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1 Section 8.1 The Arithmetic of Equations OBJECTIVES: Calculate the amount of reactants required, or product formed, in a non chemical process. 2 Section 8.1 The Arithmetic of Equations OBJECTIVES: Interpret balanced chemical equations in terms of interacting moles, representative particles, masses, and gas volume at STP. 3 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 4 Stoichiometry Greek for “measuring elements” The calculations of quantities in chemical reactions based on a balanced equation. We can interpret balanced chemical equations several ways. 5 1. In terms of Particles Element- made of atoms Molecular compound (made of only nonmetals) = molecules Ionic Compounds (made of a metal and nonmetal parts) = formula units (ions) 6 2H2 + O2 2H2O Two molecules of hydrogen and one molecule of oxygen form two molecules of water. 2 Al2O3 Al + 3O2 2 formula units Al2O3 and 3 molecules form 4 atoms Al O2 2Na + 2H2O 2NaOH + H2 7 Look at it differently 2H2 + O2 2H2O 2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water. 2 x (6.02 x 1023) molecules of hydrogen and 1 x (6.02 x 1023) molecules of oxygen form 2 x (6.02 x 1023) molecules of water. 2 moles of hydrogen and 1 mole of oxygen form 2 moles of water. 8 2. In terms of Moles 2 Al2O3 Al + 3O2 2Na + 2H2O 2NaOH + H2 The coefficients tell us how many moles of each substance 9 3. In terms of Mass The Law of Conservation of Mass applies We can check using moles 2H2 + O2 2H2O 2 moles H2 2.02 g H2 1 mole H2 = 4.04 g H2 1 mole O2 32.00 g O2 1 mole O2 = 32.00 g O2 36.04g gHH +O 36.04 +O 2 2 22 10 In terms of Mass 2H2 + O2 2H2O 2 moles H2O 18.02 g H2O 1 mole H2O = 36.04 g H2O 2H2 + O2 2H2O 36.04 g H2 + O2 = 36.04 g H2O 11 4. In terms of Volume 2H2 + O2 2H2O At STP, 1 mol of any gas = 22.4 L (2 x 22.4 L H2) + (1 x 22.4 L O2) (2 x 22.4 L H2O) NOTE: mass and atoms are always conservedhowever, molecules, formula units, moles, and volumes will not necessarily be conserved! 12 Practice: Show that the following equation follows the Law of Conservation of Mass: 2 Al2O3 Al + 3O2 13 Section 8.2 Chemical Calculations OBJECTIVES: Construct mole ratios from balanced chemical equations, and apply these ratios in mole-mole stoichiometric calculations. 14 Section 8.2 Chemical Calculations OBJECTIVES: Calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, representative particles, and volumes of gases at STP. 15 Mole to Mole conversions 2 Al2O3 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 possible conversion factors 16 Mole to Mole conversions How many moles of O2 are produced when 3.34 moles of Al2O3 decompose? 2 Al2O3 Al + 3O2 3.34 mol Al2O3 3 mol O2 2 mol Al2O3 = 5.01 mol O2 17 Practice: 2C2H2 + 5 O2 4CO2 + 2 H2O • If 3.84 moles of C2H2 are burned, how many moles (9.6 mol) of O2 are needed? •How many moles of C2H2 are needed to produce (8.95 mol) 8.95 mole of H2O? •If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed? (4.94 mol) 18 How do you get good at this? 19 Mass-Mass Calculations We do not measure moles directly, so what can we do? We can convert grams to moles Then do the math with the mole ratio Use the Periodic Table for mass values Balanced equation gives mole ratio! Then turn the moles back to grams Use Periodic table values 20 For example... If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form? 2Fe + 3CuSO4 Fe2(SO4)3 + 3Cu Answer = 17.2 g Cu 21 More practice... How many liters of CO2 at STP will be produced from the complete combustion of 23.2 g C4H10 ? Answer = 35.8 L CO2 What volume of Oxygen would be required? Answer = 58.2 L O2 22 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 17.5 L O2 1 mol O2 22.4 L O2 1 mol CH4 2 mol O2 22.4 L CH4 1 mol CH4 = 8.75 L CH4 23 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. 24 Shortcut for Volume-Volume: How many liters of H2O at STP are produced by completely burning 17.5 L of CH4 ? CH4 + 2O2 CO2 + 2H2O 17.5 L CH4 2 L H2O 1 L CH4 = 35.0 L H2O Note: This only works for Volume-Volume problems. 25 Section 8.3 Limiting Reagent & Percent Yield OBJECTIVES: Identify and use the limiting reagent in a reaction to calculate the maximum amount of product(s) produced, and the amount of excess reagent. 26 Section 8.3 Limiting Reagent & Percent Yield OBJECTIVES: Calculate theoretical yield, actual yield, or percent yield, given appropriate information. 27 “Limiting” Reagent 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? 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 28 How do you find out? Do two stoichiometry problems. The one that makes the least product is the limiting reagent. For example Copper reacts with sulfur to form copper ( I ) sulfide. If 10.6 g of copper reacts with 3.83 g S how much product will be formed? 29 If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed? 2Cu + S Cu2S 10.6 g Cu 3.83 g S Cu is the Limiting 1 mol Cu2S it 159.16 g Cu2S 1 mol Cu Reagent because 2less molproduct Cu 63.55g creates Cu 1 mol Cu2S = 13.3 g Cu2S 1 mol S 1 mol Cu2S 32.06g S 1 mol S 159.16 g Cu2S 1 mol Cu2S = 19.0 g Cu2S 30 Another example If 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many liters of gas will be produced? How many grams of solid? How much excess reagent remains? Limiting Reagent Simulation 31 Still another example If 10.3 g of aluminum are reacted with 51.7 g of CuSO4 how much copper will be produced? How much excess reagent will remain? 32 % 33 Yield The amount of product made in a chemical reaction. There are three types: 1. Actual yield- what you 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 34 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? What is the percent yield? 35 % Yield Problem AY is 6.78 g of Cu TY – Use the balanced equation and do a mass to mass calculation 2Al + 3 CuSO4 Al2(SO4)3 + 3Cu 3.92 g 3.92 g Al xg 1 mol Al 3mol Cu 63.55 g Cu 26.98 g Al 2 mol Al 1 mol Cu 13.9 g Cu % Yield = (AY / TY) x 100 = (6.78g / 13.9 g) x 100 = 48.9 % yield 36 Details Percent yield tells us how “efficient” a reaction is. Percent yield can not be bigger than 100 %. 37 The Heat of Reaction At the start of a reaction the reactants have a great heat content. Enthalpy is the amount of heat that a substance has at a given temperature and pressure (see Table 8.1 pg 190) The heat of a reaction is the heat that is released or absorbed during a chemical reaction. Heat of Reaction is represented by The symbol H 38 Heat can be added to make a reaction occur. This is an Endothermic Reaction. H will be positive. The reaction can be represented as follows: 2 NH3 N2 + 3H2 H = + 92.38 kJ or 2 NH3 + 92.38 kJ N2 + 3H2 39 Cold Pack – Endothermic Reaction 40 When heat is released by a chemical reaction, the reaction is said to be Exothermic. H will be negative Heat is most often released by a chemical reaction. The released heat can be represented by the following: Na + Cl2 2NaCl H = - 822.4 kJ or Na + Cl2 2NaCl + 822.4kJ How can you calculate the heat of a reaction??? 41 Glow Stick – Exothermic Reaction 42 Calculate the heat of reaction ( H), in kilojoules, for the reaction Use table 8.1 pg 190 4 CO2(g) 4 (-393.5) 2O2(g) + 4CO(g) 2 (0) (-1574) (0) + 4 (-110.5) + (-442) H = NRG of Products – NRG of Reactants – H = (-442 kJ) H = 1132 kJ (-1574) (+ indicates an endothermic rxn so) 4 CO2(g) + 1132kJ 2O2(g) + 4CO(g) How much energy is used to make 100g of CO(g)? 43 4 CO2(g) + 1132kJ x kJ 100 g CO 2O2(g) + 4CO(g) 100g 1 mol CO 1132 kJ 28.01 g CO 4 mol CO 1010 kJ 44