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Chapter 10 Chemical Quantities in Reactions

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Chapter 10 Slide 2 of 42 Mole Relationships in Chemical Equations Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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Chapter 10 Slide 3 of 42 The equation can be read in “moles” by placing the word “mole” or “mol” after each coefficient. 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) 4 mol Fe + 3 mol O 2 2 mol Fe 2 O 3 Moles in Equations

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Chapter 10 Slide 4 of 42 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) four molecules NH 3 react with five molecules O 2 to produce four molecules NO and six molecules H 2 O and four mol NH 3 react with five mol O 2 to produce four mol NO and six mol H 2 O Quantities in A Chemical Reaction

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Chapter 10 Slide 5 of 42 The equation can be read in “moles” by placing the word “mole” or “mol” after each coefficient. 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) 4 mol Fe + 3 mol O 2 2 mol Fe 2 O 3 Moles in Equations

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Chapter 10 Slide 6 of 42 A mole-mole factor is a ratio of the moles for two substances in an equation. 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Fe and O 2 4 mol Fe and 3 mol O 2 3 mol O 2 4 mol Fe Fe and Fe 2 O 3 4 mol Fe and 2 mol Fe 2 O 3 2 mol Fe 2 O 3 4 mol Fe O 2 and Fe 2 O 3 3 mol O 2 and 2 mol Fe 2 O 3 2 mol Fe 2 O 3 3 mol O 2 Writing Mole-Mole Factors

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Chapter 10 Slide 7 of 42 Consider the following equation: 3H 2 (g) + N 2 (g) 2NH 3 (g) 1. A mole factor for 2. A mole factor for Learning Check

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Chapter 10 Slide 8 of 42 How many moles of Fe 2 O 3 can form from 6.0 mol O 2 ? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) STEP 1 Given 6.0 mol O 2 Need: moles of Fe 2 O 3. STEP 2 moles O 2 moles Fe 2 O 3 STEP 3 3 mol O 2 = 2 mol Fe 2 O 3 3 mol O 2 and 2 mol Fe 2 O 3 2 mol Fe 2 O 3 3 mol O 2 STEP 4 Set up problem using the mol factor. 6.0 mol O 2 x 2 mol Fe 2 O 3 = 4.0 mol Fe 2 O 3 3 mol O 2 Calculations with Mole Factors

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Chapter 10 Slide 9 of 42 How many moles of Fe are needed to react with 12.0 mol O 2 ? 4Fe(s) + 3O 2 (g) 2 Fe 2 O 3 (s) 12.0 mol O 2 x 4 mol Fe = 3 mol O 2 Learning Check 16.0 mol of Fe

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Chapter 10 Slide 10 of 42 Conservation of Mass The Law of Conservation of Mass indicates No change in total mass occurs in a reaction. Mass of products is equal to mass of reactants.

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Chapter 10 Slide 11 of 42 Mass Calculations for Reactions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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Chapter 10 Slide 12 of 42 Stoichiometry Chemical Stoichiometry: using mass and quantity relationships among reactants and products in a chemical reaction to make predictions about how much product will be made.

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Chapter 10 Slide 13 of 42 In an ordinary chemical reaction, Matter cannot be created nor destroyed. The number of atoms of each element are equal. The mass of reactants equals the mass of products. H 2 (g) + Cl 2 (g) 2HCl(g) 2 mol H, 2 mol Cl = 2 mol H, 2 mol Cl 2(1.008) + 2(35.45) = 2(36.46) 72.92 g = 72.92 g Law of Conservation of Mass

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Chapter 10 Slide 14 of 42 Conservation of Mass 2 mol Ag + 1 mol S = 1 mol Ag 2 S 2 (107.9 g) + 1(32.07 g) = 1 (247.9 g) 247.9 g reactants = 247.9 g product Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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Chapter 10 Slide 15 of 42 Stoichiometric Mantra Grams A to Moles A to Moles B to Grams B 1 2 3

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Chapter 10 Slide 16 of 42 The reaction between H 2 and O 2 produces 13.1 g H 2 O. How many grams of O 2 reacted? 2H 2 (g) + O 2 (g) 2H 2 O(g) STEP 1 Given 13.1 g H 2 O Need grams O 2 STEP 2 Plan: g H 2 O mol H 2 O mol O 2 g O 2 STEP 3 1 mol O 2 = 2 mol H 2 O 1 mol H 2 O = 18.02 g 1 mol O 2 = 32.00 g O 2 STEP 4 13.1 g H 2 O x 1 mol H 2 O x 1 mol O 2 x 32.00 g O 2 18.02 g H 2 O 2 mol H 2 O 1 mol O 2 = Calculating the Mass of a Reactant 11.6 g O 2

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Chapter 10 Slide 17 of 42 How many grams of O 2 are needed to produce 0.400 mol Fe 2 O 3 ? 4Fe(s) + 3O 2 (g) 2 Fe 2 O 3 (s) mole factor molar mass 0.400 mol Fe 2 O 3 x 3 mol O 2 x 32.00 g O 2 2 mol Fe 2 O 3 1 mol O 2 = 19.2 g O 2 Learning Check

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Chapter 10 Slide 18 of 42 Learning Check Acetylene gas C 2 H 2 burns in the oxyacetylene torch for welding. How many grams of C 2 H 2 are burned if the reaction produces 75.0 g CO 2 ? 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O(g) C 12.01 g/mol H 1.008 g/mol O 16.00 g/mol

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Chapter 10 Slide 19 of 42 Limiting Reactants Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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Chapter 10 Slide 20 of 42 Limiting Reactant A limiting reactant in a chemical reaction is the substance that Is used up first. Stops the reaction. Limits the amount of product that can form.

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Chapter 10 Slide 21 of 42 Reacting Amounts In a table setting, there is 1 plate, 1 fork, 1 knife, and 1 spoon. How many table settings are possible from 5 plates, 6 forks, 4 spoons, and 7 knives? What is the limiting item? Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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Chapter 10 Slide 22 of 42 Reacting Amounts Four table settings can be made. Initially Use Left over plates 5 4 1 forks 6 4 2 spoons 4 4 0 knives 7 4 3 The limiting item is the spoon.

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Chapter 10 Slide 23 of 42 Example of Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices bread and 1 jar of peanut butter? With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item. Copyright © 2008 by Pearson Education, Inc. publishing as Benjamin Cummings

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Chapter 10 Slide 24 of 42 Example of Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices bread and 1 tablespoon of peanut butter? With 1 tablespoon of peanut butter, only 1 sandwich could be made. The peanut butter is the limiting item. Copyright © 2008 by Pearson Education, Inc. publishing as Benjamin Cummings

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Chapter 10 Slide 25 of 42 Limiting Reactants When 4.00 mol H 2 is mixed with 2.00 mol Cl 2,how many moles of HCl can form? H 2 (g) + Cl(g) 2HCl (g) 4.00 mol 2.00 mol ??? mol Calculate the moles of product from each reactant, H 2 and Cl 2. The limiting reactant is the one that produces the smaller amount of product.

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Chapter 10 Slide 26 of 42 Limiting Reactants Using Moles H 2 (g) + Cl(g) 2HCl (g) 4.00 mol 2.00 mol ??? mol HCl from H 2 4.00 mol H 2 x 2 mol HCl = 8.00 mol HCl 1 mol H 2 HCl from Cl 2 2.00 mol Cl 2 x 2 mol HCl = 4.00 mol HCl 1 mol Cl 2 The limiting reactant is Cl 2 because it produces the smaller number of moles of HCl and is used up first. (not possible) (smaller amount)

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Chapter 10 Slide 27 of 42 Checking Calculations InitiallyH 2 4.00 mol Cl 2 2.00 mol 2HCl 0 mol Reacted/ Formed -2.00 mol +4.00 mol Left after reaction 2.00 mol Excess 0 mol Limiting 4.00 mol

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Chapter 10 Slide 28 of 42 Limiting Reactants Using Mass If 4.80 mol Ca mixed with 2.00 mol N 2, which is the limiting reactant? 3Ca(s) + N 2 (g) Ca 3 N 2 (s) Moles of Ca 3 N from Ca 4.80 mol Ca x 1 mol Ca 3 N 2 = 1.60 mol Ca 3 N 2 3 mol Ca Moles of Ca 3 N 2 from N 2 2.00 mol N 2 x 1 mol Ca 3 N 2 = 2.00 mol Ca 3 N 2 1 mol N 2 All Ca is used up when 1.60 mol Ca 3 N 2 forms. Thus, Ca is the limiting reactant. N 2 is in excess. (smaller amount) (not possible)

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Chapter 10 Slide 29 of 42 Limiting Reactants Using Mass Calculate the mass of water produced when 8.00 g H 2 and 24.0 g O 2 react? 2H 2 (g) + O 2 (g) 2H 2 O(l) What do we need to do? Determine amount of water is produced for each of the reactants.

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Chapter 10 Slide 30 of 42 Limiting Reactants Using Mass Calculate the mass of water produced when 8.00 g H 2 and 24.0 g O 2 react? 2H 2 (g) + O 2 (g) 2H 2 O(l) Calculate the grams of H 2 O produced for each reactant. H 2 : 8.00 g H 2 x 1 mol H 2 x 2 mol H 2 O x 18.02 g H 2 O 2.016 g H 2 2 mol H 2 1 mol H 2 O = 71.5 g H 2 O O 2 : 24.0 g O 2 x 1 mol O 2 x 2 mol H 2 O x 18.02 g H 2 O 32.00 g O 2 1 mol O 2 1 mol H 2 O = 27.0 g H 2 O O 2 is the limiting reactant. (not possible) (smaller amount)

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Chapter 10 Slide 31 of 42 Percent Yield Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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Chapter 10 Slide 32 of 42 Theoretical, Actual, and Percent Yield Theoretical yield The maximum amount of product calculated using the balanced equation. Actual yield The amount of product obtained when the reaction takes place. Percent yield The ratio of actual yield to theoretical yield. percent yield = actual yield (g) x 100 theoretical yield (g)

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Chapter 10 Slide 33 of 42 To calculate the percent yield, the actual yield and theoretical yield are needed. You prepared cookie dough to make 5 dozen cookies. The phone rings and you answer. While talking, a sheet of 12 cookies burn and you have to throw them out. The rest of the cookies are okay. What is the percent yield of edible cookies? Theoretical yield 60 cookies possible Actual yield 48 cookies to eat Percent yield 48 cookies x 100 = 80% yield 60 cookies Calculating Percent Yield

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Chapter 10 Slide 34 of 42 Learning Check Without proper ventilation and limited oxygen, the reaction of carbon and oxygen produces carbon monoxide. 2C(g) + O 2 (g) 2CO(g) What is the percent yield if 40.0 g CO are produced when 30.0 g O 2 are used? theoretical yield of CO 30.0 g O 2 x 1 mol O 2 x 2 mol CO x 28.01 g CO 32.00 g O 2 1 mol O 2 1 mol CO = 52.5 g CO (theoretical) percent yield 40.0 g CO (actual) x 100 = 76.2 % yield 52.5 g CO (theoretical)

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Chapter 10 Slide 35 of 42 Learning Check When N 2 and 5.00 g H 2 are mixed, the reaction produces 16.0 g NH 3. What is the percent yield for the reaction? N 2 (g) + 3H 2 (g) 2NH 3 (g) 5.00 g H 2 x 1 mol H 2 x 2 mol NH 3 x 17.03 g NH 3 2.016 g H 2 3 mol H 2 1 mol NH 3 = 28.2 g NH 3 (theoretical) Percent yield = 16.0 g NH 3 x 100 = 56.7 % 28.2 g NH 3

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Chapter 10 Slide 36 of 42 Limiting Reactant What is the limiting reactant when 2.00g of Na and 2.00g of Cl 2 combine as follows: 2Na + Cl 2 2NaCl Na 22.99 g/mol Cl 35.45 g/mol

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Chapter 10 Slide 37 of 42 Problem continued…. How many grams of the remaining reactant would be left over once the reaction has run to completion? 2Na + Cl 2 2NaCl

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Chapter 10 Slide 38 of 42 Problem continued…. If the actual yield of NaCl is 2.29g, what is the percent yield?

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Chapter 10 Slide 39 of 42 Limiting Reactant Ethylene burns in air according to the following equation: C 2 H 4 (g) + 3 O 2 (g) 2 CO 2 (g) + 2 H 2 O(l) a. How many grams of CO 2 will be formed when a mixture of 2.93g of C 2 H 4 and 4.29g of O 2 combine? C 12.01 g/mol H 1.008 g/mol O 16.00 g/mol HW Due Next Class

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Chapter 10 Slide 40 of 42 Problem continued…. b. How many grams of each reactant would be left over once the reaction has gone to completion? HW Due Next Class

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Chapter 10 Slide 41 of 42 Problem continued…. c. If the percent yield of CO 2 is 72.1%, what was the actual yield? HW Due Next Class

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Chapter 10 Slide 42 of 42 Limiting Reactant Sulfur trioxide is prepared from SO 2 according to the following equation: 2SO 2 + O 2 2SO 3 In this reaction, not all SO 2 is converted to SO 3 even with excess O 2. In a given experiment, 21.2g of SO 3 is actually produced from 24.0g of SO 2. a) What is the theoretical yield of SO 3 ? b) What is the percent yield? HW S 32.07 g/mol O 16.00 g/mol Due Next Class

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Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 1 Chapter 7 Chemical Quantities 7.5 Mole Relationships in Chemical Equations.

Copyright © 2004 Pearson Education Inc., publishing as Benjamin Cummings. 1 Chapter 7 Chemical Quantities 7.5 Mole Relationships in Chemical Equations.

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