1. Chapter 9a Chemical Quantities

2. Chapter 9 Table of Contents 2 9.1 Information Given by Chemical Equations 9.2 Mole–Mole Relationships 9.3 Mass Calculations

3. Section 9.1 Information Given by Chemical Equations Return to TOC 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 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 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 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 3. 152.00 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 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 6 ---------------- ------------------ ------------------- = 1 mole C 6 H 6 1 mole C 6 H 12 84.174g C 6 H 12 78.1134 g 1 mole C 6 H 6 1 mole C 6 H 12 4.8 g C 6 H 12 How many grams of H 2 will be needed? ? 4.5 g C 6 H 6 ---------------- ---------------- ----------------- = 1 mole C 6 H 6 3 moles H 2 2.0158g H 2 78.1134 g 1 mole C 6 H 6 1 mole H 2.35 g H 2

19. Section 9.3 Return to TOC Mass Calculations 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 180.1548g s 18.015g w 6 mole w 1 mole s 10.4 g s How many grams of carbon dioxide can be produced from 12.84 g of oxygen? 12.84 g o -------------- ------------ ------------ = 1 mole o 6 mole c 44.009g c 31.998g o 6 mole o 1 mole c 17.66 g c 12.84 g?

20. Section 9.3 Return to TOC Mass Calculations 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 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) 0.875 g c) 5.24 g d) 2.62 g

22. Section 9.3 Return to TOC Mass Calculations 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 9.3 Return to TOC Mass Calculations FfFf Chapter 9b Chemical Quantities

24. Section 9.3 Return to TOC Mass Calculations 24 9.4 The Concept of Limiting Reactants 9.5 Calculations Involving a Limiting Reactant 9.6Percent Yield

25. Section 9.4 The Concept of Limiting Reactants Return to TOC 25 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 Limiting Reactant Java.

26. Section 9.4 The Concept of Limiting Reactants Return to TOC Copyright © Cengage Learning. All rights reserved 26 N 2 (g) + 3H 2 (g)  2NH 3 (g) Limiting Reactant Mixture

27. Section 9.4 The Concept of Limiting Reactants Return to TOC 27 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

28. Section 9.5 Calculations Involving a Limiting Reactant Return to TOC 28 Determine which reactant is limiting to calculate correctly the amounts of products that will be formed.

29. Section 9.5 Calculations Involving a Limiting Reactant Return to TOC 29 Methane and water will react to form products according to the equation: CH 4 + H 2 O  3H 2 + CO Limiting Reactants

30. Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 30 H 2 O molecules are used up first, leaving two CH 4 molecules unreacted. Limiting Reactants

31. Section 9.5 Calculations Involving a Limiting Reactant Return to TOC 31 The amount of products that can form is limited by the water. Water is the limiting reactant. Methane is in excess. Limiting Reactants

32. Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 32 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

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

34. Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 34 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

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 Example – Continued

36. Section 9.5 Calculations Involving a Limiting Reactant Return to TOC 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?  Convert known masses of reactants to moles. 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?  Determine which reactant is limiting. react with all of the A  Only 0.500 mol B is available, so B is the limiting reactant. Example – Continued

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

39. Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 39 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

40. Section 9.5 Calculations Involving a Limiting Reactant Return to TOC 40 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

41. Section 9.5 Calculations Involving a Limiting Reactant Return to TOC 41 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 ? 2 2.00 g ? 2.00 g H 2 -------------- ------------- ------------------- = 1 mole H 2 2 mole HI 127.9124g HI 2.0158 g 1 mole H 2 1 mole HI 254 g HI 2.00g I 2 ------------- --------------- ----------------- = 1 mole I 2 2 moles HI 127.9124g HI 253.810 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.

42. Section 9.5 Calculations Involving a Limiting Reactant Return to TOC 42 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 ? 3 2.65g1.44 g ? 1.44 g C 3 H 8 ---------------- ---------------- ------------------- = 1 mole C 3 H 8 3 mole CO 2 44.009 g CO 2 44.0962 g 1 mole C 3 H 8 1 mole CO 2 4.31 g CO 2 2.65g O 2 ------------- ----------------- ------------------- = 1 mole O 2 3 moles CO 2 44.009 g CO 2 31.998 g 5 mole O 2 1 mole CO 2 2.19 g CO 2 45 O 2 is the limiting reactant and C 3 H 8 is in XS.

43. Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 43 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.

44. Section 9.5 Calculations Involving a Limiting Reactant Return to TOC Copyright © Cengage Learning. All rights reserved 44 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.

45. Section 9.5 Calculations Involving a Limiting Reactant Return to TOC 45 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 3 3.00 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

46. Section 9.6 Percent Yield Return to TOC 46 An important indicator of the efficiency of a particular laboratory or industrial reaction. Percent Yield

47. Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 47 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).

48. Section 9.6 Percent Yield Return to TOC 48 The actual amount of a given product as the percentage of the theoretical yield. Percent Yield 16.0gCaCO 3 x

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. 7.23 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

50. Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 50 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. 7.23 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

51. Section 9.6 Percent Yield Return to TOC Copyright © Cengage Learning. All rights reserved 51 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. 7.23 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

52. Section 9.6 Percent Yield Return to TOC 52 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% 1.00 g O 2 (1mole O 2 /32g O 2 )(2 mole SO 3 /3 mole O 2 ) (80g SO 3 /1 mole SO 3 ) = 1.67g SO 3 (1.50g SO 3 actual)/(1.67g SO 3 theoretical) x 100%= 89.8% yield

53. Section 9.6 Percent Yield Return to TOC 53 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 (85.0g PF 3 actual)/(X g theoretical ) x 100 = 64.9% X g PF 3 theoretical = 85.0x 100/64.9 = 131g PF 3 131.0g PF 3 (1 mole PF 3 /88g PF 3 )(1 mole P 4 /4 mole PF 3 ) (124g P 4 /1 mole P 4 ) = 46.1g P 4

54. Section 9.6 Percent Yield Return to TOC 54 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 2 143 g= 18.02 g H 2 O 1 mol H 2 O x 1 mol H 2 2.02 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%

55. Section 9.6 Percent Yield Return to TOC 55 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 2 227 g= 17.04 g NH 3 1 mol NH 3 x Step 3: Calculate % yield 40.5 g NH 3 227 g NH 3 = % yield = x 100%17.8%= actual theoretical x 100%

56. Section 9.6 Percent Yield Return to TOC 56 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 2 62 g= 18.02 g H 2 O 1 mol H 2 O x 1 mol H 2 2.02 g H 2 x 58 g H 2 O 62 g H 2 O = % yield = x 100%93 %= actual theoretical x 100% 2 mol H 2 O 1 mol O 2 x # g H 2 O=60 g O 2 68 g= 18.02 g H 2 O 1 mol H 2 O x 1 mol O 2 32 g O 2 x

57. Section 9.6 Percent Yield Return to TOC 57 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? 2. 107 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

58. Section 9.6 Percent Yield Return to TOC 58 More Percent Yield Questions 4. Iron pyrites (FeS 2 ) reacts with oxygen according to the following equation: 4FeS 2 + 11O 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? 5. 70 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

59. Section 9.6 Percent Yield Return to TOC 59 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 12.4 g= 32 g O 2 1 mol O 2 x 1 mol H 2 O 18.02 g H 2 O x Finally, calculate % yield 12.3 g O 2 12.4 g O 2 = % yield = x 100%98.9%= actual theoretical x 100%

60. Section 9.6 Percent Yield Return to TOC 60 Q2 2.107 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 3 118 g= 32 g O 2 1 mol O 2 x 1 mol KClO 3 122.55 g KClO 3 x Finally, calculate % yield 107 g O 2 118 g O 2 = % yield = x 100%91.1%= actual theoretical x 100%

61. Section 9.6 Percent Yield Return to TOC 61 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 264 g= 87.91 g FeS 1 mol FeS x Finally, calculate % yield 220 g O 2 264 g O 2 = % yield = x 100%83.4%= actual theoretical x 100%

62. Section 9.6 Percent Yield Return to TOC 62 4.4FeS 2 + 11O 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 2 181 g Fe 2 O 3 = 159.7 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 2 200. g Fe 2 O 3 = 159.7 g Fe 2 O 3 1 mol Fe 2 O 3 x 1 mol FeS 2 119.97 g FeS 2 x 143 g Fe 2 O 3 200. g Fe 2 O 3 = % yield = x 100%78.8%= actual theoretical x 100%

63. Section 9.6 Percent Yield Return to TOC 63 5.70 g of MnO 2 + 3.5 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 62. g Cl 2 = 71 g Cl 2 1 mol Cl 2 x 1 mol MnO 2 x 70 g MnO 2 57 g Cl 2 = 70.9 g Cl 2 1 mol Cl 2 x 1 mol MnO 2 86.94 g MnO 2 x x g Cl 2 57.08 g Cl 2 = % yield = x 100%42%= actual theoretical x 100% x g Cl 2 42% x 57 g Cl 2 100% = 24 g Cl 2 = For more lessons, visit www.chalkbored.com www.chalkbored.com