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Chapter 12 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 A. Proportional Relationships 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. I have 5 eggs. How many cookies can I make? Ratio of eggs to cookies 5 eggs 5 doz. 2 eggs = 12.5 dozen cookies 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. In terms of Particles Element- made of atoms Molecular compound (made of only non- metals) = molecules Ionic Compounds (made of a metal and non-metal parts) = formula units (ions) 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 form 4 atoms Al and 3 molecules O2 2Na + 2H2O 2NaOH + H2 Looking 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. In terms of Moles Al + 3O2 2Na + 2H2O 2NaOH + H2 The coefficients tell us how many moles of each substance are needed. 2 Al2O3 In terms of Mass The Law of Conservation of Mass applies We can check using moles 2H2 + O2 2H2O 2.02 g H2 2 moles H2 = 4.04 g H2 1 moles H2 32.00 g O2 1 moles O2 = 32.00 g O2 1 moles O2 36.04 g H2+O2 In terms of Mass 2H2 + O2 2H2O 18.02 g H2O 2 moles H2O = 36.04 g H2O 1 mole H2O 2H2 + O2 2H2O 4.04 g H2 + 32.00 O2 = 36.04 g H2O A. Proportional Relationships Stoichiometry • mass relationships between substances in a chemical reaction • based on the mole ratio Mole Ratio • indicated by coefficients in a balanced equation 2 Mg + O2 2 MgO B. Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. •• Mole molesmoles moles Mole ratio ratio -moles • Molar mass moles grams • Molarity moles liters soln. • Molar volume moles liters gas Core step in all stoichiometry problems!! 4. Check answer. Mole to Mole conversions 2 Al2O3 Al + 3O2 every time we use 2 moles of Al2O3 we make 3 moles of O2 2 moles Al2O3 3 mole O2 or 3 mole O2 2 moles Al2O3 These are two possible conversion factors 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 moles O2 = 5.01 moles O2 2 moles Al2O3 Your Turn 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) How do you get good at this? Hydrogen Fuel Cell DVD Mass in Chemical Reactions How much do you make? How much do you need? Mass-Mass Calculations We do not measure moles directly, so what can we do? We can convert grams to moles • Use the Periodic Table for mass values Then do the math with the mole ratio • Balanced equation gives mole ratio! Then turn the moles back to grams • Use Periodic table values Mass-Mass Conversion If 10.1 g of Fe+3 are added to a solution of Copper (II) Sulfate, how much solid copper would form? Fe + CuSO4 Fe2(SO4)3 + Cu 2Fe + 3CuSO4 Fe2(SO4)3 + 3Cu 10.1 g Fe 1 mol Fe 55.85 g Fe = 0.181 mol Fe 2Fe + 3CuSO4 Fe2(SO4)3 + 3Cu 3 mol Cu 0.181 mol Fe = 0.272 mol Cu 2 mol Fe 63.55 g Cu 0.272 mol Cu = 17.2 g Cu 1 mol Cu All Together It Looks Like 10.1 g Fe 1 mol Fe 3 mol Cu 63.55 g Cu 55.85 g Fe 2 mol Fe 1 mol Cu = 17.2 g Cu Examples To make silicon for computer chips they use this reaction SiCl4 + 2Mg 2MgCl2 + Si How many grams of Mg are needed to make 9.3 g of Si? How many grams of SiCl4 are needed to make 9.3 g of Si? How many grams of MgCl2 are produced along with 9.3 g of silicon? Gases and Reactions We Can Also Change Movie Liters of a gas to moles If we are at STP, Standard Temperature and Pressure, which is 0ºC and 1 atmosphere pressure At STP 22.4 L of a gas = 1 mole of any gas molecules Molar Volume at STP 1 mol of a gas=22.4 L at STP Standard Temperature & 0°C and 1 atm. Pressure LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS Molar Mass (g/mol) 6.02 MOLES 1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION NUMBER OF PARTICLES 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 Gas Stoichiometry Problems How many grams of KClO3 are req’d to produce 9.00 L of O2 at STP? 2KClO3 2KCl + 3O2 ?g 9.00 L 9.00 L O2 1 mol O2 2 mol KClO3 122.55 g KClO3 22.4 L O2 3 mol O2 1 mol KClO3 = 32.8 g KClO3 Stoichiometry in the Real World Limiting Reactants and Yield Limiting Reactants Available Ingredients • 4 slices of bread • 1 jar of peanut butter • 1/2 jar of jelly Limiting Reactant • bread Excess Reactants • peanut butter and jelly Limiting Reactants Limiting Reactant • used up in a reaction • determines the amount of product Excess Reactant • Not completely used up • added to ensure that the other reactant is completely used up • cheaper & easier to recycle Limiting Reactants 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates, • the limiting reactant • and the amount of product formed. Limiting Reactants 79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP? Zn + 2HCl 0.90 L 79.1 g 2.5M ZnCl2 + H2 ?L Limiting Reactants Zn + 2HCl 79.1 g 0.90 L 2.5M 79.1 g Zn 1 mol Zn 65.39 g Zn ZnCl2 + H2 ?L 1 mol H2 22.4 L H2 1 mol Zn 1 mol H2 = 27.1 L H2 Limiting Reactants Zn + 2HCl 79.1 g 0.90 L 2.5M 0.90 L 2.5 mol HCl 1L 1 mol H2 2 mol HCl ZnCl2 + H2 ?L 22.4 L H2 = 25 L 1 mol H2 H2 Limiting Reactants Zn: 27.1 L H2 HCl: 25 L H2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25 L H2 left over zinc Another Example 1. Do two stoichiometry problems. 2. The one that makes the least product is the limiting reagent. Copper reacts with sulfur to form copper (I) sulfide. If 10.6 g of copper reacts with 3.83 g S how many grams of product will be formed? If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed? Cu is 2Cu + S Cu2S Limiting 1 mol Cu2S 159.16 g Cu2S 1 mol Cu Reagent 10.6 g Cu 63.55g Cu 2 mol Cu 1 mol Cu2S = 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 reactants How many grams of H2O is produced if 20.0 grams of H2 react with 20.0 grams of O2? Which reactant is limiting? Which reactant is excess? How much excess remains? 2H2 + O2 2H2O 20. 0 g H2 Start with one of your givens 20.0 g O2 Start with your other given 1 mol H2 2.02 g H2 2 mol H2O 1 mol O2 2 mol H2O 32.00 g O2 2 mol H2 1 mol O2 18.02 g H2O = 178 g H2O 1 mol H2O 18.02 g H2O = 22.5 g H2O 1 mol H2O To find the answer for how many grams of H2O produced compare both of your answers, the lowest number is the correct answer. Limiting Reactants: How to find excess To find out how much excess is left over you take what you produce and determine excess reactant actually used. 22.5 g H2O 1 mol H2O 2 mol H2 18.02g H2O 2 mol H2O 2.02 g H2 1 mol H2 = 2.52 g H2 This is how many Based on limiting reactant grams you used up. Now take the answer and subtract from your given amount of excess. The answer that you find is how much excess is left over. 20.0g – 2.52g = 17.5g H2 Your Turn 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 will be produced? How many atoms of Cl are in the solid? How much excess reagent remains? Your Turn II 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? Yield The amount of product made in a chemical reaction. There are three types Actual yield- what you get in the lab when the chemicals are mixed Theoretical yield- what the balanced equation tells you you should make. Percent yield Percent Yield measured in lab actual yield % yield 100 theoretical yield calculated on paper Percent Yield When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g ?g actual: 46.3 g Percent Yield K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g ?g actual: 46.3 g Theoretical Yield: 45.8 g K2CO3 1 mol K2CO3 138.21 g K2CO3 2 mol KCl 74.55 g KCl 1 mol K2CO3 = 49.4 1 mol g KCl KCl Percent Yield K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g 49.4 g actual: 46.3 g Theoretical Yield = 49.4 g KCl % Yield = 46.3 g 49.4 g 100 = 93.7% Details Percent yield tells us how “efficient” a reaction is. Percent yield can not be bigger than 100 %. Another 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? 13.9 g Cu 48.8 % What is the percent yield? Chapter 16 Energy in Chemical Reactions How Much? In or Out? Energy Energy is measured in Joules or calories Every reaction has an energy change associated with it Exothermic reactions release energy, usually in the form of heat. Endothermic reactions absorb energy Energy is stored in bonds between atoms 58 Energy C + O2 CO2+ 395 kJ C + O2 395kJ C + O2 Reactants Products 59 In terms of bonds C O O O C O Breaking this bond will require energy O C O C O O Making these bonds gives you energy In this case making the bonds gives you more energy than breaking them 60 Exothermic The products are lower in energy than the reactants Releases energy 61 Energy CaCO CaO CaCO CaO + CO+2 CO2 3 + 176 3 kJ CaO + CO2 176 kJ CaCO3 Reactants Products 62 Endothermic The products are higher in energy than the reactants Absorbs energy 63 Chemistry Happens in MOLES An equation that includes energy is called a thermochemical equation CH4 + 2 O2 CO2 + 2 H2O + 802.2 kJ 1 mole of CH4 makes 802.2 kJ of energy. When you make 802.2 kJ you make 2 moles of water 64 CH4 + 2 O2 CO2 + 2 H2O + 802.2 kJ If 10. 3 grams of CH4 are burned completely, how much heat will be produced? 10. 3 g CH4 1 mol CH4 16.05 g CH4 802.2 kJ 1 mol CH4 =514 kJ 65 CH4 + 2 O2 CO2 + 2 H2O + 802.2 kJ How many liters of O2 at STP would be required to produce 23 kJ of heat? How many grams of water would be produced with 506 kJ of heat? 66 Heats of Reaction Enthalpy The heat content a substance has at a given temperature and pressure Can’t be measured directly because there is no set starting point The reactants start with a heat content The products end up with a heat content So we can measure how much enthalpy changes 68 Enthalpy Symbol is H Change in enthalpy is DH delta H If heat is released the heat content of the products is lower DH is negative (exothermic) If heat is absorbed the heat content of the products is higher DH is negative (endothermic) 69 Energy Change is down DH is <0 Reactants Products 70 Energy Change is up DH is > 0 Reactants Products 71 Heat of Reaction The heat that is released or absorbed in a chemical reaction Equivalent to DH C + O2(g) CO2(g) +393.5 kJ C + O2(g) CO2(g) DH = -393.5 kJ In thermochemical equation it is important to say what state H2(g) + 1/2O2 (g) H2O(g) DH = -241.8 kJ H2(g) + 1/2O2 (g) H2O(l) DH = -285.8 kJ 72 Heat of Combustion The heat from the reaction that completely burns 1 mole of a substance C2H4 + 3 O2 2 CO2 + 2 H2O C2H6 + O2 CO2 + H2O 2 C2H6 + 5 O2 2 CO2 + 6 H2O C2H6 + (5/2) O2 CO2 + 3 H2O 73 Standard Heat of Formation The DH for a reaction that produces 1 mol of a compound from its elements at standard conditions Standard conditions 25°C and 1 atm. 0 Symbol is DH f The standard heat of formation of an element is 0 This includes the diatomics 74 What good are they? There are tables (pg. 190) of heats of formations The heat of a reaction can be calculated by subtracting the heats of formation of the reactants from the products 0 0 DH = DH f (reactants) - DH f (products) 75 Examples CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) 0 DH f CH4 (g) = -74.86 kJ 0 DH f O2(g) = 0 kJ 0 DH f CO2(g) = -393.5 kJ 0 DH f H2O(g) = -241.8 kJ DH= [-393.5 + 2(-241.8)]-[-74.68 +2 (0)] DH= 802.4 kJ 76 Examples 2 SO3(g) 2SO2(g) + O2(g) 77 Why Does It Work? If H2(g) + 1/2 O2(g) H2O(g) DH=-285.5 kJ then H2O(g) H2(g) + 1/2 O2(g) DH =+285.5 kJ If you turn an equation around, you change the sign 2 H2O(g) H2(g) + O2(g) DH =+571.0 kJ If you multiply the equation by a number, you multiply the heat by that number. 78 Why does it work? You make the products, so you need there heats of formation You “unmake” the products so you have to subtract their heats. How do you get good at this 79