Chapter 3 Stoichiometry QUESTION Copyright Houghton Mifflin Company

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Chapter 3 Stoichiometry

Chapter 3 Stoichiometry

QUESTION Copyright © Houghton Mifflin Company. All rights reserved. 2

QUESTION Copyright © Houghton Mifflin Company. All rights reserved. 2

ANSWER 1) 35 protons, 44 neutrons, 35 electrons. Section 2. 5 The Modern View

ANSWER 1) 35 protons, 44 neutrons, 35 electrons. Section 2. 5 The Modern View of Atomic Structure: An Introduction (p. 49) The 79 in bromine-79 is the mass number (the number of protons plus neutrons). The atomic number for bromine is 35. The number of neutrons is 79 – 35 = 44. The number of electrons equals the number of protons. Copyright © Houghton Mifflin Company. All rights reserved. 3

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QUESTION Copyright © Houghton Mifflin Company. All rights reserved. 4

ANSWER 2) 187 Re 75 Section 3. 2 Atomic Masses (p. 78) (0. 371

ANSWER 2) 187 Re 75 Section 3. 2 Atomic Masses (p. 78) (0. 371 ´ 185) + (0. 629 ´ M) = 186. 2. Solve for M (the isotopic mass). Since neutrons are slightly heavier than protons, this calculation is only an approximation, but is accurate enough to determine the correct answer. Copyright © Houghton Mifflin Company. All rights reserved. 5

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ANSWER 55. 85 amu Section 3. 2 Atomic Masses (p. 78) (. 0582 ´

ANSWER 55. 85 amu Section 3. 2 Atomic Masses (p. 78) (. 0582 ´ 53. 940 amu) + (. 9166 ´ 55. 935 amu) + (. 0219 ´ 56. 935 amu) + (. 0033 ´ 57. 933 amu) = 55. 85 amu Copyright © Houghton Mifflin Company. All rights reserved. 7

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ANSWER 5) – 22 1. 06 ´ 10 g Section 3. 3 The Mole

ANSWER 5) – 22 1. 06 ´ 10 g Section 3. 3 The Mole (p. 82) A mole of copper atoms has a mass of 63. 55 g. The mass of 1 copper atoms is 23 – 22 63. 55 g/(6. 022 ´ 10 ) = 1. 06 ´ 10 g. Copyright © Houghton Mifflin Company. All rights reserved. 9

ANSWER 3) 46. 07 Section 3. 4 Molar Mass (p. 86) The molar mass

ANSWER 3) 46. 07 Section 3. 4 Molar Mass (p. 86) The molar mass is the sum of masses of all the atoms in the molecule. 2 ´ 12. 01 + 6 ´ 1. 008 + 1 ´ 16. 00 = 46. 07 Copyright © Houghton Mifflin Company. All rights reserved. 10

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ANSWER 3) CH 3 Cl Section 3. 4 Molar Mass (p. 86) The molar

ANSWER 3) CH 3 Cl Section 3. 4 Molar Mass (p. 86) The molar mass has units of g/mol. (12. 8 g/0. 256 mol) = 50. 0 g/mol. The molecule with the closest molar mass is CH 3 Cl. Copyright © Houghton Mifflin Company. All rights reserved. 13

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ANSWER 1) 3. 10 g Section 3. 4 Molar Mass (p. 86) 6. 41

ANSWER 1) 3. 10 g Section 3. 4 Molar Mass (p. 86) 6. 41 g of S 8 converts to 0. 0250 mol. Therefore 0. 0250 mol of phosphorus is needed to have the same number of atoms as sulfur. 0. 0250 mol P 4 ´ (123. 88 g/mol P 4) = 3. 10 g. Copyright © Houghton Mifflin Company. All rights reserved. 15

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ANSWER 2) 278 g Section 3. 4 Molar Mass (p. 86) The molar mass

ANSWER 2) 278 g Section 3. 4 Molar Mass (p. 86) The molar mass of sodium hydroxide, Na. OH, is 22. 99 g/mol + 16. 00 g/mol + 1. 008 g/mol = 40. 00 g/mol. Convert to grams: 6. 94 mol ´ (40. 00 g/mol) = 278 g. Copyright © Houghton Mifflin Company. All rights reserved. 17

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ANSWER 3) 4. 04 g Section 3. 5 Percent Composition of Compounds (p. 89)

ANSWER 3) 4. 04 g Section 3. 5 Percent Composition of Compounds (p. 89) The molar mass of K 2 Cr. O 7 is 2 ´ 39. 10 + 52. 00 + 7 ´ 16. 00 = 242. 2. The mass fraction of potassium is (2 ´ 39. 10)/242. 2 = 0. 3229 ´ 12. 5 g = 4. 04 g. Copyright © Houghton Mifflin Company. All rights reserved. 19

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ANSWER 2) coefficients of compounds Section 3. 8 Balancing Chemical Equations (p. 98 )

ANSWER 2) coefficients of compounds Section 3. 8 Balancing Chemical Equations (p. 98 ) Changing the subscripts of a compound or the addition of a compound to a chemical equation changes the reaction. Only the coefficients can be adjusted without altering the reaction. Copyright © Houghton Mifflin Company. All rights reserved. 21

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ANSWER 2) 50. 0 g Section 3. 10 Calculations Involving a Limiting Reactant (p.

ANSWER 2) 50. 0 g Section 3. 10 Calculations Involving a Limiting Reactant (p. 106) This is a limiting reactant problem. The first step is determining which reactant will run out first. Copyright © Houghton Mifflin Company. All rights reserved. 23

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ANSWER (continued) Copyright © Houghton Mifflin Company. All rights reserved. 24

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ANSWER 5) All of these Section 3. 5 Percent Composition of Compounds (p. 89)

ANSWER 5) All of these Section 3. 5 Percent Composition of Compounds (p. 89) The ratio of C to H in C 8 H 8 is 1: 1. This is the same ratio found for each of the compounds, so all have the same percent composition by mass. Copyright © Houghton Mifflin Company. All rights reserved. 26

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ANSWER 2) C 8 H 8 Section 3. 6 Determining the Formula of a

ANSWER 2) C 8 H 8 Section 3. 6 Determining the Formula of a Compound (p. 91) The mass of CH is 13. 018 divides into 104. 1 about 8 times. Therefore there are 8 CH groups in this compound. C 8 H 8 is the molecular formula. Copyright © Houghton Mifflin Company. All rights reserved. 28

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ANSWER 2) Numbers of atoms are conserved in chemical change. Section 3. 8 Balancing

ANSWER 2) Numbers of atoms are conserved in chemical change. Section 3. 8 Balancing Chemical Equations (p. 98 ) This is another way of stating the Law of Conservation of Mass. Copyright © Houghton Mifflin Company. All rights reserved. 30

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QUESTION (continued) Copyright © Houghton Mifflin Company. All rights reserved. 32

ANSWER 2) b Section 3. 8 Balancing Chemical Equations (p. 98) Coefficients can be

ANSWER 2) b Section 3. 8 Balancing Chemical Equations (p. 98) Coefficients can be fractions, but usually fractions are multiplied by the appropriate number to make them whole numbers. Atoms are conserved according to the Law of Conservation of Mass. Copyright © Houghton Mifflin Company. All rights reserved. 33

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ANSWER 4) 13 Section 3. 8 Balancing Chemical Equations (p. 98) O 2 should

ANSWER 4) 13 Section 3. 8 Balancing Chemical Equations (p. 98) O 2 should be balanced last since it contains only one type of element and balancing it will not cause an imbalance in another element. Copyright © Houghton Mifflin Company. All rights reserved. 35

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ANSWER 1) 1 Section 3. 8 Balancing Chemical Equations (p. 98) The phosphate ion

ANSWER 1) 1 Section 3. 8 Balancing Chemical Equations (p. 98) The phosphate ion has a charge of – 3, while the aluminum ion takes a charge of +3. Therefore the formula of aluminum phosphate is Al. PO 4. Copyright © Houghton Mifflin Company. All rights reserved. 37

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ANSWER 2) 16. 4 g Section 3. 9 Stoichiometric Calculations: Amounts of Reactants and

ANSWER 2) 16. 4 g Section 3. 9 Stoichiometric Calculations: Amounts of Reactants and Products (p. 102) The balanced equation is Ca(OH)2 + 2 HNO 3 ® Ca(NO 3)2 + 2 H 2 O. Water is the other product for an acid/base reaction. Copyright © Houghton Mifflin Company. All rights reserved. 39

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ANSWER 5) none of these Section 3. 10 : Calculations Involving a Limiting Reactant

ANSWER 5) none of these Section 3. 10 : Calculations Involving a Limiting Reactant (p. 106) The limiting reactant in a reaction is the reactant that produces the least number of grams of any product. Copyright © Houghton Mifflin Company. All rights reserved. 41

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ANSWER 1) True Section 3. 6 Determining the Formula of a Compound (p. 91)

ANSWER 1) True Section 3. 6 Determining the Formula of a Compound (p. 91) The molecular formula contains all the information about the compound. Copyright © Houghton Mifflin Company. All rights reserved. 43