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1 Chapter 5 Chemical Reactions Chemical Changes Chemical Equations Balancing a Chemical Equation.

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1 1 Chapter 5 Chemical Reactions Chemical Changes Chemical Equations Balancing a Chemical Equation

2 2 The physical properties of a substance are the characteristics we can observe or measure without changing the substance. Physical Properties

3 3 In a physical change, The identity and composition of the substance do not change. The state can change or the material can be torn into smaller pieces. In a chemical change, New substances form with different compositions and properties. A chemical reaction takes place. Physical and Chemical Change

4 4

5 5 Some Examples of Chemical and Physical Changes

6 6 Classify each of the following as a 1) physical change or 2) chemical change A. ____ Burning a candle. B. ____ Ice melting on the street. C. ____ Toasting a marshmallow. D. ____ Cutting a pizza. E. ____ Polishing a silver bowl. Learning Check

7 7 Classify each of the following as a 1) physical change or 2) chemical change A. 2 Burning a candle. B. 1 Ice melting on the street. C. 2 Toasting a marshmallow. D. 1 Cutting a pizza. E. 2 Polishing a silver bowl. Solution

8 8 Chemical Reaction In a chemical reaction, a chemical change produces one or more new substances. During a reaction, old bonds are broken and new bonds are formed.

9 9 In a chemical reaction, atoms in the reactants are rearranged to form one or more different substances. In this reaction, Fe and O 2 react to form rust (Fe 2 O 3 ). 4Fe + 3O 2 2Fe 2 O 3 Chemical Reaction

10 10 A chemical equation Shows the chemical formulas of the reactants to the left of an arrow and the products on the right. Reactants Products MgO + CCO + Mg Can be read in words. Magnesium oxide reacts with carbon to form carbon monoxide and magnesium. Writing a Chemical Equation

11 11 Symbols Used in Equations Symbols used in equations show the states of the reactants and products and the reaction conditions.

12 12 4 NH O 2 4 NO + 6 H 2 O Four molecules of NH 3 react with five molecules of O 2 to produce four molecules of NO and six molecules of H 2 O. or Four moles of NH 3 react with 5 moles of O 2 to produce four moles of NO and six moles of H 2 O. Quantities in A Chemical Reaction

13 13 In any ordinary chemical reaction, matter is not created nor destroyed. + + H 2 + Cl 2 2 HCl Total atoms=Total atoms 2 H, 2 Cl2H, 2 Cl Total Mass=Total Mass 2(1.0) + 2(35.5)2(36.5) 73.0 g=73.0 g Law of Conservation of Mass

14 14 Balancing a Chemical Equation A chemical equation is balanced when there are the same numbers of each type of atom on both sides of the equation. Al + S Al 2 S 3 Not Balanced 2Al + 3S Al 2 S 3 Balanced

15 15 To balance an equation, place coefficients in front of the appropriate formulas. 4 NH O 2 4 NO + 6 H 2 O Check the balance by counting the atoms of each element in the reactants and the products. 4 N (4 x 1N) = 4 N (4 x 1N) 12 H (4 x 3H) = 12 H (6 x 2H) 10 O (5 x 2O)= 10 O(4O + 6O) Using Coefficients to Balance

16 16 Balance one element at a time. Use only coefficients to balance. Fe 3 O 4 + H 2 Fe + H 2 O Fe:Fe 3 O 4 + H 2 3Fe + H 2 O O:Fe 3 O 4 + H 2 3Fe + 4H 2 O H: Fe 3 O 4 + 4H 2 3Fe + 4H 2 O Steps in Balancing an Equation

17 17 Check the balance of atoms in the following: Fe 3 O H 2 3 Fe + 4 H 2 O A. Number of H atoms in products. 1) 22) 43) 8 B. Number of O atoms in reactants. 1) 22) 43) 8 C. Number of Fe atoms in reactants. 1) 12) 33) 4 Learning Check

18 18 Fe 3 O H 2 3 Fe + 4 H 2 O A. Number of H atoms in products. 3) 8 (4H 2 O) B. Number of O atoms in reactants. 2) 4 (Fe 3 O 4 ) C. Number of Fe atoms in reactants. 2) 3 (Fe 3 O 4 ) Solution

19 19 Balancing with Polyatomic Ions Polyatomic ions can be balanced as a unit when they appear on both sides. Pb(NO 3 ) 2 + NaCl NaNO 3 + PbCl 2 Balance NO 3 - as a unit Pb(NO 3 ) 2 + NaCl 2NaNO 3 + PbCl 2 2 NO 3 - = 2 NO 3 - Balance Na (or Cl) Pb(NO 3 ) 2 + 2NaCl 2NaNO 3 + PbCl 2 2Na + = 2Na + 2Cl - = 2Cl -

20 20 Balance each equation. The coefficients in the answers are read from left to right. A. __Mg + __N 2 __Mg 3 N 2 1) 1, 3, 2 2) 3, 1, 2 3) 3, 1, 1 B.__Al + __Cl 2 __AlCl 3 1) 3, 3, 22) 1, 3, 1 3) 2, 3, 2 Learning Check

21 21 A. 3) 3, 1, 1 3 Mg + 1 N 2 1 Mg 3 N 2 B. 3) 2, 3, 2 2 Al + 3 Cl 2 2 AlCl 3 Solution

22 22 A. __Fe 2 O 3 + __C __Fe + __CO 2 1) 2, 3, 2,3 2) 2, 3, 4, 3 3) 1, 1, 2, 3 B. __Al + __FeO __Fe + __Al 2 O 3 1) 2, 3, 3, 1 2) 2, 1, 1, 1 3) 3, 3, 3, 1 C. __Al + __H 2 SO 4 __Al 2 (SO 4 ) 3 + __H 2 1) 3, 2, 1, 2 2) 2, 3, 1, 3 3) 2, 3, 2, 3 Learning Check

23 23 A. 2) 2, 3, 4, 3 2 Fe 2 O C 4 Fe + 3 CO 2 B. 1) 2, 3, 3, 1 2 Al + 3 FeO 3 Fe + 1 Al 2 O 3 C. 2) 2, 3, 1, 3 2 Al + 3 H 2 SO 4 1 Al 2 (SO 4 ) H 2 Solution

24 24 Types of Reactions Chapter 5 Chemical Reactions

25 25 Chemical reactions are classified into general types: Combination Decomposition Single Replacement Double Replacement Combustion Types of Reactions

26 26 In a combination reaction, two or more elements or simple compounds combine to form one product. A + B AB Examples H 2 + Cl 2 2HCl 2S + 3O 2 2SO 3 4Fe + 3O 2 2Fe 2 O 3 Combination Reactions

27 27 In a combination reaction, magnesium and oxygen react to form magnesium oxide. 2Mg + O 2 2MgO Combination Reactions Mg O2O2 MgO

28 28 In a decomposition reaction, one substance is broken down (split) into two or more simpler substances. ABA + B 2HgO2Hg + O 2 2KClO 3 2KCl + 3 O 2 Decomposition Reactions

29 29 Classify the following reactions as 1) combination or 2) decomposition: ___A. H 2 + Br 2 2HBr ___B. Al 2 (CO 3 ) 3 Al 2 O 3 + 3CO 2 ___C. 4 Al + 3C Al 4 C 3 Learning Check

30 30 Classify the following reactions as 1) combination or 2) decomposition: 1 A. H 2 + Br 2 2HBr 2 B. Al 2 (CO 3 ) 3 Al 2 O 3 + 3CO 2 1 C. 4 Al + 3C Al 4 C 3 Solution

31 31 In a single replacement, one element takes the place of an element in a reacting compound. A + BC AC + B Zn(s) + HCl(aq) ZnCl 2 (aq) + H 2 (g) Single Replacement Zn HCl H2H2 ZnCl 2

32 32 In a double replacement, the positive ions in the reacting compounds switch places. AB + CD AD + CB AgNO 3 + NaCl AgCl + NaNO 3 ZnS + 2HCl ZnCl 2 +H 2 S Double Replacement

33 33 Example of a Double Replacement When solutions of sodium sulfate and barium chloride are mixed, solid BaSO 4 is produced. BaCl 2 + Na 2 SO 4 BaSO 4 + 2NaCl BaSO 4

34 34 Classify each of the following reactions as a 1) single replacement or 2) double replacement __A. 2Al + 3H 2 SO 4 Al 2 (SO 4 ) 3 + 3H 2 __B. Na 2 SO 4 + 2AgNO 3 Ag 2 SO 4 + 2NaNO 3 __C. 3C + Fe 2 O 3 2Fe + 3CO Learning Check

35 35 Classify each of the following reactions as a 1) single replacement or 2) double replacement 1 A. 2Al + 3H 2 SO 4 Al 2 (SO 4 ) 3 + 3H 2 2 B. Na 2 SO 4 + 2AgNO 3 Ag 2 SO 4 + 2NaNO 3 1 C. 3C + Fe 2 O 3 2Fe + 3CO Solution

36 36 In a combustion reaction, a reactant often containing carbon reacts with oxygen O 2. C + O 2 CO 2 CH 4 + 2O 2 CO 2 + 2H 2 O C 3 H 8 + 5O 2 3CO 2 + 4H 2 O Many combustion reactions utilize fuels that are burned in oxygen to produce CO 2, H 2 O, and energy. Combustion

37 37 Balance the combustion equation: ___C 5 H 12 + ___O 2 ___CO 2 + ___H 2 O Learning Check

38 38 Balance the combustion equation: 1 C 5 H O 2 5 CO H 2 O Solution

39 39 Summary of Reaction Types

40 40 Learning Check Identify each reaction as 1) combination 2) decomposition 3) combustion 4) single replacement 5) double replacement A. 3Ba + N 2 Ba 3 N 2 B. 2Ag + H 2 S Ag 2 S + H 2 C. SiO 2 + 4HF SiF 4 + 2H 2 O D. PbCl 2 + K 2 SO 4 2KCl + PbSO 4 E. K 2 CO 3 K 2 O + CO 2 F. C 2 H 4 + 3O 2 2CO 2 + 2H 2 O

41 41 Solution Identify each reaction as 1) combination 2) decomposition 3) combustion 4) single replacement 5) double replacement 1 A. 3Ba + N 2 Ba 3 N 2 4 B. 2Ag + H 2 S Ag 2 S + H 2 5 C. SiO 2 + 4HF SiF 4 + 2H 2 O 5 D. PbCl 2 + K 2 SO 4 2KCl + PbSO 4 2 E. K 2 CO 3 K 2 O + CO 2 3 F. C 2 H 4 + 3O 2 2CO 2 + 2H 2 O

42 42 Chapter 5 Chemical Reactions Oxidation-Reduction Reactions

43 43 Oxidation and reduction Are an important type of reaction. Provide us with energy from food. Provide electrical energy in batteries. Occur when iron rusts. 4Fe + 3O 2 2Fe 2 O 3 Oxidation and Reduction

44 44 An oxidation-reduction reaction involves the transfer of electrons from one reactant to another. In oxidation, electrons are lost. Zn Zn e - (loss of electrons) In reduction, electrons are gained. Cu e - Cu (gain of electrons) Electron Loss and Gain

45 45 Half-Reactions for Oxidation- Reduction In the oxidation-reduction reaction of zinc and copper(II) sulfate, the zinc is oxidized and the Cu 2+ (from Cu 2+ SO 4 2- ) is reduced. Zn Zn e - oxidation Cu e - Cureduction

46 46 Identify each of the following as an 1) oxidation or a 2) reduction: __A. Sn Sn e- __B. Fe e - Fe 2+ __C. Cl 2 + 2e - 2Cl - Learning Check

47 47 Identify each of the following as an 1) oxidation or a 2) reduction: 1 A. Sn Sn e- 2 B. Fe e - Fe 2+ 2 C. Cl 2 + 2e - 2Cl - Solution

48 48 In a balanced oxidation-reduction equation, the loss of electrons is equal to the gain of electrons. Zn + Cu 2+ Zn 2+ + Cu The loss and gain of two electrons is shown in the separate oxidation and reduction reactions. Zn Zn e - oxidation Cu e - Cu reduction Balanced Red-Ox Equations

49 49 In light-sensitive sunglasses, UV light initiates an oxidation-reduction reaction. uv light Ag + + Cl - Ag + Cl A. Which reactant is oxidized? 1) Ag + 2) Cl - 3) Ag B. Which reactant is reduced? 1) Ag + 2) Cl - 3) Cl Learning Check

50 50 In light-sensitive sunglasses, UV light initiates an oxidation-reduction reaction. uv light Ag + + Cl - Ag + Cl A. Which reactant is oxidized 2) Cl - Cl - Cl + e - B. Which reactant is reduced? 1) Ag + Ag + + e- Ag Solution

51 51 Write the separate oxidation and reduction reactions for the following equation. 2Cs + F 2 2CsF Learning Check

52 52 Write the separate oxidation and reduction reactions for the following equation. 2Cs + F 2 2CsF Cs Cs + + 1e - oxidation F + 1e - F - reduction Solution

53 53 An early definition of oxidation is the addition of oxygen O 2 to a reactant. A metal or nonmetal is oxidized while the O 2 is reduced to O 2-. 4K + O 2 2K 2 O C + O 2 CO 2 2SO 2 + O 2 2SO 3 Oxidation with Oxygen

54 54 In organic and biological reactions, oxidation involves the loss of hydrogen atoms and reduction involves a gain of hydrogen atoms. oxidation = Loss of H reduction = Gain of H CH 3 OH H 2 CO + 2H (loss of H) Methanol Formaldehyde Gain and Loss of Hydrogen

55 55 Summary

56 56 Learning Check Identify the substances that are oxidized and reduced in the following reactions. A. 4Fe + 3O 2 2Fe 2 O 3 B. 6Na + N 2 2Na 3 N C. 2K + I 2 2KI

57 57 Solution A. 4Fe + 3O 2 2Fe 2 O 3 Fe oxidized; O 2 reduced B. 6Na + N 2 2Na 3 N Na oxidized: N 2 reduced C. 2K + I 2 2KI K oxidized: I 2 reduced

58 58 The Mole

59 59 A collection term indicates a specific number of items. For example, 1 dozen doughnuts contains 12 doughnuts. 1 ream of paper means 500 sheets. 1 case is 24 cans. Collection Terms

60 60 A mole contains 6.02 x particles, which is the number of carbon atoms in g of carbon. 1 mole C = 6.02 x C atoms The number 6.02 x is known as Avogadros number. One mole of any element contains Avogadros number of atoms. 1 mole Na = 6.02 x Na atoms 1 mole Au= 6.02 x Au atoms A Mole

61 61 Avogadros number is also the number of molecules and formula units in one mole of a compound. One mole of a covalent compound contains Avogadros number of molecules. 1 mole CO 2 = 6.02 x CO 2 molecules 1 mole H 2 O = 6.02 x H 2 O molecules One mole of an ionic compound contains Avogadros number of formula units. 1 mole NaCl = 6.02 x NaCl formula units A Mole of Molecules

62 62 Samples of One Mole Quantities

63 63 Avogadros number is written as conversion factors x particles and 1 mole 1 mole 6.02 x particles The number of molecules in 0.50 mole of CO 2 molecules is calculated as 0.50 mole CO 2 molecules x 6.02 x CO 2 molecules 1 mole CO 2 molecules = 3.0 x CO 2 molecules Avogadros Number

64 64 A. Calculate the number of atoms in 2.0 moles of Al. 1) 2.0 Al atoms 2) 3.0 x Al atoms 3) 1.2 x Al atoms B. Calculate the number of moles of S in 1.8 x S. 1) 1.0 mole S atoms 2) 3.0 mole S atoms 3) 1.1 x mole S atoms Learning Check

65 65 A. Calculate the number of atoms in 2.0 moles of Al. 3) 1.2 x Al atoms 2.0 moles Al x 6.02 x Al atoms 1 mole Al B. Calculate the number of moles of S in 1.8 x S. 2) 3.0 mole S atoms 1.8 x S atoms x 1 mole S 6.02 x S atoms Solution

66 66 The mass of one mole is called molar mass. The molar mass of an element is the atomic mass expressed in grams. Molar Mass

67 67 Give the molar mass to the nearest 0.1 g. A. 1 mole of K atoms =________ B. 1 mole of Sn atoms =________ Learning Check

68 68 Give the molar mass to the nearest 0.1 g. A. 1 mole of K atoms =39.1 g B. 1 mole of Sn atoms =118.7 g Solution

69 69 Molar Mass of CaCl 2 For a compound, the molar mass is the sum of the molar masses of the elements in the formula. We calculate the molar mass of CaCl 2 to the nearest 0.1 g as follows. ElementNumber of Moles Atomic MassTotal Mass Ca140.1 g/mole40.1 g Cl g/mole71.0 g CaCl g

70 70 Molar Mass of K 3 PO 4 Determine the molar mass of K 3 PO 4 to 0.1 g. ElementNumber of Moles Atomic MassTotal Mass in K 3 PO 4 K339.1 g/mole g P131.0 g/mole 31.0 g O416.0 g/mole 64.0 g K 3 PO g

71 71 One-Mole Quantities 32.1 g 55.9 g 58.5 g g g

72 72 A. 1 mole of K 2 O = ______g B. 1 mole of antacid Al(OH) 3 = ______g Learning Check

73 73 A. 1 mole of K 2 O 2 moles K (39.1 g/mole) + 1 mole O (16.0 g/mole) 78.2 g g= 94.2 g B. 1 mole of antacid Al(OH) 3 1 mole Al (27.0 g/mole) + 3 moles O (16.0 g/mole) + 3 moles H (1.0 g/mole) 27.0 g g g = 78.0 g Solution

74 74 Prozac, C 17 H 18 F 3 NO, is an antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac? 1) 40.0 g/mole 2) 262 g/mole 3) 309 g/mole Learning Check

75 75 Prozac, C 17 H 18 F 3 NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac? 3) 309 g/mole 17C (12.0) + 18H (1.0) + 3F (19.0) + 1N (14.0) + 1 O (16.0) = Solution

76 76 Methane CH 4 known as natural gas is used in gas cook tops and gas heaters. 1 mole CH 4 = 16.0 g The molar mass of methane can be written as conversion factors g CH 4 and 1 mole CH 4 1 mole CH g CH 4 Molar Mass Factors

77 77 Acetic acid C 2 H 4 O 2 gives the sour taste to vinegar. Write two molar mass conversion factors for acetic acid. Learning Check

78 78 Acetic acid C 2 H 4 O 2 gives the sour taste to vinegar. Write two molar mass factors for acetic acid. 1 mole of acetic acid = 60.0 g acetic acid 1 mole acetic acid and 60.0 g acetic acid 60.0 g acetic acid 1 mole acetic acid Solution

79 79 Mole factors are used to convert between the grams of a substance and the number of moles. Calculations with Molar Mass Grams Mole factor Moles

80 80 Aluminum is often used for the structure of lightweight bicycle frames. How many grams of Al are in 3.00 moles of Al? 3.00 moles Al x 27.0 g Al = 81.0 g Al 1 mole Al mole factor for Al Calculating Grams from Moles

81 81 The artificial sweetener aspartame (Nutri-Sweet) C 14 H 18 N 2 O 5 is used to sweeten diet foods, coffee and soft drinks. How many moles of aspartame are present in 225 g of aspartame? Learning Check

82 82 Calculate the molar mass of C 14 H 18 N 2 O 5. (14 x 12.0) + (18 x 1.0) + (2 x 14.0) + (5 x 16.0) = 294 g/mole Set up the calculation using a mole factor. 225 g aspartame x 1 mole aspartame 294 g aspartame mole factor(inverted) = mole aspartame Solution

83 83 Conservation of Mass In a chemical reaction, the mass of the reactants is equal to the mass of the products. 2 moles Ag + 1 mole S = 1 mole Ag 2 S 2 (107.9 g) + 1(32.1 g) = 1 (247.9 g) g reactants = g product

84 84 We can read the equation in moles by placing the word moles between each coefficient and formula. 4 Fe + 3 O 2 2 Fe 2 O 3 4 moles Fe + 3 moles O 2 2 moles Fe 2 O 3 Moles in Equations

85 85 A mole-mole factor is a ratio of the coefficients for two substances. 4 Fe + 3 O 2 2 Fe 2 O 3 Fe and O 2 4 mole Fe and 3 mole O 2 3 mole O 2 4 mole Fe Fe and Fe 2 O 3 4 mole Fe and 2 mole Fe 2 O 3 2 mole Fe 2 O 3 4 mole Fe O 2 and Fe 2 O 3 3 mole O 2 and 2 mole Fe 2 O 3 2 mole Fe 2 O 3 3 mole O 2 Writing Mole-Mole Factors

86 86 Consider the following equation: 3 H 2 + N 2 2 NH 3 A. A mole factor for H 2 and N 2 is 1) 3 mole N 2 2) 1 mole N 2 3) 1 mole N 2 1 mole H 2 3 mole H 2 2 mole H 2 B. A mole factor for NH 3 and H 2 is 1) 1 mole H 2 2) 2 mole NH 3 3) 3 mole N 2 2 mole NH 3 3 mole H 2 2 mole NH 3 Learning Check

87 87 3 H 2 + N 2 2 NH 3 A. A mole factor for H 2 and N 2 is 2) 1 mole N 2 3 mole H 2 B. A mole factor for NH 3 and H 2 is 2) 2 mole NH 3 3 mole H 2 Solution

88 88 Consider the following reaction: 4 Fe + 3 O 2 2 Fe 2 O 3 How many moles of Fe 2 O 3 are produced when 6.0 moles O 2 react? Use the appropriate mole factor to determine the moles Fe 2 O mole O 2 x 2 mole Fe 2 O 3 = 4.0 mole Fe 2 O 3 3 mole O 2 Calculations with Mole Factors

89 89 Consider the following reaction: 4 Fe + 3 O 2 2 Fe 2 O 3 How many moles of Fe are needed to react with 12.0 moles of O 2 ? 1) 3.00 moles Fe 2) 9.00 moles Fe 3) 16.0 moles Fe Learning Check

90 90 3) 16.0 moles Fe Consider the following reaction: 4 Fe + 3 O 2 2 Fe 2 O 3 How many moles of Fe are needed to react with 12.0 moles of O 2 ? 12.0 mole O 2 x 4 mole Fe = 16.0 moles Fe 3 mole O 2 Solution

91 91 Mass Calculations

92 92 How many grams of O 2 are needed to produce mole of Fe 2 O 3 ? 4 Fe + 3 O 2 2 Fe 2 O 3 1) 38.4 g O 2 2) 19.2 g O 2 3) 1.90 g O 2 Learning Check

93 93 2) 19.2 g O mole Fe 2 O 3 x 3 mole O 2 x 32.0 g O 2 2 mole Fe 2 O 3 1 mole O 2 mole factor molar mass = 19.2 g O 2 Solution

94 94 The reaction between H 2 and O 2 produces 13.1 g of water. How many grams of O 2 reacted? 2H 2 + O 2 2H 2 O ? g 13.1 g Plan: g H 2 O mole H 2 O mole O 2 g O g H 2 O x 1 mole H 2 O x 1 mole O 2 x 32.0 g O g H 2 O 2 mole H 2 O 1 mole O 2 = 11.6 g O 2 Calculating the Mass of a Reactant

95 95 Learning Check Acetylene gas C 2 H 2 burns in the oxyactylene torch for welding. How many grams of C 2 H 2 are burned if the reaction produces 75.0 g of CO 2 ? 2 C 2 H O 2 4 CO H 2 O 1) 88.6 g C 2 H 2 2) 44.3 g C 2 H 2 3) 22.2 g C 2 H 2

96 96 3) 22.2 g C 2 H 2 2 C 2 H O 2 4 CO H 2 O 75.0 g CO 2 x 1 mole CO 2 x 2 moles C 2 H 2 x 26.0 g C 2 H g CO 2 4 moles CO 2 1 mole C 2 H 2 = 22.2 g C 2 H 2 Solution

97 97 You prepared cookie dough to make 5 dozen cookies. The phone rings and you answer. While you talk, a sheet of 12 cookies burns. You have to throw them out. The rest of the cookies are okay. The results of our baking can be described as follows: Theoretical yield 60 cookies possible Actual yield 48 cookies to eat Percent yield 48 cookies x 100 = 80% yield 60 cookies Percent Yield

98 98 Percent Yield The theoretical yield is the maximum amount of product calculated using the balanced equation. The actual yield is the amount of product obtained when the reaction is run. Percent yield is the ratio of actual yield compared to the theoretical yield. Percent Yield = Actual Yield (g) x 100 Theoretical Yield (g)

99 99 Sample Exercise % Yield Without proper ventilation and limited oxygen, the reaction of carbon and oxygen produces carbon monoxide. 2C + O 2 2CO What is the percent yield if 40.0 g of CO are produced from the reaction of 30.0 g O 2 ?

100 100 Sample Exercise % Yield (cont.) 1. Calculate theoretical yield of CO g O 2 x 1 mole O 2 x 2 mole CO x 28.0 g CO 32.0 g O 2 1 mole O 2 1 mole CO = 52.5 g CO (theoretical) 2. Calculate the percent yield g CO (actual) x 100 = 76.2 % yield 52.5 g CO(theoretical)

101 101 Learning Check In the lab, N 2 and 5.0 g of H 2 are reacted and produce 16.0 g of NH 3. What is the percent yield for the reaction? N 2 (g) + 3H 2 (g) 2NH 3 (g) 1) 31.3 % 2) 56.5 % 3) 80.0 %

102 102 Solution 2) 56.5 % N 2 (g) + 3H 2 (g) 2NH 3 (g) 5.0 g H 2 x 1 mole H 2 x 2 moles NH 3 x 17.0 g NH g H 2 3 moles H 2 1 mole NH 3 = 28.3 g NH 3 (theoretical) Percent yield = 16.0 g NH 3 x 100 = 56.5 % 28.3 g NH 3


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