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General, Organic, and Biological Chemistry Fourth Edition Karen Timberlake 6.6 Mole Relationships in Chemical Equations Chapter 6 Chemical Reactions and Quantities © 2013 Pearson Education, Inc. Lectures
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© 2013 Pearson Education, Inc. Chapter 6, Section 6 2 Law of Conservation of Mass The Law of Conservation of Mass indicates that in an ordinary chemical reaction, matter can neither be created nor destroyed and that the total mass of products is equal to the total mass of reactants.
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© 2013 Pearson Education, Inc. Chapter 6, Section 6 3 Conservation of Mass
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© 2013 Pearson Education, Inc. Chapter 6, Section 6 4 Information from a Balanced Equation
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© 2013 Pearson Education, Inc. Chapter 6, Section 6 5 Consider the following equation: An equation can be read in “moles” by placing the words “moles of” between each coefficient and formula. Chemical Equations: Moles–Moles
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© 2013 Pearson Education, Inc. Chapter 6, Section 6 6 In the chemical reaction of Fe and S, the mass of the reactants is the same as the mass of the product, Fe 2 S 3. Chemical Equations: Moles–Moles
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© 2013 Pearson Education, Inc. Chapter 6, Section 6 7 A mole–mole factor is a ratio of the moles (from the coefficients) for any two substances in an equation. Mole–Mole Factors from Chemical Equations
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© 2013 Pearson Education, Inc. Chapter 6, Section 6 8 Consider the following equation. 1. A mole–mole factor for H 2 and N 2 is A.B.C. 2. A mole–mole factor for NH 3 and H 2 is A.B.C. Learning Check
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© 2013 Pearson Education, Inc. Chapter 6, Section 6 9 Consider the following equation. 1. A mole–mole factor for H 2 and N 2 is B. 2. A mole–mole factor for NH 3 and H 2 is B. Solution
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© 2013 Pearson Education, Inc. Chapter 6, Section 6 10 Guide to Calculating Quantities in a Chemical Reaction
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© 2013 Pearson Education, Inc. Chapter 6, Section 6 11 How many moles of CO 2 can be produced when 2.25 moles of C 3 H 8 react according to the following balanced reaction? Step 1 State the given and needed quantities. Analyze the Problem. Calculating Quantities in a Chemical Reaction
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© 2013 Pearson Education, Inc. Chapter 6, Section 6 12 How many moles of CO 2 can be produced when 2.25 moles of C 3 H 8 react according to the following balanced reaction? Step 2 Write a plan to convert the given to the needed quantity (moles or grams). moles of C 3 H 8 mole–mole factor moles of CO 2 Calculating Quantities in a Chemical Reaction
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© 2013 Pearson Education, Inc. Chapter 6, Section 6 13 How many moles of CO 2 can be produced when 2.25 moles of C 3 H 8 react according to the following balanced reaction? Step 3 Use coefficients to write mole–mole factors; write molar mass factors if needed. 1 mole of C 3 H 8 = 3 moles of CO 2 Calculating Quantities in a Chemical Reaction
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© 2013 Pearson Education, Inc. Chapter 6, Section 6 14 How many moles of CO 2 can be produced when 2.25 moles of C 3 H 8 react according to the following balanced reaction? Step 4 Set up the problem to give the needed quantity (moles or grams). Calculating Quantities in a Chemical Reaction
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© 2013 Pearson Education, Inc. Chapter 6, Section 6 15 How many moles of Fe are needed for the reaction of 12.0 moles of O 2 ? A. 3.00 moles of Fe B. 9.00 moles of Fe C. 16.0 moles of Fe Learning Check
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© 2013 Pearson Education, Inc. Chapter 6, Section 6 16 How many moles of Fe are needed for the reaction of 12.0 moles of O 2 ? Step 1 State the given and needed quantities. Analyze the Problem. Solution
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© 2013 Pearson Education, Inc. Chapter 6, Section 6 17 How many moles of Fe are needed for the reaction of 12.0 moles of O 2 ? Step 2 Write a plan to convert the given to the needed quantity (moles or grams). 12.0 moles of O 2 mole–mole factor moles of Fe Solution
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© 2013 Pearson Education, Inc. Chapter 6, Section 6 18 How many moles of Fe are needed for the reaction of 12.0 moles of O 2 ? Step 3 Use coefficients to write mole–mole factors; write molar mass factors if needed. 4 moles of Fe = 3 moles of O 2 Solution
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© 2013 Pearson Education, Inc. Chapter 6, Section 6 19 How many moles of Fe are needed for the reaction of 12.0 moles of O 2 ? Step 4 Set up the problem to give the needed quantity (moles or grams). Answer is C. Solution
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