Chapter 9 Stoichiometry

Chapter 9 Stoichiometry

Section 9.1 Intro to Stoichiometry

9.1 Objectives Define stoichiometry. Describe the importance of the mole ratio in stoichiometric calculations. Write a mole ratio relating two substances in a chemical equation.

Section 1 Introduction to Stoichiometry Lesson Starter Mg(s) + 2HCl(aq) MgCl 2 (aq) + H 2 (g) If 2 mol of HCl react, how many moles of H 2 are obtained? 1 mol H 2 How many moles of Mg will react with 2 mol of HCl? 1 mol Mg If 4 mol of HCl react, how many mol of each product are produced? 2 mol MgCl 2 and 2 mol H 2 How would you convert from moles of substances to masses? Molar mass

Section 1 Introduction to Stoichiometry Stoichiometry Definition Stoichiometry Relationship between quantities Composition stoichiometry The mass relationships of elements in compounds (Ch 7.3) Reaction stoichiometry The mass relationships between reactants and products in a chemical reaction

Section 1 Introduction to Stoichiometry Reaction Stoichiometry Problems Problem Type 1: Convert moles of known substance to moles of unknown substance Amount of known substance (mol) Problem Type 2: Convert moles of known substance to grams unknown substance Amount of known substance (mol) Amount of unknown substance (mol) Amount of unknown substance (mol) Mass of unknown substance (g)

Section 1 Introduction to Stoichiometry Reaction Stoichiometry Problems, continued Problem Type 3: Convert grams of known substance to moles of unknown substance Mass of known substance (g) Amount of known substance (mol) Amount of unknown substance (mol) Problem Type 4: Convert grams of known substance to grams of unknown substance Mass of known substance (g) Amount of known substance (mol) Amount of unknown substance (mol) Mass of unknown substance (g)

Section 1 Introduction to Stoichiometry Mole Ratio A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction Example: 2Al 2 O 3 (l) 4Al(s) + 3O 2 (g) Mole Ratios: 2 mol Al 2 O 3 2 mol Al 2 O 3 4 mol Al,, 4 mol Al 3 mol O 2 3 mol O 2 Set up six mole ratios for the following reaction: Al 4 C 3 (s) + 12H 2 O(l) 3CH 4 (g) + 4Al(OH) 3 (s)

Section 1 Introduction to Stoichiometry Stoichiometry Calculations

Section 1 Introduction to Stoichiometry Converting Between Amounts in Moles

Section 9.2 Ideal Stoichiometric Calculations

9.2 Objectives Calculate the amount in moles of a reactant or a product from the amount in moles of a different reactant or product. Calculate the mass of a reactant or a product from the amount in moles of a different reactant or product. Calculate the amount in moles of a reactant or a product from the mass of a different reactant or product. Calculate the mass of a reactant or a product from the mass of a different reactant or product.

Section 2 Ideal Stoichiometric Calculations Conversions of Quantities in Moles

Section 2 Ideal Stoichiometric Calculations Conversions of Quantities in Moles, continued Sample Problem A In a spacecraft, the carbon dioxide exhaled by astronauts can be removed by its reaction with lithium hydroxide, LiOH, according to the following chemical equation. CO 2 (g) + 2LiOH(s) Li 2 CO 3 (s) + H 2 O(l) How many moles of lithium hydroxide are required to react with 20 mol CO 2, the average amount exhaled by a person each day?

Section 2 Ideal Stoichiometric Calculations Conversions of Quantities in Moles, continued Sample Problem A Solution CO 2 (g) + 2LiOH(s) Li 2 CO 3 (s) + H 2 O(l) Given: amount of CO 2 = 20 mol Unknown: amount of LiOH (mol) Solution: mol ratio 20 mol CO 2 2 mol LiOH = 40 mol LiOH 1 mol CO 2

Section 2 Ideal Stoichiometric Calculations Conversions of Amounts in Moles to Mass

Section 2 Ideal Stoichiometric Calculations Conversions of Amounts in Moles to Mass, continued Sample Problem B In photosynthesis, plants use energy from the sun to produce glucose, C 6 H 12 O 6, and oxygen from the reaction of carbon dioxide and water. What mass, in grams, of glucose is produced when 3.00 mol of water react with carbon dioxide?

Section 2 Ideal Stoichiometric Calculations Conversions of Amounts in Moles to Mass, continued Sample Problem B Solution Given: amount of H 2 O = 3.00 mol Unknown: mass of C 6 H 12 O 6 produced (g) Solution: Balanced Equation: 6CO 2 (g) + 6H 2 O(l) C 6 H 12 O 6 (s) + 6O 2 (g) mol ratio molar mass 3.00 mol H 2 O 1 mol C 6 H 12 O 6 180.2 g C 6 H 12 O 6 6 mol H 2 O 1 mol C 6 H 12 O 6 = 90.1 g C 6 H 12 O 6

Section 2 Ideal Stoichiometric Calculations Conversions of Amounts in Moles to Mass, continued Sample Problem C What mass of carbon dioxide, in grams, is needed to react with 3.00 mole H 2 O in the photosynthesis reaction described in Sample Problem B?

Section 2 Ideal Stoichiometric Calculations Conversions of Amounts in Moles to Mass, continued Sample Problem C Solution Given: amount of H 2 O = 3.00 mol Unknown: mass of CO 2 consumed (g) Solution: Balanced Equation: 6CO 2 (g) + 6H 2 O(l) C 6 H 12 O 6 (s) + 6O 2 (g) mol ratio molar mass 3.00 mol H 2 O 6 mol CO 2 44.01 g CO 2 6 mol H 2 O 1 mol CO 2 = 132 g CO 2

Section 2 Ideal Stoichiometric Calculations Conversions of Mass to Amounts in Moles

Section 2 Ideal Stoichiometric Calculations Conversions of Mass to Amounts in Moles, continued Sample Problem D The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH 3 (g) + O 2 (g) NO(g) + H 2 O(g) (unbalanced) The reaction is run using 824 g NH 3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H 2 O are formed?

Section 2 Ideal Stoichiometric Calculations Conversions of Mass to Amounts in Moles, continued Sample Problem D Solution, continued Balanced Equation: 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) molar mass mol ratio a. 824 g NH 3 1 mol NH 3 4 mol NO = 48.4 mol NO 17.04 g NH 3 4 mol NH 3 b. 824 g NH 3 1 mol NH 3 6 mol H 2 O = 72.5 mol H 2 O 17.04 g NH 3 4 mol NH 3

Section 2 Ideal Stoichiometric Calculations Solving Mass-Mass Problems

Section 2 Ideal Stoichiometric Calculations Mass-Mass to Calculations, continued Sample Problem E Tin(II) fluoride, SnF 2, is used in some toothpastes. It is made by the reaction of tin with hydrogen fluoride according to the following equation. Sn(s) + 2HF(g) SnF 2 (s) + H 2 (g) How many grams of SnF 2 are produced from the reaction of 30.00 g HF with Sn?

Section 2 Ideal Stoichiometric Calculations Mass-Mass to Calculations, continued Sample Problem E Solution Given: amount of HF = 30.00 g Unknown: mass of SnF 2 produced (g) Solution: Sn(s) + 2HF(g) SnF 2 (s) + H 2 (g) molar mass mol ratio molar mass 30.00 g HF 1 mol HF 1 mol SnF 2 156.71 g SnF 2 20.01 g HF 2 mol HF 1 mol SnF 2 = 117.5 g SnF2

Section 2 Ideal Stoichiometric Calculations Solving Volume-Volume Problems

Section 2 Ideal Stoichiometric Calculations Solving Particle Problems

Section 2 Ideal Stoichiometric Calculations Mole Tunnel

Section 9.3 Limiting Reactants and Percent Yield

9.3 Objectives Describe a method for determining which of two reactants is a limiting reactant. Calculate the amount in moles or mass in grams of a product, given the amounts in moles or masses in grams of two reactants, one of which is in excess. Distinguish between theoretical yield, actual yield, and percentage yield. Calculate percentage yield, given the actual yield and quantity of a reactant.

Section 3 Limiting Reactants and Percentage Yield Limiting Reactants Limiting reactant the reactant that limits the amount of the other reactant that can combine and the amount of product that can form in a chemical reaction. The excess reactant is the substance that is not used up completely in a reaction.

Section 3 Limiting Reactants and Percentage Yield Limited Reactants, continued Sample Problem F Silicon dioxide (quartz) is usually quite unreactive but reacts readily with hydrogen fluoride according to the following equation. SiO 2 (s) + 4HF(g) SiF 4 (g) + 2H 2 O(l) If 6.0 mol HF is added to 4.5 mol SiO 2, which is the limiting reactant?

Section 3 Limiting Reactants and Percentage Yield Limited Reactants, continued Sample Problem F Solution, continued SiO 2 (s) + 4HF(g) SiF 4 (g) + 2H 2 O(l) mol ratio 4.5 mol SiO 2 1 mol SiF 4 = 4.5 mol SiF 4 produced 1 mol SiO 2 6.0 mol HF 1 mol SiF 4 4 mol HF = 1.5 mol SiF 4 produced HF is the limiting reactant.

Section 3 Limiting Reactants and Percentage Yield Percentage Yield Theoretical yield the maximum amount of product that can be produced from a given amount of reactant. Actual yield the measured amount of a product obtained from a reaction. Percentage yield is the ratio of the actual yield to the theoretical yield, multiplied by 100. actual yield percentage yield 100 theorectical yield

Section 3 Limiting Reactants and Percentage Yield Percentage Yield, continued Sample Problem H Chlorobenzene, C 6 H 5 Cl, is used in the production of many important chemicals, such as aspirin, dyes, and disinfectants. One industrial method of preparing chlorobenzene is to react benzene, C 6 H 6, with chlorine, as represented by the following equation. C 6 H 6 (l) + Cl 2 (g) C 6 H 5 Cl(l) + HCl(g) When 36.8 g C 6 H 6 react with an excess of Cl 2, the actual yield of C 6 H 5 Cl is 38.8 g. What is the percentage yield of C 6 H 5 Cl?

Section 3 Limiting Reactants and Percentage Yield Percentage Yield, continued Sample Problem H Solution, continued Theoretical yield Percentage yield C 6 H 6 (l) + Cl 2 (g) C 6 H 5 Cl(l) + HCl(g) 36.8 g C 6 H 6 1 mol C 6 H 6 1 mol C 6 H 5 Cl 112.56 g C 6 H 5 Cl = 53.0 g C 6 H 5 Cl 78.12 g C 6 H 6 1 mol C 6 H 6 1 mol C 6 H 5 Cl actual yield percentage yield C6H5Cl 100 theorectical yield 38.8 g percentage yield 100 53.0 g 73.2%