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

2. 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.

3. 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) HCl(g)
2 mol H, mol Cl = 2 mol H, 2 mol Cl
2(1.008) + 2(35.45) = (36.46)
72.92 g = g

4. Quantities in A Chemical Reaction
4NH3(g) O2(g) NO(g) H2O(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

5. Conservation of Mass 2 mol Ag + 1 mol S = 1 mol Ag2S
2 (107.9 g) (32.07 g) = (247.9 g)
247.9 g reactants = g product
Copyright © by Pearson Education, Inc. Publishing as Benjamin Cummings

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

7. Writing Mole-Mole Factors
A mole-mole factor is a ratio of the moles for two
substances in an equation.
4Fe(s) O2(g) Fe2O3(s)
Fe and O mol Fe and 3 mol O2
3 mol O mol Fe
Fe and Fe2O mol Fe and 2 mol Fe2O3
2 mol Fe2O3 4 mol Fe
O2 and Fe2O mol O and 2 mol Fe2O3
2 mol Fe2O3 3 mol O2

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

9. 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

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

11. Calculations with Mole Factors
How many moles of Fe2O3 can form from 6.0 mol O2?
4Fe(s) O2(g) Fe2O3(s)
STEP 1 Given 6.0 mol O2 Need: moles of Fe2O3.
STEP 2 moles O moles Fe2O3
STEP mol O2 = 2 mol Fe2O3
3 mol O2 and 2 mol Fe2O3
2 mol Fe2O 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

12. 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) O2(g) Fe2O3(s)
1) mol Fe
2) mol Fe
3) mol Fe

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