By Steven S. Zumdahl & Donald J. DeCoste University of Illinois Introductory Chemistry: A Foundation, 6 th Ed. Introductory Chemistry, 6 th Ed. Basic Chemistry,

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Presentation transcript:

by Steven S. Zumdahl & Donald J. DeCoste University of Illinois Introductory Chemistry: A Foundation, 6 th Ed. Introductory Chemistry, 6 th Ed. Basic Chemistry, 6 th Ed.

Chapter 9 Chemical Quantities

Information Given by Chemical Equations Section 9-1

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 4 Information Given by the Chemical Equation Balanced equations show the relationship between the relative numbers of reacting molecules and product molecules. 2 CO + O 2  2 CO 2 2 CO molecules react with 1 O 2 molecules to produce 2 CO 2 molecules

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 5 Since the information given is relative: 2 CO + O 2  2 CO CO molecules react with 100 O 2 molecules to produce 200 CO 2 molecules 2.2 billion CO molecules react with 1 billion O 2 molecules to produce 2 billion CO 2 molecules 3.2 moles CO molecules react with 1 mole O 2 molecules to produce 2 moles CO 2 molecules 4.12 moles CO molecules react with 6 moles O 2 molecules to produce 12 moles CO 2 molecules Information Given by the Chemical Equation

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 6

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 7 Mole-Mole Relationship Section 9-2

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 8 The coefficients in the balanced chemical equation show the molecules and the mole ratio of the reactants and products. Since moles can be converted to masses, we can determine the mass ratio of the reactants and products as well. Information Given by the Chemical Equation

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 9 2 CO + O 2  2 CO 2 2 moles CO = 1 mole O 2 = 2 moles CO 2 Since 1 mole of CO = g, 1 mole O 2 = g, and 1 mole CO 2 = g 2(28.01) g CO = 1(32.00) g O 2 = 2(44.01) g CO 2 Information Given by the Chemical Equation

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 10 Example #1 Determine the number of moles of carbon monoxide required to react with 3.2 moles oxygen, and determine the moles of carbon dioxide produced.

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 11 Example #1 (cont.) Write the balanced equation: 2 CO + O 2  2 CO 2 Use the coefficients to find the mole relationship: 2 moles CO = 1 mol O 2 = 2 moles CO 2

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 12 Use dimensional analysis: Example #1 (cont.)

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 13 Mass Calculations Section 9-3

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 14 Mass Calculations Remember that the coefficients in a balanced chemical equation show the relationship between moles of reactants and products. If we are given the mass of reactants and/or products, we must first change them into moles before the relationship will work!!

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 15 Example #2 Determine the grams of carbon monoxide required to react with 48.0 grams oxygen, and determine the grams of carbon dioxide produced.

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 16 Example #2 (cont.) Write the balanced equation: 2 CO + O 2  2 CO 2 Use the coefficients to find the mole relationship: 2 moles CO = 1 mol O 2 = 2 moles CO 2 Determine the molar mass of each: 1 mol CO = g 1 mol O 2 = g 1 mol CO 2 = g

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 17 Use the molar mass of the given quantity to convert it to moles. Use the mole relationship to convert the moles of the given quantity to the moles of the desired quantity: Example #2 (cont.)

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 18 Use the molar mass of the desired quantity to convert the moles to mass: Example #2 (cont.)

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 19 Calculations involving a Limiting Reactant Section 9-4

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 20 Limiting and Excess Reactants Limiting reactant: a reactant that is completely consumed when a reaction is run to completion Excess reactant: a reactant that is not completely consumed in a reaction Theoretical yield: the maximum amount of a product that can be made by the time the limiting reactant is completely consumed

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 21 Figure 9.1: A mixture of 5CH 4 (g) and 3H 2 O molecules undergoes the reaction CH 4 (g) + H 2 O(g) →3H 2 (g) + CO(g).

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 22 Limiting and Excess Reactants (cont.)

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 23 Example #3 Determine the number of moles of carbon dioxide produced when 3.2 moles oxygen reacts with 4.0 moles of carbon monoxide.

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 24 Example #3 (cont.) Write the balanced equation: 2 CO + O 2  2 CO 2 Use the coefficients to find the mole relationship: 2 moles CO = 1 mol O 2 = 2 moles CO 2

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 25 Use dimensional analysis to determine the number of moles of reactant A needed to react with reactant B: Example #3 (cont.)

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 26 Compare the calculated number of moles of reactant A to the number of moles given of reactant A. If the calculated moles is greater, then A is the limiting reactant; if the calculated moles is less, then A is the excess reactant. The calculated moles of CO (6.4 moles) is greater than the given 4.0 moles; therefore CO is the limiting reactant. Example #3 (cont.)

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 27 Use the limiting reactant to determine the moles of product: Example #3 (cont.)

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 28 Figure 9.2: A map of the procedure used to create ammonia (NH 3 ).

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 29 Example #4 Determine the mass of carbon dioxide produced when 48.0 g of oxygen reacts with 56.0 g of carbon monoxide.

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 30 Example #4 (cont.) Write the balanced equation: 2 CO + O 2  2 CO 2 Use the coefficients to find the mole relationship: 2 moles CO = 1 mol O 2 = 2 moles CO 2 Determine the molar mass of each: 1 mol CO = g 1 mol O 2 = g 1 mol CO 2 = g

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 31 Determine the moles of each reactant: Example #4 (cont.)

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 32 Determine the number of moles of reactant A needed to react with reactant B: Example #4 (cont.)

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 33 Compare the calculated number of moles of reactant A to the number of moles given of reactant A. If the calculated moles is greater, then A is the limiting reactant; if the calculated moles is less, then A is the excess reactant. The calculated moles of O 2 (1.00 moles) is less than the given 1.50 moles; therefore O 2 is the excess reactant. Example #4 (cont.)

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 34 Use the limiting reactant to determine the moles of product, then the mass of product: Example #4 (cont.)

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 35 Percent Yield Section 9-5

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 36 Percent Yield Most reactions do not go to completion. Actual yield: the amount of product made in an experiment Percent yield: the percentage of the theoretical yield that is actually made Percent Yield = Actual Yield Theoretical Yield x 100%

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 37 Example #4a If in the last problem 72.0 g of carbon dioxide is actually made, what is the percentage yield ?

Copyright © Houghton Mifflin Company. All rights reserved. 9 | 38 Example #4a (cont.) Divide the actual yield by the theoretical yield, then multiply by 100% –The actual yield of CO 2 is 72.0 g –The theoretical yield of CO 2 is 88.0g