Chapter 8 Chemical Quantities in Reactions 8.1 Mole Relationships in Chemical Equations Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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.

Law of Conservation of Mass 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. H2(g) + Cl2(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

Quantities in A Chemical Reaction 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) four molecules NH3 react with five molecules O2 to produce four molecules NO and six molecules H2O and four mol NH3 react with five mol O2 four mol NO and six mol H2O

Conservation of Mass 2 mol Ag + 1 mol S = 1 mol Ag2S 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

Moles in Equations The equation can be read in “moles” by placing the word “mole” or “mol” after each coefficient. 4Fe(s) + 3O2(g) 2Fe2O3(s) 4 mol Fe + 3 mol O2 2 mol Fe2O3

Writing Mole-Mole Factors A mole-mole factor is a ratio of the moles for two substances in an equation. 4Fe(s) + 3O2(g) 2Fe2O3(s) Fe and O2 4 mol Fe and 3 mol O2 3 mol O2 4 mol Fe Fe and Fe2O3 4 mol Fe and 2 mol Fe2O3 2 mol Fe2O3 4 mol Fe O2 and Fe2O3 3 mol O2 and 2 mol Fe2O3 2 mol Fe2O3 3 mol O2

Learning Check Consider the following equation: 3H2(g) + N2(g) 2NH3(g) A. A mole factor for H2 and N2 is 1) 3 mol N2 2) 1 mol N2 3) 1 mol N2 1 mol H2 3 mol H2 2 mol H2 B. A mole factor for NH3 and H2 is 1) 1 mol H2 2) 2 mol NH3 3) 3 mol N2 2 mol NH3 3 mol H2 2 mol NH3

Solution 3H2(g) + N2(g) 2NH3(g) A. A mole factor for H2 and N2 is 2) 1 mol N2 3 mol H2 B. A mole factor for NH3 and H2 is 2) 2 mol NH3

Copyright © 2008 by Pearson Education, Inc Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Calculations with Mole Factors How many moles of Fe2O3 can form from 6.0 mol O2? 4Fe(s) + 3O2(g) 2Fe2O3(s) STEP 1 Given 6.0 mol O2 Need: moles of Fe2O3. STEP 2 moles O2 moles Fe2O3 STEP 3 3 mol O2 = 2 mol Fe2O3 3 mol O2 and 2 mol Fe2O3 2 mol Fe2O3 3 mol O2 STEP 4 Set up problem using the mol factor. 6.0 mol O2 x 2 mol Fe2O3 = 4.0 mol Fe2O3 3 mol O2

Learning Check 4Fe(s) + 3O2(g) 2 Fe2O3(s) 1) 3.00 mol Fe How many moles of Fe are needed to react with 12.0 mol O2? 4Fe(s) + 3O2(g) 2 Fe2O3(s) 1) 3.00 mol Fe 2) 9.00 mol Fe 3) 16.0 mol Fe

Solution 3) 16.0 mol Fe Consider the following reaction: 4Fe(s) + 3O2(g) 2 Fe2O3(s) How many moles of Fe are needed to react with 12.0 mol O2? 12.0 mol O2 x 4 mol Fe = 16.0 mol Fe 3 mol O2