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Chapter 9 Chemical Quantities. Chapter 9 Table of Contents Copyright © Cengage Learning. All rights reserved 2 9.1 Information Given by Chemical Equations.

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Presentation on theme: "Chapter 9 Chemical Quantities. Chapter 9 Table of Contents Copyright © Cengage Learning. All rights reserved 2 9.1 Information Given by Chemical Equations."— Presentation transcript:

1 Chapter 9 Chemical Quantities

2 Chapter 9 Table of Contents Copyright © Cengage Learning. All rights reserved Information Given by Chemical Equations 9.2 Mole–Mole Relationships 9.3 Mass Calculations 9.4 The Concept of Limiting Reactants 9.5 Calculations Involving a Limiting Reactant 9.6Percent Yield

3 Section 9.1 Information Given by Chemical Equations Return to TOC Copyright © Cengage Learning. All rights reserved 3 A balanced chemical equation gives relative numbers (or moles) of reactant and product molecules that participate in a chemical reaction. The coefficients of a balanced equation give the relative numbers of molecules.

4 Section 9.1 Information Given by Chemical Equations Return to TOC Copyright © Cengage Learning. All rights reserved 4 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 molecule of ethanol reacts with 3 molecules of oxygen to produce 2 molecules of carbon dioxide and 3 molecules of water. 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water.

5 Section 9.1 Information Given by Chemical Equations Return to TOC Copyright © Cengage Learning. All rights reserved 5 Concept Check One of the key steps in producing pure copper from copper ores is the reaction below: Cu 2 S(s) + Cu 2 O(s) → Cu(s) + SO 2 (g) (unbalanced) After balancing the reaction, determine how many dozen copper atoms could be produced from a dozen molecules of the sulfide and two dozen of the oxide of copper(I). Also, how many moles of copper could be produced from one mole of the sulfide and two moles of the oxide? a) 1 dozen and 1 mole b) 2 dozen and 2 moles c) 3 dozen and 3 moles d) 6 dozen and 6 moles 2 6

6 Section 9.2 Return to TOC Mole–Mole Relationships Copyright © Cengage Learning. All rights reserved 6 A balanced equation can predict the moles of product that a given number of moles of reactants will yield. 2 mol of H 2 O yields 2 mol of H 2 and 1 mol of O 2. 4 mol of H 2 O yields 4 mol of H 2 and 2 mol of O 2. 2H 2 O(l) → 2H 2 (g) + O 2 (g)

7 Section 9.2 Return to TOC Mole–Mole Relationships Copyright © Cengage Learning. All rights reserved 7 The mole ratio allows us to convert from moles of one substance in a balanced equation to moles of a second substance in the equation. Mole Ratio

8 Section 9.2 Return to TOC Mole–Mole Relationships Copyright © Cengage Learning. All rights reserved 8 Consider the following balanced equation: Na 2 SiF 6 (s) + 4Na(s) → Si(s) + 6NaF(s) How many moles of NaF will be produced if 3.50 moles of Na is reacted with excess Na 2 SiF 6 ? Where are we going? We want to determine the number of moles of NaF produced by Na with excess Na 2 SiF 6. What do we know? The balanced equation. We start with 3.50 mol Na. Example

9 Section 9.2 Return to TOC Mole–Mole Relationships Copyright © Cengage Learning. All rights reserved 9 Consider the following balanced equation: Na 2 SiF 6 (s) + 4Na(s) → Si(s) + 6NaF(s) How many moles of NaF will be produced if 3.50 moles of Na is reacted with excess Na 2 SiF 6 ? How do we get there? Example

10 Section 9.2 Return to TOC Mole–Mole Relationships Copyright © Cengage Learning. All rights reserved 10 Exercise Propane, C 3 H 8, is a common fuel used in heating homes in rural areas. Predict how many moles of CO 2 are formed when 3.74 moles of propane are burned in excess oxygen according to the equation: C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O a)11.2 moles b)7.48 moles c)3.74 moles d)1.25 moles

11 Section 9.3 Return to TOC Mass Calculations Copyright © Cengage Learning. All rights reserved 11 1.Balance the equation for the reaction. 2.Convert the masses of reactants or products to moles. 3.Use the balanced equation to set up the appropriate mole ratio(s). 4.Use the mole ratio(s) to calculate the number of moles of the desired reactant or product. 5.Convert from moles back to masses. Steps for Calculating the Masses of Reactants and Products in Chemical Reactions

12 Section 9.3 Return to TOC Mass Calculations Copyright © Cengage Learning. All rights reserved 12 Stoichiometry is the process of using a balanced chemical equation to determine the relative masses of reactants and products involved in a reaction. Stoichiometry

13 Section 9.3 Return to TOC Mass Calculations Copyright © Cengage Learning. All rights reserved 13 For the following unbalanced equation: Cr(s) + O 2 (g) → Cr 2 O 3 (s) How many grams of chromium(III) oxide can be produced from 15.0 g of solid chromium and excess oxygen gas? Where are we going? We want to determine the mass of Cr 2 O 3 produced by Cr with excess O 2. Example

14 Section 9.3 Return to TOC Mass Calculations Copyright © Cengage Learning. All rights reserved 14 For the following unbalanced equation: Cr(s) + O 2 (g) → Cr 2 O 3 (s) How many grams of chromium(III) oxide can be produced from 15.0 g of solid chromium and excess oxygen gas? What do we know? The unbalanced equation. We start with 15.0 g Cr. We know the atomic masses of chromium (52.00 g/mol) and oxygen (16.00 g/mol) from the periodic table. Example

15 Section 9.3 Return to TOC Mass Calculations Copyright © Cengage Learning. All rights reserved 15 For the following unbalanced equation: Cr(s) + O 2 (g) → Cr 2 O 3 (s) How many grams of chromium(III) oxide can be produced from 15.0 g of solid chromium and excess oxygen gas? What do we need to know? We need to know the balanced equation. 4Cr(s) + 3O 2 (g) → 2Cr 2 O 3 (s) We need the molar mass of Cr 2 O g/mol Example

16 Section 9.3 Return to TOC Mass Calculations Copyright © Cengage Learning. All rights reserved 16 4Cr(s) + 3O 2 (g) → 2Cr 2 O 3 (s) Convert the mass of Cr to moles of Cr. Determine the moles of Cr 2 O 3 produced by using the mole ratio from the balanced equation. How do we get there?

17 Section 9.3 Return to TOC Mass Calculations Copyright © Cengage Learning. All rights reserved 17 4Cr(s) + 3O 2 (g) → 2Cr 2 O 3 (s) Convert moles of Cr 2 O 3 to grams of Cr 2 O 3. Conversion string: How do we get there?

18 Section 9.3 Return to TOC Mass Calculations Copyright © Cengage Learning. All rights reserved 18 C 6 H 6 + H 2  C 6 H 12 Reaction Stoichiometry Balanced? 3 How many grams of cyclohexane C 6 H 12 can be made from 4.5 grams of benzene C 6 H 6 ? Use the table method. 4.5g Moles level Grams level? 4.5 g C 6 H = 1 mole C 6 H 6 1 mole C 6 H g C 6 H g 1 mole C 6 H 6 1 mole C 6 H g C 6 H 12 How many grams of H 2 will be needed? ? 4.5 g C 6 H = 1 mole C 6 H 6 3 moles H g H g 1 mole C 6 H 6 1 mole H 2.35 g H 2

19 Section 9.3 Return to TOC Mass Calculations Copyright © Cengage Learning. All rights reserved 19 C 6 H 12 O 6 + O 2  CO 2 + H 2 O 66 6 How many grams of sugar are needed to make 6.23g water? 6.23 g ? 6.23 g w = 1 mole w 1mole s g s g w 6 mole w 1 mole s 10.4 g s How many grams of carbon dioxide can be produced from g of oxygen? g o = 1 mole o 6 mole c g c g o 6 mole o 1 mole c g c g?

20 Section 9.3 Return to TOC Mass Calculations Copyright © Cengage Learning. All rights reserved 20 Exercise Tin(II) fluoride is added to some dental products to help prevent cavities. Tin(II) fluoride is made according to the following equation: Sn(s) + 2HF(aq)  SnF 2 (aq) + H 2 (g) How many grams of tin(II) fluoride can be made from 55.0 g of hydrogen fluoride if there is plenty of tin available to react? a)431 g b)215 g c)72.6 g d)1.37 g

21 Section 9.3 Return to TOC Mass Calculations Copyright © Cengage Learning. All rights reserved 21 Exercise Consider the following reaction: PCl 3 + 3H 2 O  H 3 PO 3 + 3HCl What mass of H 2 O is needed to completely react with 20.0 g of PCl 3 ? a) 7.87 g b) g c) 5.24 g d) 2.62 g

22 Section 9.3 Return to TOC Mass Calculations Copyright © Cengage Learning. All rights reserved 22 Exercise Consider the following reaction where X represents an unknown element: 6 X(s) + 2 B 2 O 3 (s)  B 4 X 3 (s) + 3 XO 2 (g) If 175 g of X reacts with diboron trioxide to produce 2.43 moles of B 4 X 3, what is the identity of X? a)Ge b)Mg c)Si d)C

23 Section 9.4 The Concept of Limiting Reactants Return to TOC Copyright © Cengage Learning. All rights reserved 23 Contains the relative amounts of reactants that matches the numbers in the balanced equation. N 2 (g) + 3H 2 (g)  2NH 3 (g) Stoichiometric Mixture

24 Section 9.4 The Concept of Limiting Reactants Return to TOC Copyright © Cengage Learning. All rights reserved 24 N 2 (g) + 3H 2 (g)  2NH 3 (g) Limiting Reactant Mixture

25 Section 9.4 The Concept of Limiting Reactants Return to TOC Copyright © Cengage Learning. All rights reserved 25 N 2 (g) + 3H 2 (g)  2NH 3 (g) Limiting reactant is the reactant that runs out first and thus limits the amounts of product(s) that can form.  H 2 Limiting Reactant Mixture

26 Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 26 Determine which reactant is limiting to calculate correctly the amounts of products that will be formed.

27 Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 27 Methane and water will react to form products according to the equation: CH 4 + H 2 O  3H 2 + CO Limiting Reactants

28 Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 28 H 2 O molecules are used up first, leaving two CH 4 molecules unreacted. Limiting Reactants

29 Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 29 The amount of products that can form is limited by the water. Water is the limiting reactant. Methane is in excess. Limiting Reactants

30 Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 30 1.Write and balance the equation for the reaction. 2.Convert known masses of reactants to moles. 3.Using the numbers of moles of reactants and the appropriate mole ratios, determine which reactant is limiting. 4.Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product. 5.Convert from moles of product to grams of product, using the molar mass (if this is required by the problem). Steps for Solving Stoichiometry Problems Involving Limiting Reactants

31 Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 31 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? Example

32 Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 32 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. What do 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. Let’s Think About It

33 Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 33 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 Example – Continued

34 Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 34 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 How do we get there?  Convert known masses of reactants to moles. Example – Continued

35 Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 35 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 How do we get there?  Determine which reactant is limiting. react with all of the A  Only mol B is available, so B is the limiting reactant. Example – Continued

36 Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 36 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 How do we get there?  Compute the number of moles of C produced. Example – Continued

37 Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 37 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 How do we get there?  Convert from moles of C to grams of C using the molar mass. Example – Continued

38 Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 38 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. Notice

39 Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 39 H 2 + I 2  HI How many grams of HI can be formed from 2.00 g H 2 and 2.00 g of I 2 ? g ? 2.00 g H = 1 mole H 2 2 mole HI g HI g 1 mole H 2 1 mole HI 254 g HI 2.00g I = 1 mole I 2 2 moles HI g HI g 1 mole I 2 1 mole HI 2.02 g HI Limiting Reactant is always the smallest value! I 2 is the limiting reactant and H 2 is in XS.

40 Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 40 C 3 H 8 + O 2  CO 2 + H 2 O How many grams of CO 2 can be formed from 1.44 g C 3 H 8 and 2.65 g of O 2 ? g1.44 g ? 1.44 g C 3 H = 1 mole C 3 H 8 3 mole CO g CO g 1 mole C 3 H 8 1 mole CO g CO g O = 1 mole O 2 3 moles CO g CO g 5 mole O 2 1 mole CO g CO 2 45 O 2 is the limiting reactant and C 3 H 8 is in XS.

41 Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 41 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.

42 Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 42 Concept Check Consider the equation: A + 3B  4C. If 3.0 moles of A is reacted with 6.0 moles of B, which of the following is true after the reaction is complete? a) A is the leftover reactant because you only need 2 moles of A and have 3. b) A is the leftover reactant because for every 1 mole of A, 4 moles of C are produced. c) B is the leftover reactant because you have more moles of B than A. d) B is the leftover reactant because 3 moles of B react with every 1 mole of A.

43 Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 43 Exercise How many grams of NH 3 can be produced from the mixture of 3.00 g each of nitrogen and hydrogen by the Haber process? N 2 + 3H 2 → 2NH 3 a)2.00 g b)3.00 g c)3.64 g d)1.82 g 3.00 g N 2 (1mole/28g N 2 )(2 moles NH 3 /1 moles N 2 ) (17 g NH 3 /1 mole NH 3 ) = 3.64 g NH g H 2 (1mole/2g H 2 )(2 moles NH 3 /3 moles H 2 ) (17 g NH 3 /1 mole NH 3 ) = 17.0 g NH 3

44 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 44 An important indicator of the efficiency of a particular laboratory or industrial reaction. Percent Yield

45 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 45 Theoretical Yield  The maximum amount of a given product that can be formed when the limiting reactant is completely consumed. The actual yield (amount actually produced) of a reaction is usually less than the maximum expected (theoretical yield).

46 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 46 The actual amount of a given product as the percentage of the theoretical yield. Percent Yield

47 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 47 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 We determined that 8.33 g C should be produced g of C is what was actually made in the lab. What is the percent yield of the reaction? Where are we going?  We want to determine the percent yield of the reaction. Example – Recall

48 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 48 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 We determined that 8.33 g C should be produced g of C is what was actually made in the lab. What is the percent yield of the reaction? What do we know?  We know the actual and theoretical yields. Example – Recall

49 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 49 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 We determined that 8.33 g C should be produced g of C is what was actually made in the lab. What is the percent yield of the reaction? How do we get there? Example – Recall

50 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 50 Exercise Find the percent yield of product if 1.50 g of SO 3 is produced from 1.00 g of O 2 and excess sulfur via the reaction: 2S + 3O 2 → 2SO 3 a)40.0% b)80.0% c)89.8% d)92.4%

51 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 51 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? a) 29.9 g b) 46.1 g c) 184 g d) 738 g

52 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 52

53 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 53 Assignment: Read 5.6 up to sample (238-9) 1. Define the following terms: yield, theoretical yield, actual yield, percentage yield. 2. Based on your reading, give 4 reasons why the actual yield in a chemical reaction often falls short of the theoretical yield. 3. Read the sample problem on the next slide and try the practice problem on slide number 5 4. When 5.00 g of KClO 3 is heated it decomposes according to the equation: 2KClO 3  2KCl + 3O 2 a) Calculate the theoretical yield of oxygen. b) Give the % yield if 1.78 g of O 2 is produced. c) How much O 2 would be produced if the percentage yield was 78.5%?

54 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 54 Answers 1) Yield: the amount of product Theoretical yield: the amount of product we expect, based on stoichiometric calculations Actual yield: amount of product from a procedure or experiment (this is given in the question) Percent yield: = x 100% actual yield theoretical yield 2) Not all product is recovered (e.g. spattering) Reactant impurities (e.g. weigh out 100 g of chemical which has 20 g of junk) A side reaction occurs (e.g. MgO vs. Mg 3 N 2 ) The reaction does not go to completion

55 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 55 Sample problem Q - What is the % yield of H 2 O if 138 g H 2 O is produced from 16 g H 2 and excess O 2 ? Step 1: write the balanced chemical equation 2H 2 + O 2  2H 2 O Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: 2 mol H 2 O 2 mol H 2 x # g H 2 O=16 g H g= g H 2 O 1 mol H 2 O x 1 mol H g H 2 x Step 3: Calculate % yield 138 g H 2 O 143 g H 2 O = % yield = x 100%96.7%= actual theoretical x 100%

56 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 56 Practice problem Q - What is the % yield of NH 3 if 40.5 g NH 3 is produced from 20.0 mol H 2 and excess N 2 ? Step 1: write the balanced chemical equation N 2 + 3H 2  2NH 3 Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: 2 mol NH 3 3 mol H 2 x # g NH 3 =20.0 mol H g= g NH 3 1 mol NH 3 x Step 3: Calculate % yield 40.5 g NH g NH 3 = % yield = x 100%17.8%= actual theoretical x 100%

57 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 57 4) 2KClO 3  2KCl + 3O 2 a) b) c) Answers 3 mol O 2 2 mol KClO 3 x # g O 2 = (also works if you use mol O 2 ) 5.00 g KClO g= 32 g O 2 1 mol O 2 x 1 mol KClO g KClO 3 x 1.78 g O g O 2 = % yield = x 100%90.9%= actual theoretical x 100% x g O g O 2 = % yield = x 100%78.5%= actual theoretical x 100% x g O % x g O 2 100% = g O 2 =

58 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 58 Challenging question 2H 2 + O 2  2H 2 O What is the % yield of H 2 O if 58 g H 2 O are produced by combining 60 g O 2 and 7.0 g H 2 ? Hint: determine limiting reagent first 2 mol H 2 O 2 mol H 2 x # g H 2 O=7.0 g H g= g H 2 O 1 mol H 2 O x 1 mol H g H 2 x 58 g H 2 O 62.4 g H 2 O = % yield = x 100%92.9%= actual theoretical x 100% 2 mol H 2 O 1 mol O 2 x # g H 2 O=60 g O 2 68 g= g H 2 O 1 mol H 2 O x 1 mol O 2 32 g O 2 x

59 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 59 More Percent Yield Questions Note: try “shortcut” for limiting reagent problems 1. The electrolysis of water forms H 2 and O 2. 2H 2 O  2H 2 + O 2 What is the % yield of O 2 if 12.3 g of O 2 is produced from the decomposition of 14.0 g H 2 O? g of oxygen is produced by heating 300 grams of potassium chlorate. Calculate % yield.2KClO 3  2KCI + 3O 2 3. What is the % yield of ferrous sulphide if 3.00 moles of Fe reacts with excess sulfur to produce 220 grams of ferrous sulphide? Fe + S  FeS

60 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 60 More Percent Yield Questions 4. Iron pyrites (FeS 2 ) reacts with oxygen according to the following equation: 4FeS O 2  2Fe 2 O 3 + 8SO 2 If 300 g of iron pyrites is burned in 200 g of O 2, 143 grams of ferric oxide is produced. What is the percent yield of ferric oxide? grams of manganese dioxide is mixed with 3.5 moles of hydrochloric acid. How many grams of Cl 2 will be produced from this reaction if the % yield for the process is 42%? MnO 2 + 4HCI  MnCl 2 + 2H 2 O + Cl 2

61 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 61 Q1 1.The electrolysis of water forms H 2 & O 2. 2H 2 O  2H 2 + O 2 Give the percent yield of O 2 if 12.3 g O 2 is produced from the decomp. of 14 g H 2 O? Actual yield is given: 12.3 g O 2 Next, calculate theoretical yield 1 mol O 2 2 mol H 2 O x # g O 2 = 14.0 g H 2 O g= 32 g O 2 1 mol O 2 x 1 mol H 2 O g H 2 O x Finally, calculate % yield 12.3 g O g O 2 = % yield = x 100%98.9%= actual theoretical x 100%

62 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 62 Q g of oxygen is produced by heating 300 grams of potassium chlorate. 2KClO 3  2KCI + 3O 2 Actual yield is given: 107 g O 2 Next, calculate theoretical yield 3 mol O 2 2 mol KClO 3 x # g O 2 = 300 g KClO g= 32 g O 2 1 mol O 2 x 1 mol KClO g KClO 3 x Finally, calculate % yield 107 g O g O 2 = % yield = x 100%91.1%= actual theoretical x 100%

63 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 63 Q3 3.What is % yield of ferrous sulfide if 3 mol Fe produce 220 grams of ferrous sulfide? Fe + S  FeS Actual yield is given: 220 g FeS Next, calculate theoretical yield 1 mol FeS 1 mol Fe x # g FeS=3.00 mol Fe g= g FeS 1 mol FeS x Finally, calculate % yield 220 g O g O 2 = % yield = x 100%83.4%= actual theoretical x 100%

64 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved FeS O 2  2Fe 2 O 3 + 8SO 2 If 300 g of FeS 2 is burned in 200 g of O 2, 143 g Fe 2 O 3 results. % yield Fe 2 O 3 ? First, determine limiting reagent 2 mol Fe 2 O 3 11 mol O 2 x 200 g O g Fe 2 O 3 = g Fe 2 O 3 1 mol Fe 2 O 3 x 1 mol O 2 32 g O 2 x 2 mol Fe 2 O 3 4 mol FeS 2 x # g Fe 2 O 3 = 300 g FeS g Fe 2 O 3 = g Fe 2 O 3 1 mol Fe 2 O 3 x 1 mol FeS g FeS 2 x 143 g Fe 2 O g Fe 2 O 3 = % yield = x 100%78.8%= actual theoretical x 100%

65 Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved g of MnO mol HCl gives a 42% yield. How many g of Cl 2 is produced? MnO 2 + 4HCI  MnCl 2 + 2H 2 O + Cl 2 1 mol Cl 2 4 mol HCl x 3.5 mol HCl g Cl 2 = 71 g Cl 2 1 mol Cl 2 x 1 mol MnO 2 x # g Cl 2 = 70 g MnO g Cl 2 = 70.9 g Cl 2 1 mol Cl 2 x 1 mol MnO g MnO 2 x x g Cl g Cl 2 = % yield = x 100%42%= actual theoretical x 100% x g Cl 2 42% x g Cl 2 100% = 24 g Cl 2 = For more lessons, visit


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