5 Steps to a 5 AP Chemistry

Reaction Stoichiometry

As we have discussed previously, the balanced chemical equation not only indicates which chem-

ical species are the reactants and the products, but also indicates the relative ratio of reactants and

products. Consider the balanced equation of the Haber process for the production of ammonia:

N 2 (g) +3H 2 (g) →2 NH 3 (g)

This balanced equation can be read as: 1 nitrogen molecule reacts with 3 hydrogen mole-

cules to produce 2 ammonia molecules.But as indicated previously, the coefficients can stand

not only for the number of atoms or molecules (microscopic level), they can also stand for

the number of molesof reactants or products. The equation can also be read as: 1 mol of

nitrogen molecules reacts with 3 mol of hydrogen molecules to produce 2 mol of ammonia mol-

ecules.And if the number of moles is known, the number of grams or molecules can be cal-

culated. This is stoichiometry, the calculation of the amount (mass, moles, particles) of one

substance in a chemical reaction through the use of another. The coefficients in a balanced

chemical equation define the mathematical relationship between the reactants and products,

and allow the conversion from moles of one chemical species in the reaction to another.

Consider the Haber process above. How many moles of ammonia could be produced

from the reaction of 20.0 mol of nitrogen with excess hydrogen?

Before any stoichiometry calculation can be done, you must have a balanced chemical equation!

You are starting with moles of nitrogen and want moles of ammonia, so we’ll convert

from moles of nitrogen to moles of ammonia by using the ratio of moles of ammonia to

moles of nitrogen as defined by the balanced chemical equation:

The ratio of 2 mol NH 3 to 1 mol N 2 is called the stoichiometric ratio and comes from the

balanced chemical equation.

Suppose you also wanted to know how many moles of hydrogen it would take to fully

react with the 20.0 mol of nitrogen. Just change the stoichiometric ratio:

Notice that this new stoichiometric ratio also came from the balanced chemical equation.

Suppose instead of moles you had grams and wanted an answer in grams. How many grams

of ammonia could be produced from the reaction of 85.0g of hydrogen gas with excess nitrogen?

In working problems that involve something other than moles, you will still need moles.

And you will need the balanced chemical equation.

In this problem we will convert from grams of hydrogen to moles of hydrogen to moles of

ammonia using the correct stoichiometric ratio, and finally to grams of ammonia. And we

will need the molar mass of H 2 (2.0158 g/mol) and ammonia (17.0307 g/mol):

Actually, you could have calculated the actual number of ammonia molecules produced if

you had gone from moles of ammonia to molecules (using Avogadro’s number):

85 0

1

1

2 0158

. 2 2 6 022

.

gH molH.

g

mol NH

3molH

2 3

2

×× ×

××

=×

10

1 693 10

23

25

molecules NH

1molNH

molecules

3

3

.NNH 3

85 0

1

. gH 2 1 molH^2 17 03.
2.0158 g

mol NH

3molH

2 3

2

×× ×

007

1

478 8. 3

g

mol NH

gNH

3

=

20 0

1

3

1

60 0

.

.

mol N mol H

mol N

(^22) mol H
2

×= 2

20 0 2

40 0

.

.

mol N

1

mol NH

1molN

(^2) mol NH
3
2

×= 3

Stoichiometry  91

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