Chemistry - A Molecular Science

Ratios of atoms or molecules are equal to ratios of moles of atoms or molecules, and
chemists usually use moles rather than numbers

of individual particles in the laboratory.

The reaction 2H

• O 2

2 →

2H

O shows us that 2 mol H 2

react with every 1 mol O 2

and are 2

required to produce every 2 mol H

O. The ratio of the coefficients is the conversion 2

factor, called the

stoichiometric factor

, that allows us to convert from one substance to

another in a balanced chemical equation. The following stoichiometric factors and their reciprocals relate to the reaction of hydrogen and oxygen:

22 222 2

2 mol H

2 mol H

1 mol O

; ;

1 mol O

2 mol H O 2 mol H O

Example 1.

How many moles of H

are required to react with 8.0 g O 2

to produce H 2

O? 2

First, write the balanced chemic

al equation for the reaction.

2H

+ O 2

→ 2

2H

O 2

The first thing that must always be done when solving a stoichiometry problem is to covert the masses into moles by dividing

by the molar mass, which is 32.0 g

.mol

-1 for O

. 2

8.0 g O

× 2

1 mol O

2

32.0 g O

= 0.25 mol O 2

2

The given number of moles is then converted

into the chemically equivalent number of

desired moles with the stoichiometric factor

that contains both substances. The balanced

equation indicates that 2 mol H

are required for every 1 mol O 2

, so we write 2

0.25 mol O

× 2

2 mol H

2

1 mol O

= 0.50 mol H 2

required 2

This is an example of

reaction

stoichiometry

, the topic of Appendix D.

Chemists seldom add chemicals in the exact

stoichiometric ratio, so the amount of

product that forms depends upon the amount of the reactant, called the

limiting reactant

,^

that is consumed first. Consider the reaction of 5 mol S and 6 mol O

to produce SO 2

. 3

There are 5 mol S, and each mole of SO

requires one mole of S, so there is enough S to 3

produce 5 mol SO

. There are 12 mol O atoms in 6 mol O 3

, and each mole of SO 2

requires 3

3 mol of O, so there is enough O to make only 4 mol SO

. After 4 mol SO 3

are produced, 3

all of the O

is gone, and no more SO 2

can be made. Making 4 mol SO 3

requires only 4 3

mol S, so there is 1 mol S left over. Thus, O

is the limiting reactant, while S is in excess. 2

Chapter 1 The Early Experiments

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