Stoichiometry – Ch. 11 I. Stoichiometric Calculations Background on things you NEED to know how to do: 1. 2. 3. 4. 5. Name/write correct chemical formula Write chemical equations Balance chemical equations Predict Products Mole/mass conversions Stoichiometry o o o o Stoichiometry uses ratios to determine relative amounts of reactants or products. For example If you were to make a bicycle, you would need one frame and two tires. 1 frame + 2 tires 1 bicycle If I had 74 tires, what is the most # of bicycles I could make? 74 tires 1 bicycle 2 tires = 37 bicycles 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 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 • can be used to determine expected amounts of products given amounts of reactants. 2 Mg + O2 2 MgO Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. • Mole ratio - moles molesmoles moles • Molar mass - moles grams • Molarity moles liters soln • Molar volume - moles liters gas Core step in all stoichiometry problems!! 4. Check answer. Molar Volume at STP 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 Mole – Mole Conversions The first type of problems we encounter will go between moles and moles. For this we need to use mole ratios. Ex: Write and balance the reaction between lead (II) nitrate and potasium iodide. Pb(NO3)2 + 2KI 2 KNO3 + PbI2 Mole ratio of potasium iodide to lead (II) iodide: 2 moles KI 1 mole PbI2 Mole to Mole Problems How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas? 2KClO3 2KCl + 3O2 ? mol 9 mol O2 2 mol KClO3 3 mol O2 9 mol = 6 mol KClO3 Mole to Mass We can also convert from moles to mass, and mass to moles For Example: • 4 Al + 3 O2 2Al2O3 If you know how many grams of Al you start with, we can write a flow chart to show how to calculate the # of moles of oxygen need to fully react with the Al. g Al moles Al moles of oxygen Mass to Moles: 4 Al + 3 O2 2Al2O3 If the reaction starts with .84 moles of aluminum, how many grams of aluminum oxide are produced? .84 mol Al 2 mol Al2O3 101.9 grams Al O 2 3 1 mol Al2O3 4 mol Al = 42.8 grams Al2O3 0.92 g of Aluminum oxide are produced from the reaction. How much aluminum was used up? .92 g Al2O3 1 mol Al2O3 4 mol Al 26.9 g Al 101.9 g Al2O3 2 mol Al2O3 1 mol Al = .49 grams Al Mass to Mass How many grams of silver will be formed from 12.0 g copper reacting with silver nitrate? Cu + 2AgNO3 2Ag + Cu(NO3)2 12.0 g ?g 12.0 1 mol 2 mol 107.87 g Cu Cu Ag g Ag = 40.7 g 63.55 1 mol 1 mol Ag g Cu Cu Ag Stoichiometry Problems – Mole/Mass In photosynthesis, carbon dioxide and water react to form glucose, C6H12O6 and oxygen gas. ___CO 6 2 6 2O ___C6H12O6 + ___O 6 2 + ___H If 15.6 grams of carbon dioxide react, how many moles of glucose will be produced? 15.6 g CO2 1 mol CO2 1 mol C2 H12O6 44.01 g CO2 6 mol CO2 = 0.0591 mol C2H12O6 How many grams of carbon dioxide must react to produce 0.25 moles of glucose? 0.25 mol C2 H12O6 6 mol CO2 44.01 g CO2 1 mol C2 H12O6 1 mol CO 2 = 66 g CO2 Stoichiometry with Gases If the pressure and temperature are constant, the ratio of moles in the balanced equation is the ratio of liters in an all gas reaction. Molar Volume at STP At STP 1 mol of a gas=22.4 L Standard Temperature & 0°C and 1 atm Pressure Molar Volume Hydrogen and chlorine gas react to produce hydrochloric acid. If 7.00 L of hydrogen gas react, how many liters of HCl gas are formed? H2 (g) + Cl2 (g) 2 HCl (g) 7.00 L H2 2.0 L HCl 1.0 L H2 = 14.0 L HCl *only in all gas reactions! Molar Volume In the following reaction, if 17 g of Mg react, how many L of H2 forms? Mg (s) + 2HCl (aq) MgCl2 (aq) + H2 (g) 17.0 g Mg 1 mol Mg 1 mol = 15.7 L H2 22.4 L H2 24.31 g 1 mol Mg Mg 1 mol H2 H2 Molar Volume Problems many grams of KClO3 are req’d to produce 9.00 L of O2 at STP? How 2KClO3 2KCl + 3O2 ?g 9.00 L 9.00 L O2 1 mol O2 2 mol 122.55 KClO3 g KClO3 22.4 L O2 3 mol O2 1 mol KClO3 = 32.8 g KClO3 Molar Volume Problems How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3? Cu + 2AgNO3 2Ag + Cu(NO3)2 ?g 1.5L 0.10M 1.5 .10 mol 1 mol 63.55 L AgNO3 Cu g Cu 1L = 4.8 g 2 mol 1 mol Cu AgNO3 Cu Molarity Molarity is the number of moles of solute dissolved in one liter of solution. Units are moles per liter or moles of solute per liter of solution. Molarity abbreviated by a capital M Molarity = moles of solute liter of solution Molarity Problems As an example, suppose we dissolve 23 g of ammonium chloride (NH4Cl) in enough water to make 145 mL of solution. What is the molarity of ammonium chloride in this solution? 23 g NH4Cl 1 mole NH4Cl = .43 mol NH4Cl 53.5 g NH4Cl 145 mL 1 L = .145 L 1000 mL .43 mol NH4Cl .145 L = 2.97 M NH4Cl Molarity Problems Now, suppose we have a beaker with 175 mL of a 5.5 M HCl solution. How many moles of HCl is in this beaker? 175 mL 1 L 1000 mL 5.5 mol HCl = .96 mol HCl 1 L Suppose you had 70 grams of NaCl and you dissolved it in exactly 2.00 L of solution. What would be the molarity of the solution? 1.2 mol NaCl 70 g 1 mol 2.0 L NaCl NaCl = 1.2 mol NaCl 58.44 g NaCl = 0.6 M NaCl Volume of Solutions A 10.% HCl solution (soln) means: 10 g HCl (pure) 100 g HCl soln A solution with a density of 1.5 g/mL means: 1.5 g soln 1 mL soln Impure Substances To say a substance is 75% NaCl by mass means: 75 g NaCl (pure) 100 g of NaCl solution or 100 g of NaCl solution 75 g of NaCL (pure) Or, Iron ore that is 15% iron by mass means: 15 g Fe 100 g ore or 100 g of ore 15 g of Fe Energy & Stoichiometry Exothermic and Endothermic process – heat is released into the surroundings • Exo = Exit HEAT Exothermic Process – heat is absorbed from the surroundings • Endo = Into HEAT Endothermic Thermochemical Equations In a thermochemical equation, the energy of change for the reaction can be written as either a reactant or a product Enthalpy: the heat content of a system at constant pressure (ΔH) Endothermic (positive ΔH) 2NaHCO3 + 129kJ Na2CO3 + H2O + CO2 Exothermic (negative ΔH) CaO + H2O Ca(OH)2 + 65.2kJ Write the thermochemical equation for the oxidation of Iron (III) if its ΔH= -1652 kJ Exo 4Fe(s) + 3 O2(g)→ 2 Fe2O3(s) + 1652 kJ How much heat is evolved when 10.00g of Iron is reacted with excess oxygen? 10.00g Fe 1 mol Fe 55.85g Fe 1652 kJ 4 mol Fe =73.97 kJ of heat Write the thermochemical equation for the decomposition of sodium bicarbonate, with a ΔH = + 129 kJ: Endo 2 NaHCO3 + 129kJ → Na2CO3(s) + H2O + CO2 How much heat is required to break down 50.0g of sodium bicarbonate? 50.0 g NaHCO3 1 mol NaHCO3 129 kJ 83.9 g NaHCO3 2 mol NaHCO3 =38.4 kJ of heat Write the thermochemical equation for the synthesis of calcium oxide and water with a ΔH= - 65.2 kJ: Exo CaO + H2O → Ca(OH)2 + 65.2kJ How much energy is released when 100 g of calcium oxide reacts? 100 g CaO 1 mol CaO 65.2 kJ 56.07 g CaO 1 mol CaO =116 kJ of heat Write the thermochemical equation for the decomposition of magnesium oxide with a ΔH= + 61.5 kJ: Endo 2 MgO + 61.5 → 2 Mg + O2 How many grams of oxygen are produced when magnesium oxide is decomposed by adding 420 kJ of Energy? 420 kJ 1 mol O2 61.5 kJ 31.98 g O2 =218 g of O 2 1 mol O2 Stoichiometry – Ch. 11 Stoichiometry in the Real World 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 Available Ingredients • 24 graham cracker squares • 1 bag of marshmallows • 12 pieces of chocolate Limiting Reactant • chocolate Excess Reactants • Marshmallows and graham crackers Limiting Reactants Limiting Reactant • one that is used up in a reaction • determines the amount of product that can be produced Excess Reactant • added to ensure that the other reactant is completely used up • cheaper & easier to recycle Limiting Reactant Steps 1. Write the balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates: • limiting reactant • amount of product actually possible Limiting Reactants 79.1 g of zinc react with 68.1 g HCl. Identify the limiting and excess reactants. How many grams of hydrogen can be formed? Zn + 2HCl 79.1 g 68.1 g ZnCl2 + H2 ?g Limiting Reactants Zn + 2HCl 79.1 g 68.1 g 68.1 1 mol g HCl HCl ZnCl2 + H2 ?g 1 mol 2.02 g H2 H2 = 1.89 36.46 2 mol 1 mol g H2 g HCl HCl H2 Limiting Reactants Zn + 2HCl 79.1 g 68.1 g ZnCl2 + H2 ?g 79.1 1 mol 1 mol 2.02 g g Zn Zn H2 H2 = 2.44 g 65.39 1 mol 1 mol H2 g Zn Zn H2 Limiting Reactants Zn: 2.44 g H2 HCl: 1.89 g H2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 1.89 g H2 left over zinc Limiting Reactants #2 5.42 g of magnesium ribbon react with 4.00 g of oxygen gas. Identify the limiting and excess reactants. How many grams of magnesium oxide are formed? 2Mg + O2 5.42 g 4.00 g 2MgO ?g Limiting Reactants #2 2Mg + O2 5.42 g 4.00 g 2MgO ?g 5.42 1 mol 2 mol 40.31 g g Mg Mg MgO MgO = 8.99 g 24.31 2 mol 1 mol MgO g Mg Mg MgO Limiting Reactants #2 2Mg + O2 5.42 g 4.00 g O2 1 mol O2 32.00 g O2 4.00 g 2MgO ?g 2 mol 40.31 MgO g MgO = 10.1 g 1 mol 1 mol MgO O2 MgO A. Limiting Reactants #2 Mg: 8.99 g MgO O2: 10.1 g MgO Limiting reactant: Mg Excess reactant: O2 Product Formed: 8.99 g MgO Excess oxygen Limiting Reactants What other information could you find in these problems? • How much of each reactant is used – in grams, liters, moles • How much of excess reactant is left over – in grams, liters, moles 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 + 2 HCl 2 KCl + H2CO3 45.8 g 1 mol 2 mol 74.55 g = 49.4 K CO 2 3 KCl KCl K2CO3 grams KCl 1 mol 1 mol 138.2 g KCl K2CO3 K2CO3 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. 46.3 grams KCl x 100 49.4 grams KCl % yield = 93.7 %