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Stoichiometry_Presentation_v_1.0.notebook March 19, 2014 Stoichiometric Calculations Stoichiometry is the quantitative determination of reactants or products using a balanced chemical equation. Stoichiometric Calculations We can interpret a balanced equation in several ways. One way is to interpret it as indicating the number of moles of each substance. For instance, this formula 2 H2 + O2 → 2 H2O can be read as: 2 moles of H2 plus 1 mole of O2 yields 2 moles of H2O The coefficients in the balanced equation give the ratio of moles of reactants and products. 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 2 H2 + O2 → 2 H2O 1. Use the equation to set up a ratio of the substances of interest. 2 mol H2O 2 mol H2 2. Set that equal to the ratio of the known to unknown quantities of the same substances. 2 mol H2O n mol H2O = 8 mol H2 2 mol H2 3. Solve for the unknown. n = 8.0 moles Stoichiometry Calculations with Moles Stoichiometry Calculations with Moles The same approach can tell you how many moles of one reactant is needed to completely react with another reactant. For instance, use the below formula to determine the number of moles of O2 that will be necessary to completely react with 12 moles of H2. 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 O2 2 mol H2 2. Set that equal to the ratio of the known to unknown quantities of the same substances. n mol O2 1 mol O2 = 12 mol H2 2 mol H2 3. Solve for the unknown. n = 6 mol 1 Stoichiometry_Presentation_v_1.0.notebook What is the largest number of moles of Al2O3 that could result from reacting 6 moles of O2? 1 March 19, 2014 How many moles of O2 would be required to create 12 moles of Al2O3? 2 4 Al (s) + 3 O2 (g) → 2 Al2O3 (s) 4 Al (s) + 3 O2 (g) → 2 Al2O3 (s) How many moles of O2 would be required to completely react with 8 moles of Al? 3 4 Al (s) + 3 O2 (g) → 2 Al2O3 (s) 4 When iron rusts in air, iron (III) oxide is produced. How many moles of oxygen react with 2.4 mol Fe in this reaction? 4 Fe (s) + 3 O2 (g) > 2 Fe2O3 (s) 5 How many moles of aluminum are needed to react completely with 1.2 mol FeO? 6 2 Al (s) + 3 FeO (s) > 3 Fe (s) + Al2O3 (s) How many moles of calcium metal are produced from the decomposition of 8 mol of calcium chloride? CaCl2 (s) > Ca (s) + Cl2 (g) 2 Stoichiometry_Presentation_v_1.0.notebook 7 How many moles (total) of calcium metal and chlorine gas are produced from the decomposition of 8 mol of calcium chloride? March 19, 2014 Stoichiometric Calculations A balanced chemical equation is needed to perform any stoichiometric calculations. CaCl2 (s) > Ca (s) + Cl2 (g) 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 Stoichiometric Calculations Sample Problem 1 Stoichiometric Calculations Solution 1A N2 + 3H2 > 2NH3 How many moles of hydrogen gas are needed to produce 17 mol ammonia? (assuming an unlimited amount of nitrogen gas available.) 1 mol N2 3 mol H2 1 mol N2 3 mol H2 2 mol NH3 2 mol NH3 How many moles of hydrogen gas are needed to produce 17 mol ammonia? Method A Create a mole ratio that involves the "wanted" quantity in the numerator and the "given" quantity in the denominator. b mol of "wanted" substance W a mol of "given" substance G Consider the quantity that is given (ammonia) and the one that is asked for (hydrogen). The second ratio is the one we shall use because it incorporates both of these substances. a and b are coefficients aG bW given quantity wanted quantity Stoichiometric Calculations Stoichiometric Calculations Solution 1B Solution 1A How many moles of hydrogen gas are needed to produce 17 mol ammonia? Method A Always write the given quantity first. Then, multiply it by the mole ratio you created. b mol W x b x mol G x = mol W a mol G a Method B Use a mole ratio that involves the given quantity and wanted quantity. Set up a proportion and solve by cross mutliplying. 3 mol H2 = 2 mol NH3 given mole ratio calculated 17 mol NH3 x 3 mol H 2 2 mol NH3 = 25.5 mol H2 X = 17 mol NH3 x x mol H2 17 mol NH3 3 mol H2 2 mol NH3 = 25.5 mol H2 3 Stoichiometry_Presentation_v_1.0.notebook March 19, 2014 Stoichiometric Calculations Warm Up How many moles of lithium nitride can be produced from 5 moles of lithium reacting with an unlimited amount of nitrogen gas. 6 Li (s) + N2 (g) → 2 Li3N (s) MoleMole Practice 1 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? B. From the amount of nitric acid given in Part A, how many moles of silver nitrate will be produced? C. From the amount of nitric acid given in Part A, how many moles of water will be produced? D. From the amount of nitric acid given in Part A, how many moles of nitrogen monoxide will be made? Stoichiometric Calculations MoleMole Practice 2 2 N2H4(l) + N2O4(l) à 3 N2(g) + 4 H2O(g) Stoichiometry Calculations with Particles A second interpretation of a balanced equation is in terms of the particles (which could be molecules, atoms, or formula units). A. How many moles of dinitrogen tetrahydride are required to produce 57 moles of nitrogen? Dividing 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. Then this B. How many moles of dinitrogen tetroxide are required to produce 57 moles of nitrogen? can be read as: C. How many moles of water are produced when 57 moles of nitrogen are made? 2 H2 + O2 → 2 H2O 2 molecules of H2 plus 1 molecule of O2 yields 2 molecules of H2O. Not very different. But, while you can get answers that are fractions of moles...particles must be whole numbers. 8 What is the largest number of of Li3N formula units that could result from reacting 6 N2 molecules? 6 Li (s) + N2 (g) → 2 Li3N (s) 9 How many N2 molecules would be required to create 4 Li3N formula units? 6 Li (s) + N2 (g) → 2 Li3N (s) 4 Stoichiometry_Presentation_v_1.0.notebook March 19, 2014 Stoichiometry Calculations with Gases 10 How many Li atoms would be required to completely react with 3 N2 molecules? A third interpretation applies if there are only gases involved. If all the gases are at the same temperature and pressure, the coefficients give the relative volumes of gas. 6 Li (s) + N2 (g) → 2 Li3N (s) 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 This approach only works when comparing two gases in a formula. It cannot be used to compare a gas to a liquid or solid. [ * at STP conditions: 1 x 22.4 L + 3 x 22.4 L → 1 x 22.4 L + 2 x 22.4 L ] Stoichiometry Calculations with Gases Stoichiometry Calculations with Gases CS2 (g) + 3 O2 (g)→ CO2 (g) + 2 SO2 (g) Using this interpretation we can directly answer questions about the relative volumes of gas reactants and products. 1. Use the formula to set up a useful ratio. 2 L of SO2 3 L of O2 For instance, how many liters of SO2 will be produced when 6 liters of O2 are reacted as shown below? 2. Set that equal to the ratio of the known to unknown quantities of the same substances. CS2 (g) + 3 O2 (g)→ CO2 (g) + 2 SO2 (g) 2 L of SO2 V of SO2 = 3 L of O2 6 L of O2 3. Solve for the unknown. V of SO2 = 6 (2/3) L = 4 L The coefficients in the balanced equation give the ratio of moles of reactants and products. 11 What volume of methane is needed to completely react with 500 L of O2 at STP? __ CH4 + ___ O2 −−> ___ CO2 + ___ H2O 12 What volume of methane is needed to completely react with 500 L of O2 at STP? __ CH4 + ___ O2 −−> ___ CO2 + ___ H2O 5 Stoichiometry_Presentation_v_1.0.notebook March 19, 2014 14 How many liters of H2O (g) will be The equation below shows the created from reacting 8.0 L of H2 (g) with a sufficient amount of O2 (g)? 13 decomposition of lead nitrate. How many liters of oxygen are produced when 12 L NO2 are formed? 2 H2 (g) + O2 (g) → 2 H2O (g) 2Pb(NO3)2(s) → 2PbO(s) + 4NO2(g) + O2(g) V of H2O (g) = 8 L of H2 (g) 2 L of H2O (g) 2 L of H2 (g) V of H2O (g) = 8L(2/2) = 8 L Stoichiometry Calculations in General How many liters of NO (g) will be 2 15 created from reacting 36 L of O 2 (g) with a sufficient amount of NH3 (g)? If there is a mixture of gases and nongases in a formula, we can't directly determine the volume of gas generated. 4 NH3 (g) + 7 O2 (g) → 4 NO2 (g) + 6 H2O (g) 4 L of NO2 (g) V of NO2 (g) = 7 L of O2 (g) 36 L of O2 (g) However, we can determine that by first calculating the number of moles of gas generated and then converting that to a Volume. V of NO2 (g) = 36 L (4/7) = 21 L At STP, we know that each mole of gas will fill a volume of 22.4 L. In these general cases, we may need to convert the given masses, moles and/or volumes back to either moles or masses to solve the problem. The coefficients in the balanced equation give the ratio of moles of reactants and products. Stoichiometry Stoichiometry Coefficients in a balanced equation can represent either moles or particles (atoms, molecules, or formula units). As we have seen, we can interpret a balanced equation in several ways. By multiplying by MW, the formula coefficients relate masses. 2H2 + O2 −−> 2H2O 2H2 + O2 −−> 2H2O 2 molecules H2 + 1 molecule O2 −−> 2 molecules H2O 2 molecules H2 + 1 molecule O2 −−> 2 molecules H2O 2 mol H2 + 1 mol O2 −−> 2 mol H2O 2 mol H2 + 1 mol O2 −−> 2 mol H2O 4.0 amu H2 + 32.0 amu O2 −−> 36.0 amu H2O 4.0 amu H2 + 32.0 amu O2 −−> 36.0 amu H2O 4.0 g H2 + 32.0 g O2 −−> 36.0 g H2O 4.0 g H2 + 32.0 g O2 −−> 36.0 g H2O For formulas with only gases, the coefficients also represent volumes. The coefficients in the balanced equation give the ratio of moles of reactants and products. 2 L H2 (g) + 1 L O2 (g) → 2 L H2O (g) 6 Stoichiometry_Presentation_v_1.0.notebook March 19, 2014 Stoichiometry Which of the following statements is 3H2 + N2 à 2NH3 16 true for this equation showing the complete combustion of pentane? 3H2 N2 2NH3 6 atoms of hydrogen 2 atoms of nitrogen 2 atoms of nitrogen and 6 atoms of hydrogen 3 molecules of hydrogen 1 molecule of nitrogen 2 molecule of ammonia x 10= 30 molecule of H2 x10=10 molecules of N2 x10= 20 molecule of NH3 3 x 6.02.1023 molecule of H2 1 x 6.02.1023 molecule of N2 2 x 6.02.1023 molecule of NH3 3 mole of H2 1 mole of N2 2 mole of NH3 3 x 2 = 6g of hydrogen 1 x 28g = 28g of N2 2 x (14+3)= 34g of ammonia 3 x 22.4 L of hydrogen 1 x 22.4L of nitrogen 2 x 22.4L of ammonia 17 Which of these is an incorrect interpretation of this balanced equation? C5H12 (l) + 8 O2 (g) > 5 CO2 (g) + 6 H2O (g) 18 In any chemical reaction the quantities that are preserved are _____. 2 S (s) + 3 O2 (g) > 2 SO3 (g) Warm Up How many moles of lithium nitride can be produced from 5 moles of lithium reacting with an unlimited amount of nitrogen gas. 6 Li (s) + N2 (g) → 2 Li3N (s) Stoichiometric Calculations with Mass So far, we have seen stoichiometry problems that can be solved in one step, using only coefficients. These are: molemole problems, • particleparticle problems and • volumevolume problems. • When mass is involved, we must now take into consideration the fact that all particles have a different molar mass. Coefficients do not represent mass in grams. 7 Stoichiometry_Presentation_v_1.0.notebook March 19, 2014 MassMass Calculations MassMass 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 it’s a product) or used (if it’s a reactant). aA bB Given Sample Problem 1 Calculate the mass of ammonia, NH3, produced by the reaction of 5.4 g hydrogen gas with an excess of nitrogen. N2 + 3H2 > 2NH3 Find Grams of substance A Grams of substance B Using Method A, we proceed using these steps: a) b) c) g H2 > mol H2 > mol NH3 > g NH3 Use molar mass of A Use molar mass of B Use coefficients of A and B from balanced equation Moles of substance A a) convert 5.4 g H2 to moles H2 b) convert moles H2 to moles NH3 c) convert moles NH3 to grams NH3 Moles of substance B Massmass Calculations MassMass Calculations Sample Problem 2 Sample Problem 1 mol H2 −−> mol NH3 −−> g NH3 gH2 −−> How many grams of water can be produced from the combustion of 1.00 g of glucose? C6H12O6 + 6 O2 > 6 CO2 + 6 H2O Method A 1 mol H2 5.40 g H2 X 2.0 g H2 2 mol NH3 X 3 mol H2 X 17.0 g NH3 1 mol NH3 = 31 g NH3 c) convert moles NH3 to grams NH3 a) convert 5.4 g H2 to moles H2 b) convert moles H2 to moles NH3 Step (a): convert mass of glucose to moles 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 use the molar mass of water • Massmass Calculations Sample Problem 2 How many grams of water can be produced from the combustion of 1.00 g of glucose? C6H12O6 + 6 O2 > 6 CO2 + 6 H2O no direct calculation 1.00 g glucose X (c) 18.0 g water X 180.0 g glucose 5.56 x 103 mol glucose X 6 mol water Sample Problem 2 How many grams of water can be produced from the combustion of 1.00 g of glucose? 0.600 g water (a) 1mol glucose Massmass Calculations C6H12O6 + 6 O2 > 6 CO2 + 6 H2O There is an alternative way to solve this problem that requires interpreting the balanced equation in another way. 1 mol water 3.33 x 102 mol water 1 mol glucose (b) for every 1 mol glucose you get 6 mols of water 8 Stoichiometry_Presentation_v_1.0.notebook March 19, 2014 Massmass Calculations Massmass Calculations Sample Problem 2 Sample Problem 2 How many grams of water can be produced from the combustion of 1.00 g of glucose? How many grams of water can be produced from the combustion of 1.00 g of glucose? Method B requires first using the balanced equation to write down all the correct mass proportions of substances. Method B 6 O2 C6H12O6 6 CO2 6 H2O 1 mol C6H12O6 6 mol O2 6 mol CO2 6 mol H2O 180 g C6H12O6 192 g O2 264 g CO2 108 g H2O C6H12O6 + 6 O2 > 6 CO2 + 6 H2O C6H12O6 1 mol C6H12O6 6 O2 6 mol O2 6 H2O 6 CO2 6 mol CO2 6 mol H2O 6 x 32g = 6 x 44g = 6 x 18g = 1 x 180g = 180 g C6H12O6 192 g O2 264 g CO2 108 g H2O Massmass Calculations How many grams of water can be produced from the combustion of 1.00 g of glucose? Method B 108 g H2O = ratio formed from balanced equation x g H2O 1 g C6H12O6 Stoichiometry Calculations with Masses We now see that a fourth interpretation of a balanced equation is in terms of mass. Sample Problem 2 180 g C6H12O6 We choose to create a ratio with the highlighted quantities because these are the two substances that are part of the problem. x = 0.60 g H2O ratio using given quantity and unknown 2 H2 + O2 → 2 H2O A mole of H2 has a mass of 2 x 1.0 g = 2 g; so 2 moles has a mass of 4 g A mole of O2 has a mass of 2 x 16 g = 32 g; so 1 mole has a mass of 32 g A mole of H2O has a mass of 1 x 2 g + 1 x 16g = 18g; so 2 moles has a mass of 36 g So our equation can be read as: 4 g of H2 plus 32 g of O2 yields 36 g of H2O Stoichiometry Calculations with Masses Stoichiometric Calculations Using this interpretation we can answer questions about the relative masses of reactants and products. How many grams of Na can be produced from the decomposition of 40 g of NaN3? For instance, use the equation below formula to determine the mass of Na (s) that results from the decomposition of 40 g of NaN3 (s). 2 NaN3 (s) → 2 Na (s) + 3 N2 (g) Sample Problem 3A 2 NaN3 (s) → 2 Na (s) + 3 N2 (g) Method A a) 40 grams of NaN3 to moles NaN3 b) moles NaN3 to moles Na c) moles Na to grams Na 9 Stoichiometry_Presentation_v_1.0.notebook March 19, 2014 Massmass Calculations Massmass Calculations Sample Problem 3B Sample Problem 3B How many grams of Na can be produced from the decomposition of 40 g of NaN3? 2 NaN3 (s) → 2 Na (s) + 3 N2 (g) 130 g of NaN3 → 46 g of Na + 84 g of N2 4. Use the formula to set up a useful ratio. 46 g of Na 130 g of NaN3 1. First, determine the masses of the products and reactants. NaN3 = 1 x 23 g + 3 x 14 g = 65 g; so 2 moles = 130 g Na = 1 x 23 g = 23 g; so 2 moles = 46 g N2 = 2 x 14 g = 28 g; so 3 moles = 84 g 5. Set that equal to the ratio of the known to unknown quantities of the same substances. m of Na 2. Rewrite the formula in terms of mass. 130 g of NaN3 → 46 g of Na + 84 g of N2 40 g of NaN3 created from reacting 36 g of Al with a sufficient amount of O2? 46 g of Na 130 g of NaN3 6. Solve for the unknown. 3. Double check to see that mass of products = mass of reactants 130 g = 46 g + 84 g 19 How many grams of Al2O3 will be = m of Na = 40 g x (46g/130g) = 14 g How many grams of Mg must react in 20 order to to create 84 g of MgO? 2 Mg (s) + O2 (g) → 2 MgO (s) 4 Al (s) + 3 O2 (g) → 2 Al2O3 (s) 1. Mg = 1 x 24.3 g = 24.3 g; so 2 moles = 48.6 g O2 = 2 x 16 g = 32 g; so 1 mole = 32 g MgO = 1 x 24.3 g + 1 x 16g = 40.3 g; so 2 moles = 80.6 g 1. Al = 1 x 27 g = 27 g; so 4 moles = 108 g O2 = 2 x 16 g = 32 g; so 3 moles = 96 g Al2O3 = 2 x 27 g + 3 x 16g = 102 g; so 2 moles = 204 g 2. 48.6 g of Mg + 32 g of O2 → 80.6 g of MgO 2. 108 g of Al + 96 g of O2 → 204 g of Al2O3 3. 48.6 g + 32 g = 80.6 g; it checks 3. 108 g + 96 g = 204 g; it checks 4 & 5. 4 & 5. m of Al2O3 = 36 g of Al 204 g of Al2O3 108 g of Al m of Mg = 84 g of MgO 48.6 g of Mg 80.6 g of MgO 6. m of Mg = 84(48.6/80.6) = 51 g 6. m of Al2O3 = 36(204/108) = 68 g 21 What mass of CaO would be required to completely react with 42 g of H2O? CaO (s) + H2O (l) → Ca(OH)2 (s) 1. CaO = 1 x 40 g + 1 x 16 g = 56 g H2O = 2 x 1.0 g +1 x 16 g = 18 g Ca(OH)2 = 1 x 40 g + 2 x 16 g + 2 x 1.0 g = 74 g Massmass Calculations Practice Problem 4 __ CaCl2 (aq) + __ AgNO3 (aq) à __ Ca(NO3)2 (aq) + __ AgCl (s) Balance the equation above. How many grams of silver chloride are produced when 45 g of calcium chloride react with excess silver nitrate? 2. 56 g of CaO + 18 g of H2O → 74 g of Ca(OH)2 3. 56 g + 18 g = 74 g; it checks 4 & 5. m of CaO = 42 g of H2O 56 g of CaO 18 g of H2O 6. m of CaO = 42(56/18) = 131 g 10 Stoichiometry_Presentation_v_1.0.notebook Massmass Calculations March 19, 2014 Mixed Stoichiometry Calculations Practice Problem 5 Fe2O3 (s) + C (s) > Fe (s) + CO2 (g) Balance the equation above. How many grams of iron can be recovered from 500 kg of the oxide ore, Fe2O3? Thus far, our problemsolving 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 Stoichiometric Calculations Every type of stoichiometry calculation may be solved by following this map. Mixed Problem 1A (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. 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 Method A (3) (1) represenative particles of G X 1 mol G x 6.02 x1023 6.02 x 1023 representative = particles of W 1 mol W Since the "Given" quantity is alread in moles, we can skip to step (2). (2) mass 1 mol G of G X mass G b mol W mol G X mol W mass W x 1 mol W a mol G 1 mol G volume of G X at STP 22.4 L G x 22.4 L W 1 mol W mass = of W Overview: mol N2 > mol NH3 > grams NH3 Volume of = W at STP Stoichiometric Calculations Stoichiometric Calculations Mixed Problem 1A Mixed Problem 1B 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 Method A We can approach this same problem using another method. Method B requires first writing down all the relative proportions of substances using the balanced equation. N2 10 mol N2 x How many grams of ammonia can be produced by using 10 mol of nitrogen gas and an unlimited or excess amount of hydrogen gas? 2 mol NH3 17 g NH3 x 1 mol N2 1 mol NH3 = 340 g NH3 3H2 2NH3 1 mol N2 3 mol H2 2 mol NH3 28 g N2 6 g H2 34 g NH3 67.2 L H2 44.8 L NH3 22.4 L N2 Any two of these quantities may be used to create a ratio. 11 Stoichiometry_Presentation_v_1.0.notebook Stoichiometric Calculations Mixed Problem 1B How many grams of ammonia can be produced by using 10 mol of nitrogen gas and an unlimited or excess amount of hydrogen gas? Method B We choose the two highlighted quantities to create a ratio because they are the quantities that are involved in the problem. N2 3H2 2NH3 1 mol N2 3 mol H2 2 mol NH3 28 g N2 6 g H2 34 g NH3 67.2 L H2 44.8 L NH3 22.4 L N2 Stoichiometric Calculations Mixed Problem 2 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? March 19, 2014 Stoichiometric Calculations Mixed Problem 1B How many grams of ammonia can be produced by using 10 mol of nitrogen gas and an unlimited or excess amount of hydrogen gas? Method B 34 g NH3 x g NH3 = 1 mol N2 x = 340 g NH3 10 mol N2 ratio formed from balanced equation ratio using given quantity and unknown Stoichiometric Calculations Mixed Problem 2 2 NaN3 (s) → 2 Na (s) + 3 N2 (g) 1. Convert 36 L of N2 to moles: 36 L (1 mole / 22.4 L) = 1.6 moles 2. Use the formula to determine the needed moles of NaN3. 2 mol NaN3 n mol NaN3 = 3 mol N2 1.6 mol N2 3. Solve for the unknown. In this case, let's convert both the quantities of NaN3 that we're solving for and the volume of N2 we're being given to moles. Then we can directly use the above equation. This approach will always yield a result. n = 1.6 mol (2/3) = 1.1 mol NaN3 4. Determine the mass of 1 mol NaN3 (M) and multiply by n. M for NaN3 = 1 x 23 g + 3 x 14 g = 65 g/mol m = nM = (1.1 mol)(65 g/mol) = 72 g How many liters of O2 (g) at STP are 22 required to create 90 g of Al2O3 (s)? Stoichiometric Calculations 4 Al (s) + 3O2 (g) → 2 Al2O3 (s) Mixed Problem 3 2 Sb + 3 Cl2 à 2 SbCl3 M for Al2O3 = 2 x 27 g + 3 x 16 g = 102 g/mol How many grams of Cl2 are needed to react with 1 mol of Sb? n = m/M = (90 g)(102 g/mol) = 0.88 mol 3 mol O2 n mol O2 = 0.88 mol Al2O3 2 mol Al2O3 n mol O2 = m/M = (0.88 mol)(3/2) = 1.32 mol O2 V = nVm = (1.32 mol)(22.4 mol/L) = 29.57 L O2 12 Stoichiometry_Presentation_v_1.0.notebook Stoichiometric Calculations Mixed Problem 4 3MnO2(s) + 4 Al(s) à 2 Al2O3(s) + 3Mn(s) A. How many manganese atoms are produced if 54.8 moles of Mn3O4 react with excess aluminum. March 19, 2014 Stoichiometric Calculations Mixed Problem 4 3MnO2(s) + 4 Al(s) à 2 Al2O3(s) + 3Mn(s) C. How many moles of manganese oxide will react with 5.33 x 1025 atoms of aluminum? B. How many moles of aluminum oxide are made if 3580 g of manganese oxide are consumed? D. If 4.37 moles of aluminum are consumed, how many molecules of aluminum oxide are produced? Stoichiometric Calculations Mixed Problem 5 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: Stoichiometric Calculations Mixed Problem 5 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? 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? D. Find the mass of tristearin required to produce 55.56 moles of water. B. At STP, what volume of carbon dioxide is produced in Part A? How Many Cookies Can I Make? Limiting Reagent & Excess Reagent Percent Yield & Theoretical Yield • 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). 13 Stoichiometry_Presentation_v_1.0.notebook How Many Cookies Can I Make? • In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make. March 19, 2014 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. 23 When two substances react to form products, the reactant which is used up is called the Limiting Reactants Limiting reagent problems are different from those done previously in that two quantities of the reactants 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. Two different methods are presented here. Limiting Reactants Consider the reaction between hydrogen and oxygen to yield water. 2 H2 (l) + O2 (g) > 2 H2O (l) Limiting Reactant & Excess Reactant In this example, ____ is the limiting reagent and ____is the excess reagent. Starting with 10 molecules of H2 and 7 molecules of O2, which reactant will run out first? Before reaction 10H2 and 7 O2 After reaction 10 H2O and 2O2 Before reaction 10H2 and 7 O2 After reaction 10 H2O and 2O2 14 Stoichiometry_Presentation_v_1.0.notebook Limiting Reactants Method A overview 1. Write a balanced equation. 2. Use the amount of each reactant given to calculate the amount of product that could be formed. 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 creates the smaller amount of product is thus the limiting reagent. Limiting Reactants 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. March 19, 2014 Limiting Reactants Limiting reagent problems are different from those done previously in that two quantities are given, instead of just one. There are various methods used to determine which reactant is the limiting one. Consider the previous example. Method A Example 1. Write a balanced equation. 2 H2 (l) + O2 (g) > 2 H2O (l) 2. Use the amount of each reactant given to calculate the amount of product that could be formed. 10 molecules H2 would theoretically yield 10 molecules water 7 molecules O2 would theoretically yield 14 molecules water Limiting Reactants Method B overview 1. Write a balanced chemical equation. In actuality, 10 molecules of water are formed. The larger of the two amounts is irrelevant and meaningless. 2. Write down both given quantities under the equation (include units). It is impossible to produce 14 molecules of water, since there is not enough hydrogen gas. 3. Convert all quantities to moles. 4. Whichever reactant creates the smaller amount of product is thus the limiting reagent. 4. Divide each # of moles by the coefficient in front of each reactant. The smallest value is the limiting reagent. Hydrogen (10 molecules) is the limiting reagent. The theoretical yield, or maximum amount of product that could be made, is 10 molecules of water. Limiting Reactants Method B Example: How much ammonia can be produced from 65 g nitrogen and 25 g hydrogen? 1. Write a balanced chemical equation. Write down both given quantities under the equation (include units). N2 + 3H2 > 2NH3 65 g 25 g 2. Convert all quantities to moles. 65 g N2 = 2.3 mol N2 25 g H2 = 12.5 mol H2 5. Use the given quantitiy of the LR to determine the theoretical yield. Limiting Reactants Method B Example: What mass of ammonia can be produced from 65 g nitrogen and 25 g hydrogen? N2 + 3H2 > 2NH3 3. Divide each # of moles by the coefficient in front of each reactant. The smallest value corresponds to the limiting reagent. 2.3 mol N2 / 1 = 2.3 12.5 mol H2 / 3 = 4.2 4. Use the given quantitiy of the LR to determine the theoretical yield. Since N2 is the LR, convert from 2.3 mol N2 to mass of NH3 using the stoichiometric ratio. Answer: 78 g NH3 15 Stoichiometry_Presentation_v_1.0.notebook 24 Which of the following would be the limiting reagent in the reaction: 2H2(g) + O2(g) → 2H2O(g) March 19, 2014 25 Which of the following is NOT a true statement concerning limiting and excess reagents? Limiting Reagent & Theoretical Yield Limiting Reagent & Theoretical Yield Sample Problem 1: Find the theoretical yield of AlCl3, if 27g Al and 71g Cl2 react. Here we will apply Method A. Sample Problem 2: Find the theoretical yield of AlCl3, if 100g of each Al and Cl2 react. Here, we will apply Method B. 1. See how much AlCl3 can be made from each reactant 2. Smaller amount identifies the LR 2Al(s) + 3Cl2(g) à 2AlCl3 ? g 71 g 27 g 1. Find the limiting reagent 2. Use the LR to determine the theoretical yield of AlCl3 2Al(s) + 3Cl2(g) à 2AlCl3 ? g 100 g 100 g 27g Al will yield _______ g AlCl3 71g Cl2 will yield _______ g AlCl3 100g Al /27g/mol 100g Cl2/71g/mol Therefore, the LR is ________ and the theoretical yield of AlCl3 is _________ grams. =3.70 mol Al = 1.41 mol Cl2 3.70/2 = 1.85 1.41/3 = 0.47 smallest number Limiting Reagent = Cl2 Limiting Reagent & Theoretical Yield Limiting Reagent & Theoretical Yield Practice Problem 1 2Al(s) + 3Cl2(g) à 2AlCl3 100 g 100 g ? g Now that we know that the Limiting Reagent is Cl2, we can use the amount of Cl2 to determine the theoretical yield of AlCl3. 3 x 71.0g 100g of Cl2 = 2 x 133.35g X g of AlCl3 1.41 mol Cl2 x 2 mol AlCl3 a) Calculate the theoretical yield of water if 6 g H2 and 160 g O2 are combined in a reaction vessel. OR b) How many grams of the excess reagent will be remaining in the vessel? 133.5 g AlCl3 x 3 mol Cl2 ____H2 + ____O2 > ____H2O 1 mol AlCl3 = 16 Stoichiometry_Presentation_v_1.0.notebook March 19, 2014 Limiting Reagent & Theoretical Yield Limiting Reagent & Theoretical Yield Warm Up Practice Problem 2 Solid copper reacts with a solution of silver nitrate. Write a balanced chemical reaction. What is the coefficient for silver nitrate in the balanced chemical reaction? Cu(s) + 2 AgNO3(aq) à Cu(NO3)2(aq) + 2 Ag(s) 26 Answer? Determine the limiting reactant if 100 g of copper reacts with 200 g of silver nitrate. Cu(s) + 2 AgNO3(aq) à Cu(NO3)2(aq) + 2 Ag(s) 27 A Cu B AgNO3 Limiting Reagent & Theoretical Yield Limiting Reagent & Theoretical Yield Practice Problem 2 Practice Problem 3 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? 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. Cu(s) + 2 AgNO3(aq) à Cu(NO3)2(aq) + 2 Ag(s) 28 Answer? Limiting Reagent & Theoretical Yield Practice Problem 4 ___N2H4(l) + ___N2O4(l) à ___N2(g) + ___H2O(g). A. Determine the limiting reagent if you begin with 400 g of N2H4 and 900 g of N2O4. ___N2H4(l) + ___N2O4(l) à ___N2(g) + ___H2O(g). 400 g 900 g C. What volume of nitrogen is produced at STP, assuming the reaction goes to completion? B. Find the number of liters of water produced at STP, assuming the reaction goes to completion (i.e. one of the reactants gets completely used up). 17 Stoichiometry_Presentation_v_1.0.notebook March 19, 2014 Limiting Reagent & Theoretical Yield ___N2H4(l) + ___N2O4(l) à ___N2(g) + ___H2O(g). 400 g 900 g D. Find the mass of excess reactant left over at the conclusion of the reaction. 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? Theoretical Yield vs. Actual Yield Theoretical yield Actual yield the maximum amount of product that can be made, based on the stoichiometry (i.e. balanced equation) the amount of product one actually produces and measures in the laboratory; usually less than the TY Credit to Tom Greenbowe Chemical Education Group at Iowa State University 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%. 29 For a given chemical reaction, the theoretical yield is _____ greater than the actual yield. • Percent yield is the ratio comparing the amount actually obtained (actual yield) to the maximum amount that was possible (theoretical yield). Actual Yield x 100 Percent Yield = Theoretical Yield 18 Stoichiometry_Presentation_v_1.0.notebook 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? • • • • March 19, 2014 Stoichiometry Practice Problem 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? 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? Stoichiometry Practice Problem 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) Definitions • 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. 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 Skills Practice 33 Answers 1. a. O2 b. 351 g CO2, 180 g H2O 2. 322 g CaF2, 322 g Al2(CO3)3 • Actual yield the amount of product that forms when a reaction is carried out in the laboratory 3. a. 18.6 g Ca3O2 left over b. 130 g • 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 4. 280 g K3P, 142 g Al2S3 19 Stoichiometry_Presentation_v_1.0.notebook March 19, 2014 20 Attachments Skill Practice 3234 answers.docx Stoichiometry Unit Review 2013.doc Schedule 225 2013.doc ChemQuest 3234 answers.docx micromole rockets entire lab 1011.pdf micromole rockets.notebook micromole rockets.pdf Skill Practice 3034.doc Lab limiting reagent smore.doc