Chapter 3. Stoichiometry

Transcription

1 Chapter 3 Stoichiometry

2 Chapter 3 Chemical Stoichiometry Stoichiometry The study of quantities of materials consumed and produced in chemical reactions. Copyright Cengage Learning. All rights reserved 2

3 Section 3.1 Counting by Weighing Objects behave as though they were all identical. Atoms are too small to count. Need average mass of the object. Copyright Cengage Learning. All rights reserved 3

4 Section 3.1 Counting by Weighing EXERCISE! A pile of marbles weigh g. 10 marbles weigh g. How many marbles are in the pile? Avg. Mass of 1 Marble = g 10 marbles = 3.76 g / marble g = 105 marbles 3.76 g Copyright Cengage Learning. All rights reserved 4

5 Section 3.2 Atomic Masses 12 C is the standard for atomic mass, with a mass of exactly 12 atomic mass units (u). The masses of all other atoms are given relative to this standard. Elements occur in nature as mixtures of isotopes. Carbon = 98.89% 12 C 1.11% 13 C < 0.01% 14 C Copyright Cengage Learning. All rights reserved 5

6 Section 3.2 Atomic Masses Average Atomic Mass for Carbon 98.89% of 12 u % of u = exact number (0.9889)(12 u) + (0.0111)( u) = u Copyright Cengage Learning. All rights reserved 6

7 Section 3.2 Atomic Masses Average Atomic Mass for Carbon Even though natural carbon does not contain a single atom with mass 12.01, for stoichiometric purposes, we can consider carbon to be composed of only one type of atom with a mass of This enables us to count atoms of natural carbon by weighing a sample of carbon. Copyright Cengage Learning. All rights reserved 7

8 Section 3.2 Atomic Masses Schematic Diagram of a Mass Spectrometer Copyright Cengage Learning. All rights reserved 8

9 Section 3.2 Atomic Masses EXERCISE! An element consists of 62.60% of an isotope with mass u and 37.40% of an isotope with mass u. Calculate the average atomic mass and identify the element u Rhenium (Re) Copyright Cengage Learning. All rights reserved 9

10 Section 3.3 The Mole The number equal to the number of carbon atoms in exactly 12 grams of pure 12 C. 1 mole of something consists of units of that substance (Avogadro s number). 1 mole C = C atoms = g C Copyright Cengage Learning. All rights reserved 10

11 Section 3.3 The Mole EXERCISE! Calculate the number of iron atoms in a 4.48 mole sample of iron Fe atoms Copyright Cengage Learning. All rights reserved 11

12 Section 3.4 Molar Mass Mass in grams of one mole of the substance: Molar Mass of N = g/mol Molar Mass of H 2 O = g/mol ( g) g Molar Mass of Ba(NO 3 ) 2 = g/mol g + ( g) + ( g) Copyright Cengage Learning. All rights reserved 12

13 Section 3.4 Molar Mass CONCEPT CHECK! Which of the following is closest to the average mass of one atom of copper? a) g b) g c) g d) g e) x g Copyright Cengage Learning. All rights reserved 13

14 Section 3.4 Molar Mass CONCEPT CHECK! Calculate the number of copper atoms in a g sample of copper Cu atoms Copyright Cengage Learning. All rights reserved 14

15 Section 3.4 Molar Mass CONCEPT CHECK! Which of the following g samples contains the greatest number of atoms? a) Magnesium b) Zinc c) Silver Copyright Cengage Learning. All rights reserved 15

16 Section 3.4 Molar Mass EXERCISE! Rank the following according to number of atoms (greatest to least): a) g of silver b) 70.0 g of zinc c) 21.0 g of magnesium b) a) c) Copyright Cengage Learning. All rights reserved 16

17 Section 3.4 Molar Mass EXERCISE! Consider separate gram samples of each of the following: H 2 O, N 2 O, C 3 H 6 O 2, CO 2 Rank them from greatest to least number of oxygen atoms. H 2 O, CO 2, C 3 H 6 O 2, N 2 O Copyright Cengage Learning. All rights reserved 17

18 Section 3.5 Learning to Solve Problems Conceptual Problem Solving Where are we going? Read the problem and decide on the final goal. How do we get there? Work backwards from the final goal to decide where to start. Reality check. Does my answer make sense? Is it reasonable? Copyright Cengage Learning. All rights reserved 18

19 Section 3.6 Percent Composition of Compounds Mass percent of an element: mass of element in compound mass % = 100% mass of compound For iron in iron(iii) oxide, (Fe 2 O 3 ): mass % Fe = 2(55.85 g) 2(55.85 g)+3(16.00 g) = g g n 100% = 69.94% Copyright Cengage Learning. All rights reserved 19

20 Section 3.6 Percent Composition of Compounds EXERCISE! Consider separate gram samples of each of the following: H 2 O, N 2 O, C 3 H 6 O 2, CO 2 Rank them from highest to lowest percent oxygen by mass. H 2 O, CO 2, C 3 H 6 O 2, N 2 O Copyright Cengage Learning. All rights reserved 20

21 Section 3.7 Determining the Formula of a Compound Formulas Empirical formula = CH Simplest whole-number ratio Molecular formula = (empirical formula) n [n = integer] Molecular formula = C 6 H 6 = (CH) 6 Actual formula of the compound Copyright Cengage Learning. All rights reserved 21

22 Section 3.7 Determining the Formula of a Compound Analyzing for Carbon and Hydrogen Device used to determine the mass percent of each element in a compound. Copyright Cengage Learning. All rights reserved 22

23 Section 3.7 Determining the Formula of a Compound EXERCISE! The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol. What is the empirical formula? C 3 H 5 O 2 What is the molecular formula? C 6 H 10 O 4 Copyright Cengage Learning. All rights reserved 23

24 Section 3.7 Determining the Formula of a Compound A representation of a chemical reaction: C 2 H 5 OH + 3O 2 reactants 2CO 2 + 3H 2 O products Reactants are only placed on the left side of the arrow, products are only placed on the right side of the arrow. Copyright Cengage Learning. All rights reserved 24

25 Section 3.8 Chemical Equations C 2 H 5 OH + 3O 2 2CO 2 + 3H 2 O The equation is balanced. All atoms present in the reactants are accounted for in the products. 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water. Copyright Cengage Learning. All rights reserved 25

26 Section 3.8 Chemical Equations The balanced equation represents an overall ratio of reactants and products, not what actually happens during a reaction. Use the coefficients in the balanced equation to decide the amount of each reactant that is used, and the amount of each product that is formed. Copyright Cengage Learning. All rights reserved 26

27 Section 3.9 Balancing Chemical Equations Writing and Balancing the Equation for a Chemical Reaction 1. Determine what reaction is occurring. What are the reactants, the products, and the physical states involved? 2. Write the unbalanced equation that summarizes the reaction described in step Balance the equation by inspection, starting with the most complicated molecule(s). The same number of each type of atom needs to appear on both reactant and product sides. Do NOT change the formulas of any of the reactants or products. Copyright Cengage Learning. All rights reserved 27

28 Section 3.9 Balancing Chemical Equations To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 28

29 Section 3.9 Balancing Chemical Equations EXERCISE! Which of the following correctly balances the chemical equation given below? There may be more than one correct balanced equation. If a balanced equation is incorrect, explain what is incorrect about it. CaO + C CaC 2 + CO 2 I. CaO 2 + 3C CaC 2 + CO 2 II. 2CaO + 5C 2CaC 2 + CO 2 III. CaO + (2.5)C CaC 2 + (0.5)CO 2 IV. 4CaO + 10C 4CaC 2 + 2CO 2 Copyright Cengage Learning. All rights reserved

30 Section 3.9 Balancing Chemical Equations CONCEPT CHECK! Which of the following are true concerning balanced chemical equations? There may be more than one true statement. I. The number of molecules is conserved. II. The coefficients tell you how much of each substance you have. III. Atoms are neither created nor destroyed. IV. The coefficients indicate the mass ratios of the substances used. V. The sum of the coefficients on the reactant side equals the sum of the coefficients on the product side. Copyright Cengage Learning. All rights reserved 30

31 Section 3.9 Balancing Chemical Equations Notice The number of atoms of each type of element must be the same on both sides of a balanced equation. Subscripts must not be changed to balance an equation. A balanced equation tells us the ratio of the number of molecules which react and are produced in a chemical reaction. Coefficients can be fractions, although they are usually given as lowest integer multiples. Copyright Cengage Learning. All rights reserved 31

32 Section 3.10 Stoichiometric Calculations: Amounts of Reactants and Products Stoichiometric Calculations Chemical equations can be used to relate the masses of reacting chemicals. Copyright Cengage Learning. All rights reserved 32

33 Section 3.10 Stoichiometric Calculations: Amounts of Reactants and Products Calculating Masses of Reactants and Products in Reactions 1. Balance the equation for the reaction. 2. Convert the known mass of the reactant or product to moles of that substance. 3. Use the balanced equation to set up the appropriate mole ratios. 4. Use the appropriate mole ratios to calculate the number of moles of the desired reactant or product. 5. Convert from moles back to grams if required by the problem. Copyright Cengage Learning. All rights reserved 33

34 Section 3.10 Stoichiometric Calculations: Amounts of Reactants and Products Calculating Masses of Reactants and Products in Reactions Copyright Cengage Learning. All rights reserved 34

35 Section 3.10 Stoichiometric Calculations: Amounts of Reactants and Products EXERCISE! Consider the following reaction: P 4 (s) + 5 O 2 (g) 2 P 2 O 5 (s) If 6.25 g of phosphorus is burned, what mass of oxygen does it combine with? 8.07 g O 2 Copyright Cengage Learning. All rights reserved 35

36 Section 3.10 Stoichiometric Calculations: Amounts of Reactants and Products EXERCISE! (Part I) Methane (CH 4 ) reacts with the oxygen in the air to produce carbon dioxide and water. Ammonia (NH 3 ) reacts with the oxygen in the air to produce nitrogen monoxide and water. Write balanced equations for each of these reactions. Copyright Cengage Learning. All rights reserved 36

37 Section 3.10 Stoichiometric Calculations: Amounts of Reactants and Products EXERCISE! (Part II) Methane (CH 4 ) reacts with the oxygen in the air to produce carbon dioxide and water. Ammonia (NH 3 ) reacts with the oxygen in the air to produce nitrogen monoxide and water. What mass of ammonia would produce the same amount of water as 1.00 g of methane reacting with excess oxygen? Copyright Cengage Learning. All rights reserved 37

38 Section 3.10 Stoichiometric Calculations: Amounts of Reactants and Products Let s Think About It Where are we going? To find the mass of ammonia that would produce the same amount of water as 1.00 g of methane reacting with excess oxygen. How do we get there? We need to know: How much water is produced from 1.00 g of methane and excess oxygen. How much ammonia is needed to produce the amount of water calculated above. Copyright Cengage Learning. All rights reserved 38

39 Section 3.11 The Concept of Limiting Reactant Limiting Reactants Limiting reactant the reactant that runs out first and thus limits the amounts of products that can be formed. Determine which reactant is limiting to calculate correctly the amounts of products that will be formed. Copyright Cengage Learning. All rights reserved 39

40 Section 3.11 The Concept of Limiting Reactant Limiting Reactants To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 40

41 Section 3.11 The Concept of Limiting Reactant A. The Concept of Limiting Reactants Stoichiometric mixture N 2 (g) + 3H 2 (g) 2NH 3 (g)

42 Section 3.11 The Concept of Limiting Reactant Limiting reactant mixture A. The Concept of Limiting Reactants N 2 (g) + 3H 2 (g) 2NH 3 (g)

43 Section 3.11 The Concept of Limiting Reactant A. The Concept of Limiting Reactants Limiting reactant mixture N 2 (g) + 3H 2 (g) 2NH 3 (g) Limiting reactant is the reactant that runs out first. H 2

44 Section 3.11 The Concept of Limiting Reactant Limiting Reactants The amount of products that can form is limited by the methane. Methane is the limiting reactant. Water is in excess. Copyright Cengage Learning. All rights reserved 44

45 Section 3.11 The Concept of Limiting Reactant CONCEPT CHECK! Which of the following reaction mixtures could produce the greatest amount of product? Each involves the reaction symbolized by the equation: 2H 2 + O 2 2H 2 O a) 2 moles of H 2 and 2 moles of O 2 b) 2 moles of H 2 and 3 moles of O 2 c) 2 moles of H 2 and 1 mole of O 2 d) 3 moles of H 2 and 1 mole of O 2 e) Each produce the same amount of product. Copyright Cengage Learning. All rights reserved 45

46 Section 3.11 The Concept of Limiting Reactant Notice We cannot simply add the total moles of all the reactants to decide which reactant mixture makes the most product. We must always think about how much product can be formed by using what we are given, and the ratio in the balanced equation. Copyright Cengage Learning. All rights reserved 46

47 Section 3.11 The Concept of Limiting Reactant CONCEPT CHECK! You know that chemical A reacts with chemical B. You react 10.0 g of A with 10.0 g of B. What information do you need to know in order to determine the mass of product that will be produced? Copyright Cengage Learning. All rights reserved 47

48 Section 3.11 The Concept of Limiting Reactant Let s Think About It Where are we going? To determine the mass of product that will be produced when you react 10.0 g of A with 10.0 g of B. How do we get there? We need to know: The mole ratio between A, B, and the product they form. In other words, we need to know the balanced reaction equation. The molar masses of A, B, and the product they form. Copyright Cengage Learning. All rights reserved 48

49 Section 3.11 The Concept of Limiting Reactant EXERCISE! You react 10.0 g of A with 10.0 g of B. What mass of product will be produced given that the molar mass of A is 10.0 g/mol, B is 20.0 g/mol, and C is 25.0 g/mol? They react according to the equation: A + 3B 2C Copyright Cengage Learning. All rights reserved 49

50 Section 3.11 The Concept of Limiting Reactant Percent Yield An important indicator of the efficiency of a particular laboratory or industrial reaction. Actual yield 100% Theoretical yield percent yield Copyright Cengage Learning. All rights reserved 50

51 Section 3.11 The Concept of Limiting Reactant EXERCISE! Consider the following reaction: P 4 (s) + 6F 2 (g) 4PF 3 (g) What mass of P 4 is needed to produce 85.0 g of PF 3 if the reaction has a 64.9% yield? 46.1 g P 4 Copyright Cengage Learning. All rights reserved 51

52 Section 3.11 The Concept of Limiting Reactant Agenda Balancing Equations Stoichiometry Calculations Limiting Reactants Percent yield Mass spectrometry Percent Composition

53 Section 3.11 The Concept of Limiting Reactant

54 Section 3.11 Balance the Equation The Concept of Limiting Reactant

55 Section 3.11 Balance the Equation The Concept of Limiting Reactant

56 Section 3.11 Balance the Equation The Concept of Limiting Reactant

57 Section 3.11 Balance the Equation The Concept of Limiting Reactant

58 Section 3.11 The Concept of Limiting Reactant

59 Section 3.11 The Concept of Limiting Reactant

60 Copy & balance the equation:

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63 Copy & balance the equation:

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65 Section 3.11 The Concept of Limiting Reactant

66 Section 3.11 The Concept of Limiting Reactant

67 Section 3.11 The Concept of Limiting Reactant

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73 Section 3.11 The Concept of Limiting Reactant Gas Ma ss Mole Molari ty Particl es

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76 Section 3.11 The Concept of Limiting Reactant Gas Ma ss Mole Molari ty Particl es

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78 Section 3.11 Memorize The Concept of Limiting Reactant

79 Stoichiometry Problems Example 1 How many moles of KClO 3 must decompose in order to produce 9 moles of oxygen gas? Problem: X molkclo 3 9mol O 2 2KClO 3 2KCl + 3O 2 Balanced : 2 molkclo 3 3mol O 2 Equation XmolKClO3 2 molkclo3 9molO2 3molO2 = 6 mol KClO 3

80 Stoichiometry Problems Example 1 How many moles of KClO 3 must decompose in order to produce 9 moles of oxygen gas? Problem: X molkclo 3 9mol O 2 2KClO 3 2KCl + 3O 2 Balanced : 2 molkclo 3 3mol O 2 Equation XmolKClO3 2 molkclo3 9molO2 3molO2 = 6 mol KClO 3

81 Example 2 Stoichiometry Problems How many grams of silver will be formed when 12 g of copper reacts with silver nitrate to produce copper (II) nitrate and silver? Problem: 12gCu Xg Ag Cu + 2 AgNO 3 2 Ag + Cu(NO 3 ) 2 Balanced: 63.5 gcu Equation 2(107.9) g Ag g XgAg 215.8gAg 12gCu 63.5gCu = 41 g Ag

82 Example 2 Stoichiometry Problems How many grams of silver will be formed when 12 g of copper reacts with silver nitrate to produce copper (II) nitrate and silver? Problem: 12gCu Xg Ag Cu + 2 AgNO 3 2 Ag + Cu(NO 3 ) 2 Balanced: 63.5 gcu Equation 2(107.9) g Ag g XgAg 215.8gAg 12gCu 63.5gCu = 41 g Ag

83 Stoichiometry Problems Example 3 If 12.0 grams of potassium chlorate decompose, how many moles of potassium chloride will be produced? Problem: 12gKClO 3 X moles KCl 2 KClO 3 2KCl + 3 O 2 Balanced: 2(122.6) g KClO 3 2 moles KCl Equation g Xmol KCl 2 molkcl 12gKClO gkclo3 = mol KCl

84 Stoichiometry Problems Example 3 If 12.0 grams of potassium chlorate decompose, how many moles of potassium chloride will be produced? Problem: 12gKClO 3 X moles KCl 2 KClO 3 2KCl + 3 O 2 Balanced: 2(122.6) g KClO 3 2 moles KCl Equation g Xmol KCl 2 molkcl 12gKClO gkclo3 = mol KCl

85 EARNING HECK Stoichiometry Problems In an experiment, red mercury (II) oxide powder is placed in an open flask and heated until it is converted to liquid mercury and oxygen gas. The liquid mercury has a mass of 92.6 g. What is the mass of oxygen formed in the reaction? Problem: 92.6 g Hg + X g O 2 2HgO 2Hg + O 2 Balanced: 2 ( 200.6) g Hg + 32 g O 2 Equation g Hg Xg O2 32g O2 92.6gHg g Hg = 7.39 g O 2

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87 EARNING HECK Stoichiometry Problems In an experiment, red mercury (II) oxide powder is placed in an open flask and heated until it is converted to liquid mercury and oxygen gas. The liquid mercury has a mass of 92.6 g. What is the mass of oxygen formed in the reaction? Problem: 92.6 g Hg + X g O 2 2HgO 2Hg + O 2 Balanced: 2 ( 200.6) g Hg + 32 g O 2 Equation g Hg Xg O2 32g O2 92.6gHg g Hg = 7.39 g O 2

88 EARNING HECK Stoichiometry Problems In an experiment, red mercury (II) oxide powder is placed in an open flask and heated until it is converted to liquid mercury and oxygen gas. The liquid mercury has a mass of 92.6 g. What is the mass of oxygen formed in the reaction? Problem: 92.6 g Hg + X g O 2 2HgO 2Hg + O 2 Balanced: 2 ( 200.6) g Hg + 32 g O 2 Equation g Hg Xg O2 32g O2 92.6gHg g Hg = 7.39 g O 2

89 Limiting Reactant

90 Bike Analogy Consider the following Analogy: 2 Wheels + 1 Body + 1 Handle bar + 1 Gear Chain = 1 bike How many bikes can be produced given, 8 wheels, 5 bodies, 6 handle bar and 5 gear chain?

91 Bike Analogy Consider the following Analogy: 2 Wheels + 1 Body + 1 Handle bar + 1 Gear Chain = 1 bike How many bikes can be produced given, 8 wheels, 5 bodies, 6 handle bar and 5 gear chain? Limiting Reactant Excess Reactant

92 Cheeseburger Analogy Consider the following Analogy: 2 Cheese + 1 burger patty + 1 bun = cheese burger Given three hamburger patties, six buns, and 12 slices of cheese, how many cheese burger can be made?

93 Cheeseburger Analogy Consider the following Analogy: 2 Cheese + 1 burger patty + 1 bun = 1 cheese burger LR Given three hamburger patties, six buns, and 12 slices of cheese, how many cheese burger can be made? ER

94 Limiting Reactant vs. Excess Reactants Limiting reactant is the reactant that runs out first In our examples, the limiting reactants will be the wheels in the bicycle analogy and the burger patty in our hamburger analogy When the limiting reactant is exhausted, then the reaction stops

95 Limiting Reactants Calculations 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. The reactant that resulted in the smallest amount of product is the limiting reactant(lr). 4. To find the amount of leftover reactant excess calculate the amount of the no LR used by the LR. 5. Subtract the calculated amount in step 4 from the original no LR amount given in the problem.

96 xample Determine how many moles of water can be formed if I 1 start with 2.75 moles of hydrogen and 1.75 moles of oxygen. Problem: 2.75 mol H 2 XmolH 2 O 2H 2 + O 2 2H 2 O Balanced: 2 mol H 2 2molH 2 O Equation XmolH2O 2molH2O 2.75 mol H2 2 mol H2 Limiting reactant =H 2 = 2.75 mol H 2 O Problem: 1.75 mol O 2 XmolH 2 O 2H 2 + O 2 2H 2 O Balanced: 1 mol O 2 2molH 2 O Equation XmolH2O 2molH2O 1.75 mol O2 1mol O2 = 3.50 mol H 2 O

97 xample Determine how many moles of water can be formed if I 1 start with 2.75 moles of hydrogen and 1.75 moles of oxygen. Problem: 2.75 mol H 2 XmolH 2 O 2H 2 + O 2 2H 2 O Balanced: 2 mol H 2 2molH 2 O Equation XmolH2O 2molH2O 2.75 mol H2 2 mol H2 Limiting reactant =H 2 = 2.75 mol H 2 O Problem: 1.75 mol O 2 XmolH 2 O 2H 2 + O 2 2H 2 O Balanced: 1 mol O 2 2molH 2 O Equation XmolH2O 2molH2O 1.75 mol O2 1mol O2 = 3.50 mol H 2 O

98 xample 2 If 2.0 mol of HF are exposed to 4.5 mol of SiO 2, which is the limiting reactant? Problem: 2.0 mol HF XmolH 2 O SiO 2 (s) + 4HF(g) SiF 4 (g) + 2H 2 O(l) Balanced: 4 mol HF 2molH 2 O Equation XmolH2O 2 molh2o 2.0mol HF 4.0mol HF = 1.0 mol H 2 O Problem: 4.5 mol SiO 2 XmolH 2 O SiO 2 (s) + 4HF(g) SiF 4 (g) + 2H 2 O(l) Balanced: 1 mol SiO 2 2molH 2 O Equation XmolH2O 2 molh2o 4.5molSiO2 1.0molSiO2 = 9.0 mol H 2 O Limiting reactant =HF

99 xample 2 If 2.0 mol of HF are exposed to 4.5 mol of SiO 2, which is the limiting reactant? Problem: 2.0 mol HF XmolH 2 O SiO 2 (s) + 4HF(g) SiF 4 (g) + 2H 2 O(l) Balanced: 4 mol HF 2molH 2 O Equation XmolH2O 2 molh2o 2.0mol HF 4.0mol HF = 1.0 mol H 2 O Problem: 4.5 mol SiO 2 XmolH 2 O SiO 2 (s) + 4HF(g) SiF 4 (g) + 2H 2 O(l) Balanced: 1 mol SiO 2 2molH 2 O Equation XmolH2O 2 molh2o 4.5molSiO2 1.0molSiO2 = 9.0 mol H 2 O Limiting reactant =HF

100 EARNING CHECK If 36.0 g of H 2 O is mixed with 167 g of Fe, which is the limiting reactant? Problem: 36.0 g H 2 O XgFe 2 O 3 2Fe(s) + 3H 2 O(g) Fe 2 O 3 (g) + 3H 2 (g) Balanced: 54 g H 2 O 159.6gFe 2 O 3 Equation XgFe2O gFe2O3 36.0g H2O 54.0gH2O Problem: 167 g Fe XgFe 2 O 3 = 106 g Fe 2 O 3 Limiting reactant =H 2 O 2Fe(s) + 3H 2 O(g) Fe 2 O 3 (g) + 3H 2 (g) Balanced: g Fe 159.6gFe 2 O 3 Equation XgFe2O gFe2O3 167g Fe gfe = 238 g Fe 2 O 3

101 EARNING CHECK If 36.0 g of H 2 O is mixed with 167 g of Fe, which is the limiting reactant? Problem: 36.0 g H 2 O XgFe 2 O 3 2Fe(s) + 3H 2 O(g) Fe 2 O 3 (g) + 3H 2 (g) Balanced: 54 g H 2 O 159.6gFe 2 O 3 Equation XgFe2O gFe2O3 36.0g H2O 54.0gH2O Problem: 167 g Fe XgFe 2 O 3 = 106 g Fe 2 O 3 Limiting reactant =H 2 O 2Fe(s) + 3H 2 O(g) Fe 2 O 3 (g) + 3H 2 (g) Balanced: g Fe 159.6gFe 2 O 3 Equation XgFe2O gFe2O3 167g Fe gfe = 238 g Fe 2 O 3

102 Limiting Reactants Calculations 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. The reactant that resulted in the smallest amount of product is the limiting reactant(lr). 4. To find the amount of leftover reactant excess calculate the amount of the no LR used by the LR. 5. Subtract the calculated amount in step 4 from the original no LR amount given in the problem.

103 XS Limiting Reactants LR 3Fe(s) + 4H 2 O(g) Fe 3 O 3 (g) + 4H 2 (g) Limiting reactant: H 2 O Excess reactant: Fe Products Formed: 107 g Fe 3 O 3 & 4.00 g H 2 Problem: XgFe 3Fe(s) + 4H 2 O(g) alanced: g Fe quation XgFe 111.6gFe 36.0 g H 2 O 54 g H 2 O 36gH2O 54 gh2o Fe 3 O 3 (g) + 4H 2 (g) = 74.4 g Fe used 167gFe g Fe= 92.6 g Fe Original Used = Excess left over iron

104 Percent Yield So far, the masses we have calculated from chemical equations were based on the assumption that each reaction occurred 100%. The THEORETICAL YIELD the maximum amount of product that can be produced in a reaction (calculated from the balanced equation) The ACTUAL YIELD is the amount of product that is actually produced in an experiment (usually less than the theoretical yield)

105 Percent Yield Theoretical Yield the maximum amount of product that can be produced in a reaction Percent Yield The actual amount of a given product as the percentage of the theoretical yield.

106 Look back at the problem from LEARNING CHECK. We found that 106 g Fe 2 O 3 could be formed from the reactants. In an experiment, you formed 90.4 g of Fe 2 O 3. What is your percent yield? % Yield = 90.4 g x 100 = 85.3% 106 g

107 xample 1 A 10.0 g sample of ethanol, C 2 H 5 OH, was boiled with excess acetic acid, CH 3 COOH, to produce 14.8 g of ethyl acetate, CH 3 COOC 2 H 5. What percent yield of ethyl acetate is this? Problem: 10.0g C 2 H 5 OH Xg CH 3 COOC 2 H 5 CH 3 COOH + C 2 H 5 OH CH 3 COOC 2 H 5 + H 2 O Balanced: 46.0 g C 2 H 5 OH 88.0 g CH 3 COOC 2 H 5 Equation XgCH3COOC2 H5 88.0gCH3COOC2H5 10.0gC2H5OH 46.0gC2H5OH = 19.1 g CH 3 COOC 2 H 5 % Yield = 14.8 g x 100 = 77.5% 19.1g

108 xample 1 A 10.0 g sample of ethanol, C 2 H 5 OH, was boiled with excess acetic acid, CH 3 COOH, to produce 14.8 g of ethyl acetate, CH 3 COOC 2 H 5. What percent yield of ethyl acetate is this? Problem: 10.0g C 2 H 5 OH Xg CH 3 COOC 2 H 5 CH 3 COOH + C 2 H 5 OH CH 3 COOC 2 H 5 + H 2 O Balanced: 46.0 g C 2 H 5 OH 88.0 g CH 3 COOC 2 H 5 Equation XgCH3COOC2 H5 88.0gCH3COOC2H5 10.0gC2H5OH 46.0gC2H5OH = 19.1 g CH 3 COOC 2 H 5 % Yield = 14.8 g x 100 = 77.5% 19.1g

109 ARNING HECK When 36.8 g of C 6 H 6 reacts with Cl 2, what is the theoretical yield of C 6 H 5 Cl produced? If the actual is 43.7 g, determine the percentage yield of C 6 H 5 Cl. Problem: 36.8g C 5 H 5 Xg C 5 H 5 Cl 2C 6 H 6 + Cl 2 2C 6 H 5 Cl + H 2 Balanced: g C 5 H g C 5 H 5 Cl Equation XgC6H5Cl 225.0gC6H5Cl 36.8gC6H gC6H6 = 53.1 g C 6 H 5 Cl % Yield = 43.7 g x 100 = 88.3% 53.1g

110 ARNING HECK When 36.8 g of C 6 H 6 reacts with Cl 2, what is the theoretical yield of C 6 H 5 Cl produced? If the actual is 43.7 g, determine the percentage yield of C 6 H 5 Cl. Problem: 36.8g C 5 H 5 Xg C 5 H 5 Cl 2C 6 H 6 + Cl 2 2C 6 H 5 Cl + H 2 Balanced: g C 5 H g C 5 H 5 Cl Equation XgC6H5Cl 225.0gC6H5Cl 36.8gC6H gC6H6 = 53.1 g C 6 H 5 Cl % Yield = 43.7 g x 100 = 88.3% 53.1g

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117 Mass Spectrometry Activities