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Let’s make some Cookies! When baking cookies, a recipe is usually used, telling the exact amount of each ingredient. • If you need more, you can double or triple the amount Thus, a recipe is much like a balanced equation. Stoichiometry CH 2 (Vol. 2), page 38 I. Stoichiometry Greek Mr. Mole for “measuring elements” Defined: the study of quantitative relationship between the amts of reactants used & amts of products formed by a chem. rxn. • based on the Law of Conservation of Mass (mostly) REVIEW: There are 4 ways to interpret a balanced chemical equation #1. In terms of Particles An Element is made of atoms A Molecular compound (made of only nonmetals) is made up of molecules (Don’t forget the diatomic elements!!!) Ionic Compounds (made of a metal and nonmetal parts) are made of formula units a. Example: 2H2 + O2 → 2H2O Two molecules of hydrogen and one molecule of oxygen form two molecules of water. Balanced! b. Another example: 2Al2O3 Al + 3O2 2 formula units Al2O3 form 4 atoms Al and 3 molecules O2 Now read this: 2Na + 2H2O 2NaOH + H2 #2. In terms of Moles a. The coefficients tell us how many moles of each substance 2Al2O3 Al + 3O2 2Na + 2H2O 2NaOH + H2 b. A balanced equation is a Molar Ratio (more on this later) #3. In terms of Mass The Law of Conservation of Mass applies a. We can check mass by using moles (reactants). 2H2 + O2 2H2O 2 moles H2 1 mole O2 2.02 g H2 1 mole H2 32.00 g O2 1 mole O2 = 4.04 g H2 + = 32.00 g O2 36.04 gg H H22 ++ O2 36.04 reactants b. In terms of Mass (for products) 2H2 + O2 2H2O 2 moles H2O 18.02 g H2O = 36.04 g H2O 1 mole H2O 36.04 g H2 + O2 = 36.04 g H2O 36.04 grams reactant = 36.04 grams product The mass of the reactants must equal the mass of the products. #4. In terms of Volume At STP, 1 mol of any gas = 22.4 L 2H2 + O2 2H2O (2 x 22.4 L H2) + (1 x 22.4 L O2) (2 x 22.4 L H2O) A. Mole Ratio – a ratio between the numbers of moles of any two of the substances in a balanced chemical equation 1. Example: Write all possible mole ratios for: 2HgO (s) 2Hg(l) + O2(g) 2 mol HgO 2 mol HgO 2 mol Hg 1 mol O2 2 mol Hg 2 mol Hg 2 mol HgO 1 mol O2 1 mol O2 1 mol O2 2 mol HgO 2 mol Hg Practice Problem Write all the possible mole ratios for the following equation: 4 Al + 3 O2 2 Al2O3 4 Al + 3 O2 2 Al2O3 Answer: 4 mol Al 3 mol O2 3 mol O2 4 mol Al 2 mol Al2O3 4 mol Al 4 mol Al 2 mol Al2O3 3 mol O2 2 mol Al2O3 2 mol Al2O3 3 mol O2 Using Compound Masses II. Stoichiometric Calculations – p. 40 All stoichiometric calculations begin with a balanced chemical equation. Mole ratios are needed, as well as mass-to-mole conversions (using molar mass) A. stoichiometric mole-to-mole conversion 2K(s) + 2H20 (l) 2KOH (aq) + H2(g) • From the balanced equation we know that two moles of potassium yields one mole of hydrogen. • But how many moles of hydrogen is produced if we have only 0.0400 mol of K? Mole to Mole conversions 2Al2O3 Al + 3O2 • each time we use 2 moles of Al2O3 we will also make 3 moles of O2 2 moles Al2O3 3 mole O2 or Mole Ratio 3 mole O2 2 moles Al2O3 These are the two possible conversion factors to use in the solution of the problem. Mole to Mole conversions How many moles of O2 are produced when 3.34 moles of Al2O3 decompose? 2Al2O3 Al + 3O2 3.34 mol Al2O3 3 mol O2 2 mol Al2O3 = 5.01 mol O2 Conversion factor from balanced equation If you know the amount of ANY chemical in the reaction, you can find the amount of ALL the other chemicals! Steps to Calculate Stoichiometric Problems 1. Balance the equation! 2. Convert the given amount into moles (if not given in moles) 3. Set up mole ratios (to use to calculate moles of desired chemical) 4. If needed, convert to grams (using Molar Mass) Practice: 2C2H2 + 5 O2 4CO2 + 2 H2O • If 3.84 moles of C2H2 are burned, how many moles of O2 are needed? (9.60 mol) •How many moles of C2H2 are needed to produce 8.95 mole of H2O? (8.95 mol) •If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed? (4.94 mol) Practice Problem Methane gas and sulfur (S8) react to produce carbon disulfide (a liquid often used in the production of cellophane) and a colorless, poisonous gas of hydrogen sulfide (rotten egg smell). a. Write chemical equation (don’t forget to balance!) b. Calculate moles of CS2 produced when 1.50 mol S8 is used c. How many moles of H2S is produced? Practice Problem - ANSWER a. 2CH4(g) + S8(s) 2CS2(l) + 4H2S(g) b. find: moles of CS2 1.50 mol S8 2 mol CS2 = 3.00 mol CS2 1 mol S8 c. Find moles of H2S 1.50 mol S8 4 mol H2S = 6.00 mol H S 2 1 mol S8 Steps to Calculate Stoichiometric Problems 1. Balance the equation! 2. Convert to moles (if needed) 3. Set up mole ratios 4. If needed, convert to grams (using MM) B. Stoichiometric mole-to-mass calculations Suppose you know the number of moles of a reactant or product in a reaction & you want to calculate the mass of another product or reactant. Example: Determine the mass of NaCl produced when 1.25 mol of chlorine gas reacts with excess Na. 2 Na(s) + Cl2 (g) 2 NaCl (s) 1.25 mol Cl2 2 mol NaCl 1 mol Cl2 Mole Ratio!!! 58.44 g NaCl 1 mol NaCl Molar Mass!!! = 146 g NaCl Practice Problem Sodium chloride is decomposed into the elements sodium and chlorine gas by means of electrical energy. How much chlorine gas, in grams, is obtained when you have 2.50 mol NaCl? Answer: 88.6 g Cl2 Steps to Calculate Stoichiometric Problems 1. Balance the equation! 2. Convert to moles (if needed) 3. Set up mole ratios 4. If needed, convert to grams (using MM) C. Stoichiometric mass-to-mass • If you were preparing to carry out a chemical reaction in the laboratory, you would need to know how much of each reactant to use in order to produce the mass of product required. Mass-Mass Problem: 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4Al + 3O2 2Al2O3 6.50 g Al 1 mol Al 2 mol Al2O3 101.96 g Al2O3 26.98 g Al 4 mol Al 1 mol Al2O3 (6.50 x 1 x 2 x 101.96) ÷ (26.98 x 4 x 1) = = ? g Al2O3 12.3 g Al2O3 are formed How do you get good at this? Another example: If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form? 2Fe + 3CuSO4 Fe2(SO4)3 + 3Cu Answer = 17.2 g Cu Practice Problem One of the reactions used to inflate automobile air bags involves solid sodium azide (NaN3), which produces sodium and nitrogen gas. a. b. Write the chemical equation. Determine the mass of N2 produced from the decomposition of 100.0 g NaN3. Answer: 64.64 g N2 D. Volume Stoichiometry At STP, 1 mol of any gas = 22.4 L 2H2 + O2 2H2O (2 x 22.4 L H2) + (1 x 22.4 L O2) (2 x 22.4 L H2O) 67.2 Liters of reactant ≠ 44.8 Liters of product! NOTE: mass and atoms are ALWAYS conserved - however, molecules, formula units, moles, and volumes will not necessarily be conserved! Avogadro told us: Equal volumes of gas, at the same temperature and pressure contain the same number of particles. Moles are numbers of particles You can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same. 1 mole = 22.4 L @ STP Page 135 - 139 - Chapter 3: The KMT (Section entitled: “Stoichiometry Involving Gases”) Volume-Volume Calculations: How many liters of CH4 at STP are required to completely react with 17.5 L of O2 ? CH4 + 2O2 CO2 + 2H2O 1 mol O2 1 mol CH4 22.4 L CH4 17.5 L O2 22.4 L O2 2 mol O2 1 mol CH4 = 8.75 L CH4 Notice anything relating these two steps? Shortcut for Volume-Volume? How many liters of CH4 at STP are required to completely react with 17.5 L of O2? CH4 + 2O2 CO2 + 2H2O 17.5 L O2 1 L CH4 2 L O2 = 8.75 L CH4 Note: This only works for VolumeVolume problems. Calculations Grams molar mass Moles Avogadro’s number particles Everything must go through Moles!!! III. “Limiting” Reactants – p. 59 A. Why do reactions stop? If you are given one dozen loaves of bread, a gallon of mustard, and three pieces of salami, how many salami sandwiches can you make? 1. The limiting reactant is the reactant you run out of first. Determines how much product you can make! 2. The excess reactant is the one you have left over. • Often referred to as limiting reagent or excess reagent Limiting Reagents - Combustion How do you find out which is limited? 3. The chemical that makes the least amount of product is the “limiting reactant”. You can recognize limiting reactant problems because they will give you 2 amounts of chemical 4. Do two stoichiometry problems, one for each reactant given If 10.6 g of copper reacts with Cugrams is the of the 3.83 g sulfur, how many Limiting product (copper (I) sulfide) will be formed? Reagent, 2Cu + S Cu2S since it 1 mol Cu2S 159.16 g Cu2S 1 mol Cu produced less 10.6 g Cu Cu 63.55g Cu 2 mol 1 mol Cu2S product. = 13.3 g Cu2S 1 mol S 3.83 g S 32.06g S 1 mol Cu2S 159.16 g Cu2S 1 mol S 1 mol Cu2S = 19.0 g Cu2S Another example: If 10.3 g of aluminum are reacted with 51.7 g of CuSO4 how much copper (grams) will be produced? 2Al + 3CuSO4 → 3Cu + Al2(SO4)3 10.3 g Al 1 mol Al 3 mol Cu 63.546 g Cu = 36.4 g Cu 26.981 g Al 2 mol Al 1 mol Cu 51.7 g CuSO4 1 mol CuSO4 3 mol Cu 63.546 g Cu = 159.60 g CuSO4 3 mol CuSO4 1 mol Cu = 20.6 g Cu the CuSO4 is limited, so Cu = 20.6 g Another example: 2Al + 3CuSO4 → 3Cu + Al2(SO4)3 10.3 g Al 1 mol Al 3 mol Cu 63.546 g Cu = 36.4 g Cu 26.981 g Al 2 mol Al 1 mol Cu 51.7 g CuSO4 1 mol CuSO4 3 mol Cu 63.546 g Cu = 159.60 g CuSO4 3 mol CuSO4 1 mol Cu = 20.6 g Cu How much excess reactant will remain? • Means how much excess Al will remain!!! 36.4 g Cu – 20.6 g Cu = 15.8 g Cu How much excess reactant will remain? 36.4 g Cu – 20.6 g Cu = 15.8 g Cu 1 mol Cu 15.8 g Cu 63.55g Cu 2 mol Al 3 mol Cu 26.981 g Al 1 mol Al = 4.47 g Al ANOTHER Example Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose and oxygen. A plant has 88.0 g of carbon dioxide and 64.0 g of water available for photosynthesis. 1. Determine the limiting reactant. 2. Determine the excess reactant. 3. Determine the amount of the reactant in excess. Write a balanced chemical reaction: 6CO2(g) + 6H2O(l) C6H12O6(aq) + 6O2(g) A plant has 88.0 g of carbon dioxide and 64.0 g of water available for photosynthesis. 1. Find limiting reactant 6CO2(g) + 6H2O(l) C6H12O6(aq) + 6O2(g) 88.0 g CO2 1 mol CO2 6 mol O2 31.988 g O2 43.99 g CO2 6 mol CO2 1 mol O2 == 64.0 64.0 gg O O22 64.O g H2O 1 mol H2O 6 mol O2 18.01 g H2O 6 mol H20 Limiting Reactant is CO2! 31.988 g O2 1 mol O2 = 114 g O2 2. Find excess reactant 6CO2(g) + 6H2O(l) C6H12O6(aq) + 6O2(g) 88.0 g CO2 1 mol CO2 6 mol O2 31.988 g O2 43.99 g CO2 6 mol CO2 1 mol O2 = 64.0 g O2 64.O g H2O 1 mol H2O 6 mol O2 18.01 g H2O 6 mol H20 EXCESS Reactant is H2O! 31.988 g O2 1 mol O2 = 113.71 g O2 6CO2(g) + 6H2O(l) C6H12O6(aq) + 6O2(g) 3. Determine the mass (of the reactant) in excess. 114 g O2 – 64.0 g O2 = 50.0 g O2 50.O g O2 1 mol O2 31.998 g O2 6 mol H20 18.02 g H2O 6 mol O2 1 mol H2O = 28.2 g H2O Practice Problem The reaction between solid sodium and iron (III) oxide yields solid sodium oxide and iron. (Another reaction, in a series, that inflates an automobile airbag). If 100.0 g of Na and 100.0 g of iron (III) oxide are used in this reaction, determine the following: a. Limiting reactant b. Excess reactant c. Mass of solid iron produced d. Mass of excess reactant that remains after the reaction is complete Practice Problem - ANSWERS 6Na(s) + Fe2O3(s) 3Na2O(s) + 2Fe(s) a. Fe2O3 b. Na c. 69.92 g Fe d. 13.6 g Na The Concept of: A little different type of yield than you had in Driver’s Education class. What is YIELD? IV. Percent Yield – p. 67 A. Yield is the amount of product made in a chemical reaction. There are three types: 1. Actual yield- what you actually get in the lab when the chemicals are mixed • Or the amount of product produced when the chemical reaction is carried out in an experiment 2. Theoretical yield- what the balanced equation tells should be made • The maximum amount of product that can be produced from a given amount of reactant 3. Percent yield: Actual x 100 Theoretical • Percent yield of the product is the ratio of the actual yield to the theoretical yield, expressed as a percent – Percent yield tells us how “efficient” a reaction is. • A way of measuring the efficiency of the reaction in producing a product – Percent yield can not be bigger than 100 %. Theoretical yield will always be larger than actual yield! – Why? Due to impure reactants, competing side reactions, loss of product in filtering or transferring between containers, measuring Example: 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. Write balanced equation: 2Al + 3 CuSO4 Al2(SO4)3 + 3Cu What is the actual yield? What is the theoretical yield? What is the percent yield? Example: 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 2Al + 3 CuSO4 Al2(SO4)3 + 3Cu What is the actual yield? = 6.78 g Cu What is the theoretical yield? 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 2Al + 3 CuSO4 Al2(SO4)3 + 3Cu What is the actual yield? = 6.78 g Cu What is the theoretical yield? 3.92 g Al 1 mol Al 3 mol Cu 63.546 g Cu 26.982 g Al 2 mol Al 1 mol Cu = 13.8481 g Cu = 13.8 g Cu 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 2Al + 3 CuSO4 Al2(SO4)3 + 3Cu What is the actual yield? = 6.78 g Cu What is the theoretical yield? = 13.8 g Cu What is the percent yield? actual yield X 100 Percent Yield = theoretical yield 6.78 g Cu 13.8 g Cu X 100 = 49.1 % Your Turn: Aluminum hydroxide is often present in antacids to neutralize stomach acid (HCl). The reaction is as follows: Al(OH)3(s) + 3HCl(aq) AlCl3(aq) + 3H2O(l) a. b. If 14.0 g of Al(OH)3 is present in an antacid tablet, determine the theoretical yield of AlCl3 produced when the tablet reacts with HCl. What is the percent yield if 20.5 g is recovered? Practice Problem - ANSWER Al(OH)3(s) + 3HCl(aq) AlCl3 (aq) + 3H2O(l) If 14.0 g of Al(OH)3 is present, determine the theoretical yield of AlCl3 produced. 14.0 g Al(OH)3 1 mol Al(OH)3 1 mol AlCl3 133.34 g AlCl3 78.004 g Al(OH)3 1 mol Al(OH)3 1 mol AlCl3 = 23.9 g AlCl3 (20.5 g / 23.9 g ) x 100% = 85.7%