1. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 1 Chapter 9 Chemical Quantities in Reactions 9.1 Mole Relationships in Chemical Equations

2. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 2 Conservation of Mass The law of conservation of mass indicates that in an ordinary chemical reaction matter cannot be created or destroyed no change in total mass occurs in a reaction mass of products is equal to mass of reactants

3. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 3 Conservation of Mass ReactantsProducts 2 mol of Ag + 1 mol of S = 1 mol of Ag 2 S 2 (107.9 g) + 1(32.1 g) = 1 (247.9 g) 247.9 = 247.9 g

4. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 4 Consider the following equation: 2Fe(s) + 3S(s) Fe 2 S 3 (s) An equation can be read in moles by placing the words “mol of” between each coefficient and formula. 2 mol of Fe + 3 mol of S 1 mol of Fe 2 S 3 Reading an Equation in Moles

5. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 5 A mole-mole factor is a ratio of the moles for two substances in an equation. 2Fe(s) + 3S(s) Fe 2 S 3 (s) Fe and S 2 mol Fe and 3 mol S 3 mol S 2 mol Fe Fe and Fe 2 S 3 2 mol Fe and 1 mol Fe 2 S 3 1 mol Fe 2 S 3 2 mol Fe S and Fe 2 S 3 3 mol S and 1 mol Fe 2 S 3 1 mol Fe 2 S 3 3 mol S Writing Mole-Mole Factors

6. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 6 Consider the following equation: 3H 2 (g) + N 2 (g) 2NH 3 (g) A. A mole factor for H 2 and N 2 is 1) 3 mol N 2 2) 1 mol N 2 3) 1 mol N 2 1 mol H 2 3 mol H 2 2 mol H 2 B. A mole factor for NH 3 and H 2 is 1) 1 mol H 2 2) 2 mol NH 3 3) 3 mol N 2 2 mol NH 3 3 mol H 2 2 mol NH 3 Learning Check

7. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 7 3H 2 (g) + N 2 (g) 2NH 3 (g) A. A mole factor for H 2 and N 2 is 2) 1 mol N 2 3 mol H 2 B. A mole factor for NH 3 and H 2 is 2) 2 mol NH 3 3 mol H 2 Solution

8. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 8 Using Mole-Mole Factors

9. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 9 How many moles of Fe are needed for the reaction of 12.0 mol of O 2 ? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) 1) 3.00 mol of Fe 2) 9.00 mol of Fe 3) 16.0 mol of Fe Calculations with Mole Factors

10. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 10 STEP 1 Write the given and needed number of moles. Given 12 mol of O 2 Need moles of Fe STEP 2 Write a plan to convert the given to the needed moles. mol of O 2 mol of Fe Calculations with Mole-Mole Factors

11. Basic Chemistry Copyright © 2011 Pearson Education, Inc. 11 STEP 3 Use coefficients to write relationship and mole-mole factors. 4 mol of Fe = 3 mol of O 2 4 mol Fe and 3 mol O 2 3 mol O 2 4 mol Fe STEP 4 Set up problem using the mole  mole factor that cancels given moles. 12.0 mol O 2 x 4 mol Fe = 16.0 mol Fe (C) 3 mol O 2 Calculations with Mole-Mole Factors (continued)