5 Steps to a 5 AP Chemistry 2019

Stoichiometry ‹ 91

Reaction Stoichiometry

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

cal 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 molecules

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 moles of 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 molecules. And

if the number of moles is known, the number of grams or molecules can be calculated. 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 equa-

tion 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:

20.0 mol N

1

2 mol  NH

1 mol N

(^23) 40.0 mol NH
2

×= 3

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:

×=

20.0 mol  N

1

3 mol  H

1 mol N

(^22) 60.0 mol  H
2
2
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 85.0 g 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):
×× ×=
85.0 g H
1
1 molH
2.0158 g
2 molNH
3 molH
17.0307 g
1 molNH
(^223) 478.8 g NH
23
3
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 g H

1

1 molH

2.0158 g

2 molNH

3 molH

6.022 10 moleculesNH

1 molNH

1.693 10 molecules NH

22 3

2

23

3

3

25

3

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