s) + 4 Al(s) à 2 Al2O3(s) + * Slide 80 / 139 3Mn(s) C. How many moles of manganese oxide will react with 5.33 x 1025 atoms of aluminum? D. If 4.37 moles of aluminum are consumed, how many formula units of aluminum oxide are produced? Slide 81 / 139 Answers: A) 3.3 x 1025 atoms Mn B) 27.5 mol Al2O3 C) 66.4 mol MnO2 D) 1.32 x 1024 formula units Al2O3 Mixed Stoichiometry Practice Slide 82 / 139 ** The compound tristearin (C57H110O6) is a type of fat which camels store in their hump. Besides being a source of energy, the fat is a source of water for the camel because when the fat is burned, the following combustion reaction occurs: 2 C57H110O6(s) + 163 O2(g) à 114 CO2(g) + 110 H2O(l) A. At STP, what volume of O2 is required to consume 0.64 moles of tristearin? B. At STP, what volume of carbon dioxide is produced in Part A? Mixed Stoichiometry Practice ** Slide 83 / 139 2 C57H110O6(s) + 163 O2(g) à 114 CO2(g) + 110 H2O(l) C. If 22.4 L of oxygen is consumed at STP, how many moles of water are produced? D. Find the mass of tristearin required to produce 55.56 moles of water. Slide 84 / 139 Answers: A) 1168 L O2 (or 1200 L) B) 817 L CO2 (or 820 L) C) 0.00414 mol H2O D) 899 g C57H110O6 Slide 85 / 139 Limiting Reagent & Excess Reagent Percent Yield & Theoretical Yield How Many Cookies Can I Make? Slide 86 / 139 · You can make cookies until you run out of one of the ingredients. · Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat). How Many Cookies Can I Make? In this example the egg would be the limiting reactant, because it will limit the amount of cookies you can make. Slide 87 / 139 Slide 88 / 139 How Many Cookies Can I Make? Recipe for Butter Cream Cookies 1 c. butter 1 c. sugar 1-3oz. pkg 1 egg yolk cream cheese 2 T. vanilla 2-1/2 c. sifted cake flour This recipe will make 5 dozen cookies. Slide 89 / 139 28 How many dozen cookies could you make with 3 c. butter and enough of all the other ingredients? A 1 doz B 3 doz C 5 doz D 10 doz E Recipe for Butter Cream Cookies 1 c. butter 1 c. sugar 1-3oz. pkg 1 egg yolk cream cheese 2 T. vanilla 2-1/2 c. sifted cake flour Yields 5 dozen cookies. 15 doz Slide 90 / 139 29 How many dozen cookies could you make with 10 c. sifted cake flour and enough of all the other ingredients? A 2 doz B 5 doz C 10 doz D 20 doz E 40 doz Recipe for Butter Cream Cookies 1 c. butter 1 c. sugar 1-3oz. pkg 1 egg yolk cream cheese 2 T. vanilla 2-1/2 c. sifted cake flour Yields 5 dozen cookies. Slide 91 / 139 30 How many dozen cookies could you make with 1 T. vanilla and enough of all the other ingredients? A 1 doz B 2 doz C 2-1/2 doz D 5 doz E Recipe for Butter Cream Cookies 1 c. butter 1 c. sugar 1-3oz. pkg 1 egg yolk cream cheese 2 T. vanilla 2-1/2 c. sifted cake flour Yields 5 dozen cookies. 10 doz Limiting Reagents & Theoretical Yield Slide 92 / 139 · 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. · Theoretical yield is the the maximum amount of product that can be made, based on the limiting reagent. Limiting Reagents Recipe for Butter Cream Cookies 1 c. butter 1 c. sugar 1-3oz. pkg 1 egg yolk cream cheese 2 T. vanilla 2-1/2 c. sifted cake flour Yields 5 dozen cookies. In the three previous Senteo questions, the amount given in the problem was the limiting reagent. · 3 c butter · 10 c. sifted cake flour · 1T. vanilla Slide 93 / 139 Slide 94 / 139 Limiting Reagents Limiting reagent Excess reagent Theoretical yield 3 c. butter all other ingredients 15 doz. 10 c. sifted cake flour all other ingredients 20 doz. 1 T. vanilla all other ingredients 2-1/2 doz. Note that in every case, the theoretical yield is determined by the limiting reagent, not the excess reagents. Slide 95 / 139 31 When two substances react, the one that is used up first is called the: A determining reagent B limiting reagent C unlimited reagent D excess reagent E reactive reagent Limiting Reagents Limiting reagent problems are different from those done previously in that two quantities are given, instead of just one. It is your job to figure out which reactant is limiting because that will determine the maximum yield. There are a variety of methods one can use to determine which reactant is the limiting one. Once you have identified the limiting reagent, then the other reactant is the excess reagent. Slide 96 / 139 Slide 97 / 139 Definitions · Limiting reagent - any reactant that is used up firsts in a chemical reaction; it determines the amount of product that can be formed in the reaction · Excess reagent - a reagent present in a quantity that is more than sufficient to react with a limiting reagent; any reactant that remains after the limiting reagent is used up in a chemical reaction · Theoretical yield - the maximum amount of product that can be made, based on the limiting reagent Limiting Reagent & Excess Reagent Slide 98 / 139 This is how you identify which is the limiting reagent: 1. Write a balanced equation. 2. Use the given amount of each reactant to calculate the amount of product that could be formed. Use the problem-solving methods for stoichiometry that you learned earlier in this unit. 3. Compare the two amounts of product that could be made. The smaller of the two amounts indicates the maximum amount of product that could be made and is called theoretical yield. The larger of the two amounts is irrelevant and meaningless. 4. Whichever reactant yields the smaller amount of product is thus the limiting reagent. Slide 99 / 139 Limiting Reagent & Excess Reagent O2 Sample Problem #10 Consider the reaction between hydrogen and oxygen to yield water. 2 H2 (l) + O2 (g) --> 2 H2O (l) Starting with 10 molecules of H2 and 7 molecules of O2, which reactant will run out first? Sample Problem #10 - Solution Use stoichiometry methods to calculate the following: 10 molecules of H2 will yield ___ molecules of H2O 7 molecules of O2 will yield ___ molecules of H2O H2 Limiting Reagent & Excess Reagent Slide 100 / 139 2 H2 (l) + O2 (g) --> 2 H2O (l) Sample Problem #10 - Solution (con't) 10 molecules H2 x 2 molecules H2O 2 molecules H2 10 molecules of H2 will yield __10_ molecules of H2O 7 molecules O2 x 2 molecules H2O 1 molecule O2 7 molecules of O2 will yield __14_ molecules of H2O Now, we compare the two amounts. Limiting Reagent & Excess Reagent Slide 101 / 139 2 H2 (l) + O2 (g) --> 2 H2O (l) Sample Problem #10 - Solution (con't) 10 molecules of H2 will yield __10_ molecules of H2O 7 molecules of O2 will yield __14_ molecules of H2O · The smaller amount is the theoretical yield: in reality, 10 molecules of water can be produced. · 10 molecules of H2 is the limiting reagent · 7 molecules of O2 is the excess reagent. · The amount of 14 molecules of water cannot in fact be made; this would require 14 molecules of H2 which is not available. The amount of 14 molecules is meaningless and serves only to compare to the 10 molecules of water that can be produced. Limiting Reagent & Excess Reagent Sample Problem #11 What mass of ammonia can be produced from 65 g nitrogen and 25 g hydrogen? Sample Problem #11 - Solution This problem asks for theoretical yield; however, we must first determine which reagent is limiting. 1. Write a balanced chemical equation. N2 + 3H2 ---> 2NH3 2. Calculate how much product can be made from the first "given" amount, 65 g nitrogen. 65 g N2 1 x 1 mol N2 28 g N2 x 2 mol NH3 1 mol N2 x 17 g NH3 = 1 mol NH3 79 g NH3 3. Calculate how much product can be made from the second "given" amount, 25 g hydrogen. 25 g H2 1 x 1 mol H2 2 g H2 x 2 mol NH3 3 mol H2 x 17 g NH3 = 1 mol NH3 142 g NH3 Slide 102 / 139 Limiting Reagent & Excess Reagent Slide 103 / 139 Sample Problem #11 What mass of ammonia can be produced from 65 g nitrogen and 25 g hydrogen? Sample Problem #11 - Solution (con't) 4. Compare the two amounts. The smaller of the two amounts is the answer to the problem. This is called the theoretical yield. Answer: 79 g of NH3 can be produced; nitrogen is the limiting reagent and hydrogen is the excess reagent. Slide 104 / 139 32 If you are provided with 75 molecules of H2 and 50 molecules of N2, what are the limiting reagent & theoretical yield for the reaction: 3 H2 (g) + N2 (g) --> 2 NH3 (g) A 75 molecules H2; 25 molecules NH3 B 75 molecules H2; 50 molecules NH3 C 50 molecules N2; 50 molecules NH3 D 50 molecules N2; 100 molecules NH3 E Not enough information is given. Slide 105 / 139 33 If you are provided with 50 molecules of H2 and 50 molecules of O2, which is the limiting reagent for the reaction: 2 H2 (g) + O2 (g) --> 2 H2O (l) A 50 molecules H2 B 50 molecules O2 C 50 molecules H2O D There is no limiting reagent. E Not enough information is given. Slide 106 / 139 34 If you are provided with 60 molecules of H2 and 20 molecules of N2, which is the limiting reagent for the reaction: 3 H2 (g) + N2 (g) --> 2 NH3 (g) A 60 molecules H2 B 20 molecules N2 C 50 molecules NH3 D There is no limiting reagent. E Not enough information is given. Slide 107 / 139 35 If you are provided with 15 molecules of H2 and 10 molecules of N2, which is the limiting reagent for the reaction: 3 H2 (g) + N2 (g) --> 2 NH3 (g) A 15 molecules H2 B 10 molecules N2 C 20 molecules NH3 D There is no limiting reagent. E Not enough information is given. Slide 108 / 139 36 Which of the following is NOT a true statement? A The amount of product is determined by the limiting reagent. B A balanced equation is necessary to determine which reagent is limiting. C Some of the excess reagent is left over after the reaction is complete. D The reactant that has the smallest given mass is the limiting reagent. E Adding more of the limiting reagent will cause more product to be made. Limiting Reagent & Theoretical Yield Slide 109 / 139 Sample Problem #12 Find the theoretical yield of AlCl3, if 27g Al and 71g Cl2 react. Sample Problem #12 - Solution 1. See how much AlCl3 can be made from each reactant 2. Smaller amount identifies the LR 2Al(s) 27 g + 3Cl2(g) 71 g à 2AlCl3 ?g Limiting Reagent & Theoretical Yield 2Al(s) 27 g + 3Cl2(g) 71 g Slide 110 / 139 2AlCl3 ?g à Sample Problem #12 - Solution (con't) 27g Al will yield _______ g AlCl3 71g Cl2 will yield _______ g AlCl3 Therefore, the LR is _____________ and the theoretical yield is ________ grams of AlCl3 Answer 27g Al will yield 134 g AlCl3 71g Cl2 will yield 89 g AlCl3 LR = 71 g Cl2 TY = 89 g AlCl3 Limiting Reagent & Theoretical Yield Sample Problem #13 Find the theoretical yield of AlCl3, if 100g of each Al and Cl2 react. Sample Problem #13 - Solution 1. See how much AlCl3 can be made from each reactant 2. Smaller amount identifies the LR 2Al(s) 100 g + 3Cl2(g) 100 g à 2AlCl3 ?g Slide 111 / 139 Limiting Reagent & Theoretical Yield 2Al(s) 100 g + 3Cl2(g) 100 g à Slide 112 / 139 2AlCl3 ?g Sample Problem #13 - Solution (con't) 100g Al will yield ______ g AlCl3 100g Cl2 will yield _______ g AlCl3 Therefore, the LR is ____________ and the theoretical yield of AlCl3 is ____________ grams. Answer 100g Al will yield 494 g AlCl3 100g Cl2 will yield 125 g AlCl3 LR = 100 g Cl2 TY = 125 g AlCl3 Excess Reagent Slide 113 / 139 You may sometimes be asked to calculate how much of the excess reagent is left over, after all of the limiting reagent has been consumed. The general strategy is rather simple, once you have identified the excess reagent. How to calculate excess reagent 1) Write down how much of it you have at the start of the reaction. 2) Use stoichiometry (i.e. a balanced equation and the LR) to calculate how much of the excess reagent gets used up. 3) Subtract the answer from Step 2 from Step 1. It does not matter if the amounts are in moles or mass (grams), as long as you are consistent. Limiting Reagent, Excess Reagent & Theoretical Yield BIG IDEA #1: The limiting reagent determines the theoretical yield. BIG IDEA #2: The limiting reagent determines how much of the excess reagent gets used up. Slide 114 / 139 Slide 115 / 139 Limiting Reagent & Excess Reagent Consider this reaction which we have seen before: 2 H2 (l) + O2 (g) --> 2 H2O (l) Sample Problem #14 If we start with 10 molecules of H2 and 7 molecules of O2, how much of the excess reagent will be left over after the reaction is complete? O2 H2 Slide 116 / 139 Limiting Reagent & Excess Reagent O2 2 H2 (l) + O2 (g) --> 2 H2O (l) H2 Sample Problem #14 - Solution We had already determined (in Sample Problem #10) that hydrogen is the limiting reagent. Therefore, oxygen is the excess reagent. and we must now complete the following: O2 we start with: ________________ O2 we use (based on LR): ________________ O2 we have left over: __________________ Limiting Reagent & Excess Reagent 2 H2 (l) + O2 (g) --> 2 H2O (l) Sample Problem #14 - Solution (con't) O2 we start with: 7 molecules O2 O2 we use(based on LR*): 5 molecules O2 10 molecules H2 Always start with the LR here! 1 x 1 molecule O2 2 molecules H2 = 5 molecules O2 get used O2 we have left over: 2 molecules O2 Slide 117 / 139 Limiting Reagent & Excess Reagent Slide 118 / 139 Try this one on your own: 4 Fe (s) + 3 O2 (g) ® 2 Fe2O3 (s) Sample Problem #15 If we start with 12 molecules of Fe and 12 molecules of O2, how much of the excess reagent will be left over after the reaction is complete? Fe is the LR. Answer O2 is the ER. You start with 12 molecules of O2 use up 9 of them, so 3 molecules are left over. Slide 119 / 139 37 If you start with 16 mol H2 and 18 mol O2, how much of the excess reagent is left over? ® 2 H 2O 2 H2 + O 2 A 2 mol H2 B 7 mol H2 C 2 mol O2 D 10 mol O2 E Nothing is left over. Slide 120 / 139 38 If you start with 9 mol H2 and 5 mol N2, how much of the excess reagent is left over? 3 H2 + N2 ® 2 NH3 A 2 mol N2 B 4 mol N2 C 4 mol H2 D 6 mol H2 E Nothing is left over. Slide 121 / 139 39 If you start with 16 mol Fe and 12 mol O2, how much of the excess reagent is left over? ® 2 Fe2O3 4 Fe + 3 O2 A 4 mol Fe B 12 mol Fe C 4 mol O2 D 9 mol O2 E Nothing is left over. Limiting Reagent & Theoretical Yield Slide 122 / 139 Practice Problem 1 ____H2 + ____O2 ---> ____H2O A. Calculate the theoretical yield (in grams) of water if 6 g H2 and 160 g O2 are combined in a reaction vessel. B. How many grams of the excess reagent will be remaining in the vessel? 6 g H2 is the LR.Answer 160 g O2 is the ER. You start with (5 mol) 160 g of O2 and use up (1.5 mol) 48 g, so (3.5 mol) 112 g are left over. Limiting Reagent & Theoretical Yield Practice Problem 2 * According to the balanced chemical equation, how many atoms of silver will be produced from combining 100 g of copper with 200 g of silver nitrate? Cu(s) + 2 AgNO3(aq) à Cu(NO3)2(aq) + 2 Ag(s) Answer 200 g AgNO3 is the LR. 7.10 x 1023 atoms Ag can be made. Slide 123 / 139 Limiting Reagent & Theoretical Yield Slide 124 / 139 Practice Problem 3 * At STP, what volume of “laughing gas” (dinitrogen monoxide) will be produced from 50 g of nitrogen gas and 75 g of oxygen gas? First, write a balanced equation. Answer 50 g N2 is the LR. 40 L N2O can be made. Limiting Reagent & Theoretical Yield Slide 125 / 139 Practice Problem 4 ___N2(g) + ___O2(g) à ___N2O5(g) A. If you begin with 400 g of N2 and 800 g of O2, how many liters of N2O5 could be produced at STP? B. Find the mass (in grams) of excess reagent left over at the conclusion of the reaction. Answer 800 g O2 is the LR. 400 g N2 is the ER. You start with (14.3 mol) 400 g of N2 and use up (10 mol) 280 g, so (4.3 mol) 120 g are left over. Limiting Reagent & Theoretical Yield Practice Problem 5 Carbon monoxide can be combined with hydrogen to produce methanol, CH3OH. Methanol is used as an industrial solvent, as a reactant in some synthesis reactions, and as a cleanburning fuel for some racing cars. If you had 152.5 kg of carbon monoxide and 24.5 kg of hydrogen gas, how many kilograms of methanol could be produced? (First, write a balanced equation.) Answer 152.5 kg CO is the LR. 5446 mol = 174 kg CH3OH can be made. Slide 126 / 139 Slide 127 / 139 Credit to Tom Greenbowe Chemical Education Group at Iowa State University Theoretical Yield & Actual Yield Slide 128 / 139 We have recently seen that whenever you are given TWO amounts of reactants, you must first determine which is the limiting reagent. BIG IDEA: The limiting reagent determines the theoretical yield. However, when experiments are carried out in the laboratory, the theoretical yield is not always easily obtained. The amount of product that is made when performing an experiment is called "Actual yield." Theoretical Yield vs. Actual Yield Theoretical yield (TY) Actual yield (AY) the maximum amount of the amount of product one product that can be made, actually produces and based on the stoichiometry measures in the laboratory; (i.e. balanced equation) and usually less than the TY limiting reagent Slide 129 / 139 Slide 130 / 139 Percent Yield · 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 · The efficiency of a reaction can be expressed using this ratio. · For example, a percent yield of 85% shows that the reaction conditions are more favorable than with a percent yield of only 55%. · Chemical manufacturers strive for high percent yields in their processes. Slide 131 / 139 Percent Yield Sample Problem #16 Use the balanced equation to determine the percent yield if 5 g Mg reacts to actually form 8.1g of MgO 2 Mg (s) + O2 (g) à 2 MgO (s) Sample Problem #16 - Solution 1) Since only one amount is given, then 5 g Mg must be the limiting reagent; oxygen is the excess reagent since there is no amount given. 2) Use 5 g to calculate the TY of MgO. 5 g Mg 1 x 1 mol Mg 24.3 g Mg x 2 mol MgO 2 mol Mg x 40.3 g MgO 1 mol MgO = 8.29 g MgO can be made Percent Yield 2 Mg(s) + O2(g) à 2 MgO(s) Sample Problem #16 - Solution (con't) 3) Use percent yield formula Actual Yield x 100 Theoretical Yield Percent Yield = 8.1 g MgO 8.3 g MgO Percent Yield = Percent Yield = 98% x 100 Slide 132 / 139 Slide 133 / 139 What is the percent yield if the theoretical yield is 57.3 g and the actual yield is 50.9 g? 40 A 0.8883% B 1.1257% C 88.83% D 112.57% E 0.117% Slide 134 / 139 What is the actual yield if the theoretical yield is 16 g and the percent yield is 94.23%? 41 A 16.98 g B 15.08 g C 100 g D 28% E 1507.68 Stoichiometry Practice Problem Practice Problem 6 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? Answer TY = 3.17 kg Ca(OH)2 PY = (2.06/3.17) x 100 = 65% Slide 135 / 139 Stoichiometry Practice Problem Slide 136 / 139 Practice Problem 7 Some underwater welding is done via the thermite reaction, in which rust (Fe2O3) reacts with aluminum to produce iron and aluminum oxide (Al2O3). In one such reaction, 258 g of aluminum and excess rust produced 464 g of iron. What was the percent yield of the reaction? Remember to first write a balanced chemical equation. Answer TY = 533 g Fe PY = (464 g Fe/533 g Fe) x 100 = 65% Stoichiometry Practice Problem Slide 137 / 139 Practice Problem 8 Use the balanced equation to find out how many liters of sulfur dioxide are actually produced at STP if 1.5 x 1027 formula units of zinc sulfide are allowed to react with excess oxygen and the percent yield is 75%. 2 ZnS(s) + 3 O2(g) à 2 ZnO(s) + 2 SO2(g) Answer TY = 5.6 x 104 L SO2 75% = (AY/5.6 x 104 L) x 100 AY = (0.75)(5.6 x 104 L) AY = 4.2 x 104 L SO2 Definitions · Limiting reagent - any reactant that is used up firsts in a chemical reaction; it determines the amount of product that can be formed in the reaction · Excess reagent - a reagent present in a quantity that is more than sufficient to react with a limiting reagent; any reactant that remains after the limiting reagent is used up in a chemical reaction · Theoretical yield - the maximum amount of product that can be made, based on the limiting reagent and a balanced chemical equation · Actual yield - the amount of product that forms when a reaction is carried out in the laboratory · 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 Slide 138 / 139 Slide 139 / 139 Outline: · General idea of stoichiometry · mole ratios · mole-mole problems · particle-particle problems · volume-volume · mole-mole-mass problems · mass-mass problems · mixed problems · LR/TY/PY · how to determine LR · using LR to obtain TY · calculate amount of ER left over · solve for PY or AY