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Chapter 12 “Stoichiometry” Mr. Mole Stephen L. Cotton Section 12.1 The Arithmetic of Equations OBJECTIVES: • Explain how balanced equations apply to both chemistry and everyday life. Section 12.1 The Arithmetic of Equations OBJECTIVES: • Interpret balanced chemical equations in terms of: a) moles, b) representative particles, c) mass, and d) gas volume (Liters) at STP. Section 12.1 The Arithmetic of Equations OBJECTIVES: • Identify the quantities that are always conserved in chemical reactions. STOICHIOMETRY E.Q.: What mathematical relationships can be determined from a balanced chemical equation? 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 is… Greek for “measuring elements” Pronounced “stoy kee ahm uh tree” Defined as: calculations of the quantities in chemical reactions, based on a balanced equation. There are 4 ways to interpret a balanced chemical equation #1. In terms of Particles An Element is made of atoms A covalent 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 Example: 2H2 + O2 → 2H2O Two molecules of hydrogen and one molecule of oxygen form two molecules of dihydrogen monoxide (water). Another example: 2Al2O3 Al + 3O2 2 formula units aluminum oxide form 4 atomsAl and 3 molecules oxygen gas Now read this: 2Na + 2H2O 2NaOH + H2 #2. In terms of Moles The coefficients tell us how many moles of each substance 2Al2O3 Al + 3O2 2Na + 2H2O 2NaOH + H2 Remember: A balanced equation is a Molar Ratio #3. In terms of Mass The Law of Conservation of Mass applies We can check mass by using moles. 2H2 + O2 2H2O 2 moles H2 2.016 g H2 = 4.032 g H2 1 mole H2 + 1 mole O2 31.998 g O2 = 31.998 g O2 1 mole O2 36.04 ggHH2 2++OO2 2 36.030 reactants In terms of Mass (for products) 2H2 + O2 2H2O 2 moles H2O 18.015 g H2O = 36.030 g H2O 1 mole H2O 36.030 g H2 + O2 = 36.030 g H2O 36.030 grams reactants = 36.030 grams products 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) 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! Practice: Show that the following equation follows the Law of Conservation of Mass (show the atoms balance, and the mass on both sides is equal) 3 2 4 + __O 2 2O3 Al __Al Section 12.2 Chemical Calculations OBJECTIVES: • Construct “mole ratios” from balanced chemical equations, and apply these ratios in mole-mole stoichiometric calculations. Section 12.2 Chemical Calculations OBJECTIVES: • Calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, representative particles, and volumes of gases at STP. 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 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. Correctly balance the equation. 2. Convert the given amount into moles. 3. Set up mole ratios. 4. Use mole ratios to calculate moles of desired chemical. 5. Convert moles back into final unit. Practice: 2C2H2 + 5 O2 4CO2 + 2 H2O • If 3.84 moles of C2H2 are burned, how many moles of O2 are needed?(9.6 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) How do you get good at this? Assignment #1 2 KClO3 ---> 2 KCl + 3 O2 1. How many moles of O2 can be produced by letting 12.00 moles of KClO3 react? 2. How many moles of KClO3 are needed to produce 5.45 moles of KCl? 3. If 10.4 moles of KCl were produced, how many moles of O2 were also produced? 4. How many moles of KCl can be produced by letting 7.5 moles of KClO3 decompose? 5. How many moles of KClO3 are needed to produce 5.5 moles of KCl? 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 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? 2_Fe + 3 _CuSO4 Fe2(SO4)3 + 3_Cu Answer = 17.2 g Cu Assignment #2 2 K + Cl2 ---> 2 KCl 1. How many grams of KCl are produced from 2.50 g of K and excess Cl2 ? 2. How many grams of KCl are produced from 1.00 g of Cl2 and excess K? 3. What mass of Cl2 is needed to produce 5.75 grams of KCl? 4. How many grams of K are needed to produce 2.57 grams of KCl? Other conversion examples…. Mass to particles – Particles to Mass 2Al2O3 Al + 3O2 1) If 2.55 grams of aluminum oxide are available to decompose, how many atoms of aluminum can be produced? 2) If 3.15 x 1025 molecules of O2 need to be produced, how many grams of Al2O3 are needed for this reaction to take place? Other conversion examples…. Mass to moles – Moles to Mass 2Al2O3 Al + 3O2 1) If 2.55 grams of aluminum oxide are available to decompose, how many moles of aluminum can be produced? 2) If 3.5 moles of O2 need to be produced, how many grams of Al2O3 are needed for this reaction to take place? Assignment #3 Na2O + H2O ---> 2 NaOH 1) How many grams of NaOH is produced from 1.20 x 102 grams of Na2O? 2) How many grams of Na2O are required to produce 1.60 x 102 grams of NaOH? 3) How many grams of Na2O are required to produce 3.25 x 1025 units of NaOH? 4) How many molecules of H2O are needed to produce 4.75 grams of NaOH? 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? 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 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 Volume-Volume problems. Section 12.3 Limiting Reagent & Percent Yield OBJECTIVES: • Identify the limiting reagent in a reaction. Section 12.3 Limiting Reagent & Percent Yield OBJECTIVES: • Calculate theoretical yield, percent yield, and the amount of excess reagent that remains unreacted given appropriate information. LIMITING & EXCESS REAGENTS E.Q.: HOW CAN A BALANCED EQUATION HELP US DETERMINE THE LIMITING AND EXCESS REAGENTS IN A CHEMICAL REACTION? LIMITING & EXCESS REAGENTS WE WILL DETERMINE LIMITING & EXCESS REAGENTS THROUGH CALCULATIONS. I WILL DESCRIBE THE STEPS NEEDED TO CALCULATE THE LIMITING REAGENT BY WRITING AN EXIT TICKET. “Limiting” Reagent If you are given one dozen loaves of bread, a gallon of mustard, and three pieces of bologna, how many bologna sandwiches can you make? The limiting reagent is the reactant you run out of first. The excess reagent is the one you have left over. The limiting reagent determines how much product you can make Limiting Reagents - Combustion How do you find out which is limited? The chemical that makes the least amount of product is the “limiting reagent”. You can recognize limiting reagent problems because they will give you 2 amounts of chemical Do two stoichiometry problems, one for each reagent you are given. If 10.6 g of copper reacts with Cugrams is the of copper 3.83 g sulfur, how many Limiting (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 Limiting Limiting Reactant 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl2 2 AlCl3 Start with Al: 10.0 g Al Reactant: Example 2 1 mol Al 27.0 g Al Now Cl2: 35.0g Cl2 1 mol Cl2 71.0 g Cl2 2 mol AlCl3 133.5 g AlCl3 2 mol Al 1 mol AlCl3 2 mol AlCl3 133.5 g AlCl3 3 mol Cl2 1 mol AlCl3 = 49.4g AlCl3 = 43.9g AlCl3 Limiting Reagent Problems…..Your Turn…. 1) If 10.3 g of aluminum are reacted with 51.7 g of CuSO4, which reactant is the limiting reagent? __Al + __CuSO4 → __Cu + __Al2(SO4)3 2) 15.0 g of sodium oxide reacts with 15.0 g of water. Calculate which reactant is limiting and how much product is made. __Na2O + __H2O ---> __NaOH 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 the CuSO4 is limited, so Cu = 20.6 g How much excess reagent will remain? Excess = 4.47 grams Limiting & Excess Reagents Example: 15.0 g of potassium reacts with 15.0 g of iodine. Calculate how much potassium iodide is produced and how much excess is left over? 2 K + I2 2 KI 15.0 g K 1 mol K 2 mol KI 39.098 g K 2 mol K 15.0 g I2 1 mol I2 253.808 g I2 2 mol KI 1 mol I2 166.002 g KI = 63.7 g KI 1 mol KI Excess Reagent 166.002 g KI 1 mol KI = 19.6 g KI Limiting Reagent Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2 2 KI We found that Iodine is the limiting reagent, and 19.6 g of potassium iodide are produced. 15.0 g I2 1 mol I2 2 mol K 39.098 g K 253.808 g I2 1 mol I2 1 mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = 10.38 g K EXCESS Given amount of excess reagent Amount of excess reagent actually used Note that we started with the limiting reagent! The Concept of: A little different type of yield than you had in Driver’s Education class. What is Yield? 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 2. Theoretical yield- what the balanced equation tells should be made 3. Percent yield = Actual x 100 Theoretical 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? What is the theoretical yield? = 13.8 g Cu What is the percent yield? = 6.78 g Cu = 49.13 % Details on Yield Percent yield tells us how “efficient” a reaction is. yield can not be bigger than 100 %. Theoretical yield will always be larger than actual yield! Percent • Why? Due to impure reactants; competing side reactions; loss of product in filtering or transferring between containers; measuring Percent Yield 2KClO3 → 2KCl + 3O2 2) If 5.00 g of KClO3 is decomposed to form potassium chloride and oxygen, what is the percent yield of the reaction if 1.25 g of O2 are produced? Ans: 63.78% Your Turn…… 2AlCl3 (s) → 2Al(s) + 3Cl2 (g) 1) If 6.25 g of AlCl3 is decomposed to form aluminum and chlorine gas, what is the percent yield of the reaction if 2.25 g of Cl2 are produced? H3PO4 + 3 KOH ---> K3PO4 + 3 H2O 2) If 49.0 g of H3PO4 is reacted with excess KOH, determine the percent yield of K3PO4 if you isolate 49.0 g of K3PO4.