Stoichiometry. The quantitative study of reactants and products in a chemical reaction. Burlingame High School Chemistry

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1 Stoichiometry The quantitative study of reactants and products in a chemical reaction 1

2 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 2

3 Stoichiometry Particles Particles Mass Moles Moles Mass Liters Liters Known Unknown 3

4 Review before starting Dimensional Analysis Conversion Factors The Mole Molar Conversions Balancing Chemical Equations 4

5 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 5

6 Mole to Mole An example problem If we have 4 moles of CO and abundant O 2 How many moles of CO 2 will be produced? 4 moles CO (g)? Moles CO 2(g) known unknown 2 CO (g) + O 2 2 CO 2(g) Notice that you need a balanced equation to start! 6

7 The Balanced Equation 2 CO (g) + O 2 --> 2CO 2(g) 2 moles 1 mole 2 moles Coefficients show relative amounts 7

8 Mole Ratio A conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction 2 CO (g) + O 2 --> 2 CO 2(g) 2 moles of CO is equivalent to 2 moles of CO 2 The mole ratio between CO and CO 2 is 2:2 or 1:1 2mol CO 2 mol CO 2 2 mol CO 2 and 2 mol CO 8

9 Mole to Mole If we have 4 moles of CO and abundant O 2. What theoretical quantity of CO 2 will be produced? Balanced equation 2 CO (g) + O 2 --> 2 CO 2(g) Converting moles CO to moles CO 2 4 moles CO 2 moles CO 2 = 4 moles CO 2 2 moles CO Mole ratio (conversion factor) 9

10 Problem Solving Strategy Known: mass 4 mols CO Unknown: mass 2 moles moles 4 mols CO 2mols CO 2 = 4 mols CO 2 2 mols CO mole bridge 2 CO (g) + O 2(g) --> 2 CO 2(g) 10

11 Mole to Mass If we have 4.00 moles of CO (g) and excess O 2 (g) what mass of CO 2(g) will theoretically form? known unknown 4.00 moles of CO (g)? mass (grams) of CO 2 11

12 Problem Solving Strategy (mole to mass) Known: mass 4.00 mols CO Unknown: grams CO 2 mass? moles moles 4.00 mols Co 4.00 mols CO 2 mole bridge 12

13 Mole to Mass Convert 4.00 mols of CO to grams CO 2 Molar Mass of CO 2 = 44.0 g/mole 4.00 mols CO 2mols CO g = 176 g 2mols CO 1 mole CO 2 13

14 Problem Solving Strategy mole to mass Known: mass 4 mols CO Unknown: Grams CO 2 mass 176 grams moles moles 4.00 mols Co 4.00 mols CO 2 mole bridge 14

15 Mass to Mole 140. grams of carbon monoxide reacts with an excess of oxygen gas to theoretically produce how many moles of carbon dioxide? 140. grams CO? moles CO 2 15

16 Problem Solving Strategy mass to mole Known: 140. g CO Unknown:? moles CO 2 mass 140. g CO mass moles moles mole bridge? Moles CO 2 16

17 Mass to Mole 140 grams of carbon monoxide reacts with an excess of oxygen gas to produce how many moles of carbon dioxide? Balance the equation 2 CO (g) + O 2 2 CO 2(g) Convert grams to moles using the molar mass of CO 140. grams CO 1 mole CO = 5.00 moles CO 28.0 grams CO Convert moles CO to moles CO 2 using a mole ratio (from balanced equation) 5.00 moles CO 2 moles CO 2 = 5.00 moles CO 2 2 moles CO 17

18 Problem Solving Strategy (mass to mole) Known: 140 g CO Unknown:? moles CO 2 mass mass 140. g CO moles 5.00 moles CO moles 5.00 Moles CO 2 mole bridge 18

19 Mass to Mass 140. grams of carbon monoxide reacts with an excess of oxygen gas to theoretically produce how many grams of carbon dioxide? 140 grams CO (g)? grams CO 2 Remember the balanced equation 2 CO (g) + O 2 2 CO 2(g) 140. g CO 1 mole CO 2 moles CO g CO 2 = 220. g CO g CO 2 moles CO 1 mole CO 2 19

20 Problem Solving Strategy (mass to mass) Known: 140 g CO Unknown:? grams CO 2 mass 140. g CO mass 220. g CO 2 moles 5 moles CO moles 5 Moles CO 2 mole bridge 20

21 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 theoretically produced? 4FeS O 2 2 Fe 2 O SO 2 What is the mole ratio of FeS 2 & SO 2? 4 moles FeS 2 : 8 moles SO 2 4:8 or 1:2 21

22 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? 4FeS O 2 2 Fe 2 O SO 2 Use the mole ratio - 4 mol FeS 2 to 8 SO 2 use the given 6 mols FeS 2 8 mols SO 2 = 12 mols SO 2 4 mols FeS 2 22

23 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 4FeS O 2 2 Fe 2 O SO 2 23

24 Use dimensional analysis g FeS 2 24

25 Convert mass of reactant to moles of reactant g FeS 2 1 mole FeS g FeS 2 25

26 Convert moles of reactant to moles of product g FeS 2 1 mole FeS 2 8 mole SO g FeS 2 4 mole FeS 2 26

27 Convert moles of product to grams of product g FeS 2 1 mole FeS 2 8 mole SO g SO g FeS 2 4 mole FeS 2 1 mole SO 2 27

28 Multiply across the top and bottom g FeS 2 1 mole FeS 2 8 mole SO g SO 2 = g FeS 2 4 mole FeS 2 1 mole SO g SO

29 Divide the top by the bottom g FeS 2 1 mole FeS 2 8 mole SO g SO 2 = g FeS 2 4 mole FeS 2 1 mole SO 2 51,200 g SO = g SO 2 29

30 Apply Sig Fig Rules g FeS 2 1 mole FeS 2 8 mole SO g SO 2 = g FeS 2 4 mole FeS 2 1 mole SO 2 12,800 g SO = g SO 2 = 107 g SO 2 30

31 Moles to Number of Particles Another Conversion Factor 1mole = 6.02 x particles Moles of substance x 6.02 x mol = Number of atoms or molecules Number of atoms or molecules x 1mol 6.02 x = Moles of substance 31

32 Number of Particles to Moles How many atoms are present in 0.35 mol of Na? 32

33 Number of Particles to Moles How many atoms are present in 0.35 mol of Na?.35 mol Na 6.02 x = 2.1 x atoms 1 mol 33

34 Moles to Number of Particles How many moles are present in 3.00 x molecules of C 2 H 6? 3.00 x molecules C 2 H 6 1 mole C 2 H 6 = 4.98 x 10-3 moles 6.02 x molecules 34

35 The Balanced Equation 2 CO (g) + O 2(g) --> 2 CO 2(g) 2 moles : 1 mole : 2 moles 2(6.02x10 23 ) : (6.02x10 23 ) : 2(6.02x10 23 ) molecules molecules molecules 56g CO : 32 g O 2 :88 g CO 2 gases 2 volumes : 1 volume :2 volumes 35

36 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 36

37 Volume Conversions A new conversion factor!! At STP 1mole of any gas occupies 22.4 Liters. 1mole gas and 22.4 L 22.4 L 1 mole gas 37

38 Volume Conversions If 5 g of magnesium is added to a solution of hydrochloric acid, what volume of hydrogen gas is produced at STP? 38

39 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) 39

40 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) MgCl 2(aq) + H 2(g) 2. Dimensional Analysis 5.0g Mg 1mol Mg 1mol H L = 4.6L 24.3 g 1 mol Mg 1 mol H 2 mass Mg mols Mg mols H 2 volume H 2 40

41 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? 41

42 Limiting Reagents Zinc & Sulfur react to form zinc (II) sulfide according to the following equation 8 Zn(s) + S 8 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 be? 42

43 Limiting Reagents Zinc & Sulfur react to form zinc (II) sulfide according to the following equation 8 Zn(s) + S 8 8ZnS(s) If 2.00 mol of Zn are heated with 1.00 mole S 8, identify the limiting reactant? 2 mol Zn 1mol 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 S 8 ) 43

44 Limiting Reagents Zinc & Sulfur react to form zinc (II) sulfide according to the following equation 8 Zn(s) + S 8 8ZnS(s) If 2.00 mol of Zn are heated with 1.00 mole S 8, identify the limiting reactant? 2 mol Zn 1mol S8 =.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 S 8 ).75 moles of S 8 44

45 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. 45

46 Percent Yield Percent Yield is the ratio of the actual yield to the theoretical yield multiplied by 100 Percent yield = actual yield x 100 theoretical yield 46

47 Percent Yield Quicklime, CaO, can be prepared by roasting limestone, CaCO 3, according to the following reaction. CaCO 3(s) CaO (s) + CO 2(g) When 2.00 x 10 3 g of CaCO 3(s) is heated the actual yield of CaO is 1.05 x 10 3 g. What is the percent yield? 47

48 Percent Yield CaCO 3(s) CaO (s) + CO 2(g) Given: 2.00 x 10 3 g of CaCO 3(s) actual yield of CaO is 1.05 x 10 3 g To solve 1. find theoretical yield (mass mass problem) 2. Find percent yield (actual/theoretical x 100) 2.00 x 10 3 g CaCO 3(s) 1 mol CaCO 3 1mol CaO 56.0 g = 1120g CaO 100.g 1 mol CaCO 3 1 mol CaO Percent Yield = 1.05 x 10 3 g x 100 = 93.8% 1.12 x 10 3 g 48

Stoichiometry CHAPTER 12

Stoichiometry CHAPTER 12 CHAPTER 12 Stoichiometry 12.1 Using Everyday Equations Stoichiometry is the calculation of quantities in chemical equations. * The balanced equation gives the ratios for the reactants and products. 3 eggs