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Copyright©2004 by Houghton Mifflin Company. All rights reserved. 1 Introductory Chemistry: A Foundation FIFTH EDITION by Steven S. Zumdahl University of.

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Presentation on theme: "Copyright©2004 by Houghton Mifflin Company. All rights reserved. 1 Introductory Chemistry: A Foundation FIFTH EDITION by Steven S. Zumdahl University of."— Presentation transcript:

1 Copyright©2004 by Houghton Mifflin Company. All rights reserved. 1 Introductory Chemistry: A Foundation FIFTH EDITION by Steven S. Zumdahl University of Illinois

2 2 Chemical Quantities Chapter 9

3 3 CHEMISTRY 100 Dr. Jimmy Hwang Sections 5 and 7 Tu & Th 9:30a.m.-10:45a.m. Textbook: Zumdahl, Introductory Chemistry, 5 th Edition Office Hours: by appointment Ext (voic )

4 4 Overview In Chapter 9, our goals are for the students to: 1. Understand the molecular and mass information given in a balanced equation. 2. Learn to use a balanced equation to determine the relationships between moles of reactants and moles of products. 3. Learn to relate masses of reactants and products in a chemical reaction. 4. Learn to recognize the limiting reactant in a reaction. 5. Learn to use the limiting reactant to do stoichiometric calculations. 6. Learn to calculate actual yield as a percentage of theoretical yield.

5 5 Information Given by the Chemical Equation Balanced equation provides 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

6 6 Information Given by the Chemical Equation 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 billion CO molecules react with 1 billion O 2 molecules to produce 20 billion CO 2 molecules 2 moles CO molecules react with 1 mole O 2 molecules to produce 2 moles CO 2 molecules 12 moles CO molecules react with 6 moles O 2 molecules to produce 12 moles CO 2 molecules

7 7 Information Given by the Chemical Equation The coefficients in the balanced chemical equation shows the molecules and 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

8 8 Information Given by the Chemical Equation 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

9 9 Fundamentals of Stoichiometry 1.Stoichiometry is the quantitative study of reactants and products in a chemical reaction. The quantitative relationships in a given chemical reaction can be obtained from a balanced chemical equation. 2.The key concept is the use of the coefficients in the balanced chemical equation to establish mole ratios of the various materials (reactants and products) involved in the reaction. 3.The mass of the various materials can then be obtained from the molar mass ratios.

10 10 Chemical Equation A representation of a chemical reaction: CH 4 (g) + O 2 (g)  CO 2 (g) + H 2 O (l) reactants products Unbalanced !

11 11 Chemical Equation CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l) The equation is balanced. 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water

12 12 1 mole CH 4 1 mole CH 4 1 mole CH 4 2 mol O 2 or 1 mol O 2 2 mol O 2 2 mol O 2 1 mol CO 2 2 mol H 2 O 2 mol H 2 O 1 mol H 2 O 1 mol CO 2 Obtaining Ratios from the Balanced Chemical Equation 16 g CH 4 16 g CH 4 16 g CH 4 32 g O 2 2 x (32g) O 2 2x(32g) O 2 44 g CO 2 2x(18g) H 2 O 18 g H 2 O 44 g CO 2

13 13 Steps for Calculating Masses of Reactants and Products in Chemical Equation Step 1.Balance the equation. Step 2.Convert mass to moles. Step 3.Set up mole ratios. Step 4.Use mole ratios to calculate moles of desired reactant or product. Step 5.Convert from moles back to masses.

14 14 ¬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 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

15 15 ®Use dimensional analysis 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

16 16 ¬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 Example #2 Determine the Number of grams of Carbon Monoxide required to react with 48.0 g Oxygen, and determine the mass of Carbon Dioxide produced

17 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 Determine the Number of grams of Carbon Monoxide required to react with 48.0 g Oxygen, and determine the mass of Carbon Dioxide produced

18 18 ±Use the molar mass of the desired quantity to convert the moles to mass Example #2 Determine the Number of grams of Carbon Monoxide required to react with 48.0 g Oxygen, and determine the mass of Carbon Dioxide produced

19 19 Limiting and Excess Reactants A reactant which is completely consumed when a reaction is run to completion is called a limiting reactant A reactant which is not completely consumed in a reaction is called an excess reactant –calculate the amount of excess reactant unused by (1) calculating the amount of excess reactant used from the limiting reactant, then (2) subtract this amount from the amount of excess reactant started with –The maximum amount of a product that can be made when the limiting reactant is completely consumed is called the theoretical yield

20 20 Limiting Reactant The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed. Fig. 4.4 The molecular state picture of the reaction of 6 molecules each of H 2 and O 2 illustrating that H 2 is the limiting reactant (L.R.).

21 21 6 green used up Limiting Reactant 6 red left over

22 22 ¬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 Example #3 Determine the Number of Moles of Carbon Dioxide produced when 3.2 moles Oxygen reacts with 4.0 moles of Carbon Monoxide

23 23 ®Use dimensional analysis to determine the number of moles of reactant A needed to react with reactant B Example #3 Determine the Number of Moles of Carbon Dioxide produced when 3.2 moles Oxygen reacts with 4.0 moles of Carbon Monoxide

24 24 ¯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 Determine the Number of Moles of Carbon Dioxide produced when 3.2 moles Oxygen reacts with 4.0 moles of Carbon Monoxide

25 25 °Use the limiting reactant to determine the moles of product Example #3 Determine the Number of Moles of Carbon Dioxide produced when 3.2 moles Oxygen reacts with 4.0 moles of Carbon Monoxide

26 26 Example #4 Determine the Mass of Carbon Dioxide produced when 48.0 g of Oxygen reacts with 56.0 g of Carbon Monoxide ¬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

27 27 ¯Determine the moles of each reactant Example #4 Determine the Mass of Carbon Dioxide produced when 48.0 g of Oxygen reacts with 56.0 g of Carbon Monoxide

28 28 °Determine the number of moles of reactant A needed to react with reactant B Example #4 Determine the Mass of Carbon Dioxide produced when 48.0 g of Oxygen reacts with 56.0 g of Carbon Monoxide

29 29 ±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 Determine the Mass of Carbon Dioxide produced when 48.0 g of Oxygen reacts with 56.0 g of Carbon Monoxide

30 30 ²Use the limiting reactant to determine the moles of product, then the mass of product Example #4 Determine the Mass of Carbon Dioxide produced when 48.0 g of Oxygen reacts with 56.0 g of Carbon Monoxide

31 31 ³Divide 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 Example #4a Determine the Mass of Carbon Dioxide produced when 48.0 g of Oxygen reacts with 56.0 g of Carbon Monoxide If 72.0 g of Carbon Dioxide is actually made, what is the Percentage Yield

32 32 Example: Methanol burns in air according to the equation 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH 3 OHmoles CH 3 OHmoles H 2 Ograms H 2 O Molar mass of reactant, CH 3 OH Mole ratio from equation Molar mass of product, H 2 O 209 g CH 3 OH 1 mol CH 3 OH 32.0 g CH 3 OH x 4 mol H 2 O 2 mol CH 3 OH x 18.0 g H 2 O 1 mol H 2 O x = 235 g H 2 O Grams of reactant

33 33 Example 4.2 In the combustion of methane, how many grams of water can be produced if sufficient hydrogen reacts with 26.0 g oxygen? (Ans.: 29.2 g) (O 2 = 32.0 g/mol; H 2 O = 18.0 g/mol) 2H 2 (g) + O 2 (g)  2 H 2 O(l) Will illustrate using the flow diagram in Fig. 4.3 and the short cut using mass ratios.

34 34 Percent Yield Most reactions do not go to completion The amount of product made in an experiment is called the actual yield The percentage of the theoretical yield that is actually made is called the percent yield Percent Yield = Actual Yield Theoretical Yield x 100%

35 35 Example 4.3 The reaction that starts a match burning occurs between potassium chlorate, KClO 3 (s), and tetraphosphorus trisulfide, P 4 S 3. The equation for the formation of P 4 S 3 from its constituent elements is, 8 P 4 (s) + 3 S 8  8 P 4 S 3 (l) If we have 153 g of S 8 and an excess of phosphorus, what mass of P 4 S 3 can be produced by this reaction? (S 8 = 256 g/mol; P 4 S 3 = g/mol) Ans.: 350 g P 4 S 3

36 36 Problem on Complete Combustion of Octane What mass of CO 2 is produced by the complete combustion of 2,659 g of C 8 H 18 ? 2 C 8 H O 2  16 CO H 2 O (C 8 H 18 = g/mol; CO 2 = g/mol) Ans.: 8,195 g CO 2

37 37 Example 4.4 A solution of HCl contains 5.22 g HCl. When it is allowed to react with 3.25 g of solid K 2 CO 3, the products are KCl, CO 2, and H 2 O. What reactant is in excess? (Ans.: HCl) (HCl = 36.5 g/mol; K 2 CO 3 = g/mol) 2 HCl + K 2 CO 3  2 KCl + CO 2 + H 2 O

38 38 Limiting Reactant Calculations What weight of molten iron is produced by 1 kg each of the reactants? Fe 2 O 3 (s) + 2 Al(s)  Al 2 O 3 (s) + 2 Fe( l ) Ratio: 6.26 < LimitingExcess The 6.26 mol Fe 2 O 3 will Disappear first

39 39 Example 4.5 In the reaction that produces P 4 S 3, one of the reactants in the combustion of a match, 8 P 4 (s) + 3 S 8  8 P 4 S 3 (l) If 28.2 g of P 4 are allowed to react with 18.3 g of S 8, which reactant is the limiting reactant? (P 4 = 124 g/mol; S 8 = 256 g/mol)

40 40 Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Its amount is Calculated using the balanced equation. Actual Yield is the amount of product actually obtained from a reaction. It is usually given.

41 41 Percent Yield Actual yield = quantity of product actually obtained Theoretical yield = quantity of product predicted by stoichiometry

42 42 Percent Yield Example 14.4 gexcess Actual yield = 6.26 g


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