Stoichiometry The quantitative study of reactants and products in a chemical reaction Stoichiometry Whether the units given for reactants or products are moles , grams , liters (for gases), or some other units, we use moles to calculate the amount of product formed in a reaction Stoichiometry Mass Mass Moles Liters Particles Known Moles Liters Particles Unknown Review before starting The Mole Molar Conversions Balancing Chemical Equations Stoichiometry Problem Types Mole to Mole Mole to Mass & Mass to Mole Mass to Mass Volume to Moles or Mass Limiting Reactants & Per Cent Yield The Balanced Equation 2 CO(g) + O2 --> 2 moles 1 mole 2CO2(g) 2 moles Coefficients show relative amounts Mole Ratio A conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction 2 CO(g) + O2 --> 2 CO2(g) 2 moles of CO is equivalent to 2 moles of CO2 The mole ratio between CO and CO2 is 2:2 or 1:1 2mol CO 2 mol CO2 2 mol CO2 and 2 mol CO Mass to Mass 140. grams of carbon monoxide reacts with an excess of oxygen gas to theoretically produce how many grams of carbon dioxide? CO(g) + O2 140. g CO 2 CO2(g) 1 mole CO 28.0 g CO 2 moles CO2 44.0 g CO 2 = 220. g CO 2 2 moles CO 1 mole CO2 Problem Solving Strategy (mass to mass) Known: 140 g CO Unknown: ? grams CO2 mass mass 140. g CO moles 5 moles CO 220. g CO2 moles 5 Moles CO2 mole bridge 20 Burlingame High School Chemistry Example Problem If a furnace burns an amount of coal containing 6.0 moles of FeS2, how many moles of SO 2 (an air pollutant) is theoretically produced? 4FeS2 + 11O2 2 Fe2O3 + 8 SO2 What is the mole ratio of FeS 2 & SO 2 ? 4 moles FeS2 : 8 moles SO 2 4:8 or 1:2 Example Problem If a furnace burns an amount of coal containing 6.0 moles of FeS 2 , how many moles of SO 2 (an air pollutant) is produced? 4FeS2 + 11O2 2 Fe2O3 + 8 SO2 Use the mole ratio - 4 mol FeS2 to 8 SO2 use the given 6 mols FeS2 8 mols SO2 4 mols FeS2 = 12 mols SO 2 Another Example If a furnace burns an amount of coal containing 100.0g of FeS 2, how many grams of SO 2 (an air pollutant) is theoretically produced? Remember the balanced equation 4FeS2 + 11O2 2 Fe2O3 + 8 SO2 Use dimensional analysis 100.0 g FeS2 Convert mass of reactant to moles of reactant. 100.0 g FeS2 1 mole FeS2 120.0 g FeS2 Convert moles of reactant to moles of product. 100.0 g FeS2 1 mole FeS2 120.0 g FeS2 8 mole SO2 4 mole FeS2 Convert moles of product to grams of product. 100.0 g FeS2 1 mole FeS2 120.0 g FeS2 8 mole SO2 64.0 g SO2 4 mole FeS2 1 mole SO2 Multiply across the top and bottom 100.0 g FeS2 1 mole FeS2 8 mole SO2 64.0 g SO2 = 120.0 g FeS2 4 mole FeS2 1 mole SO2 51200 g SO2 = 106.67 g SO 2 = 107 g SO 2 480 Moles to Number of Particles Another Conversion Factor 1 mole = 6.02 x 1023 particles Moles of substance Number of atoms or molecules 6.02 x x 1023 1mol x 1mol 6.02 x 1023 Number of = atoms or molecules = Moles of substance 31 Number of Particles to Moles How many atoms are present in 0.35 mol of Na? Number of Particles to Moles How many atoms are present in 0.35 mol of Na? .35 mol Na 6.02 x 1023 = 2.1 x 1023 atoms 1 mol Moles to Number of Particles How many moles are present in 3.00 x 10 21 molecules of C 2 H 6 ? 3.00 x 1021 molecules C2H6 1 mole C2H6 = 4.98 x 1 0-3 moles 6.02 x 1 023 molecules The Balanced Equation 2 CO(g) + 2 moles : O2(g) --> 1 mole 2 CO2(g) : 2 moles 2(6.02x1023) : (6.02x1 023): 2(6.02x1 023) molecules molecules molecules 56g CO : gases 2 volumes : 32 g O2 1 volume :88 g CO2 :2 volumes Volume Conversions Volume of a gas is dependent on the temperature & pressure In this unit we use Standard Temperature & Pressure (STP) for our problems STP = 0 0 Celsius & 1 atm of pressure Volume Conversions A new conversion factor!! At SIP 1 mole of any gas occupies 22.4 Liters. 1 mole gas 22.4 L and 22.4 L 1 mole gas Volume Conversions If 5 g of magnesium is added to a solution of hydrochloric acid, what volume of hydrogen gas is produced at STP? Volume Conversions If 5.0 g of magnesium is added to a solution of hydrochloric acid, what volume of hydrogen gas is produced at STP? 1. Start with a balanced equation Mg(s) + 2HCl(aq) MgCl 2(aq) + H 2(g) Volume Conversions If 5.0 g of magnesium is added to a solution of hydrochloric acid, what volume of hydrogen gas is produced at STP? 1. Balanced equation Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) 1. Dimensional Analysis 5.0g Mg mass Mg 1 mol Mg 24.3 g mols Mg 1 mol H2 1 mol Mg mols H2 22.4 L = 4.6L 1 mol H2 volume H2 Limiting Reagents To make a dozen brownies the recipe calls for 2 cups flour, 112 grams chocolate, 25O ml water . You have 2 cups flour 50 grams chocolate, & 250 ml water If you want to make quality brownies you will make less than a dozen and have flour & water left over! What is the limiting reagent ? Limiting Reagents Zinc & Sulfur react to form zinc (II) sulÞde according to the following equation 8 Zn(s) + S8 8ZnS(s) If 2.00 mol of Zn are heated with 1.00 mole S 8, identify the limiting reactant. How many moles of excess reactant will there Limiting Reagents Zinc & Sulfur react to form zinc (II) sulfide according to the following equation 8 Zn(s) + S8 8ZnS(s) If 2.00 mol of Zn are heated with 1 .00 mole S8, identify the limiting reactant? 2 mol Zn 1 mol S 8 = .25 mole S 8 8 mol Zn How many moles of excess reactant? Zn is limiting (there isn’t enough to react with all the S8) Limiting Reagents Zinc & Sulfur react to form zinc (II) sulfide according to the following equation 8 Zn(s) + S8 8ZnS(s) If 2.00 mol of Zn are heated with 1 .00 mole S8, identify the limiting reactant? 2 mol Zn 1 mol S8 = .25 mole S 8 8 mol Zn How many moles of excess reactant? .75 moles of S8 Zn is limiting (there isn’t enough to react with all the S8) Percent Yield So far we have been doing stoichiometry problems that represent theoretical yields Actual Yield - the measured amount of product that you really get in the reaction. Percent Yield Percent Yield is the ratio of the actual yield to the theoretical yield multiplied by 100 Percent yield = actual yield theoretical yield x 100 Percent Yield Quicklime, CaO, can be prepared by roasting limestone, CaCO3, according to the following reaction. CaCO3(s) CaO(s) + CO2(g) When 2.00 x 1 03g of CaCO3(s) is heated the actual yield of CaO is 1.05 x 1 03g. What is the percent yield? Percent Yield CaCO3(s) CaO(s) + CO2(g) Given: 2.00 x 1 03g of CaCO 3(s) actual yield of CaO is 1.05 x 1 03g To solve 1. Find theoretical yield (mass mass problem) 2. Find percent yield (actual/theoretical x 1 00) 2.00 x 1 0 3g CaCO 3(s) 1 mol CaCO3 100.g 1 mol CaO 56.0 g 1 mol CaCO 3 1 mol CaO = 11 20g CaO Percent Yield = 1.05 x 10 3g x 100 = 93.8% 1.12 x 103g Quest 1. Na2SIO3 (s) + 8 HF(aq) - H2SiF6 (aq) + 2 NaF(aq) + 3 H2O (l) How many moles of HF are needed to react with 0.300 mol of Na2SiO3? How many grams of NaF form when 0.500 mol of HF reacts with excess Na2SiO3? How many grams of Na2SiO3 can react with 0.800 g of HF? Quest Quest Quest Quest Quest