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By the end of this section, you will be able to:
  • Explain the concept of stoichiometry as it pertains to chemical reactions
  • Use balanced chemical equations to derive stoichiometric factors relating amounts of reactants and products
  • Perform stoichiometric calculations involving mass, moles, and solution molarity

A balanced chemical equation provides a great deal of information in a very succinct format. Chemical formulas provide the identities of the reactants and products involved in the chemical change, allowing classification of the reaction. Coefficients provide the relative numbers of these chemical species, allowing a quantitative assessment of the relationships between the amounts of substances consumed and produced by the reaction. These quantitative relationships are known as the reaction’s stoichiometry    , a term derived from the Greek words stoicheion (meaning “element”) and metron (meaning “measure”). In this module, the use of balanced chemical equations for various stoichiometric applications is explored.

The general approach to using stoichiometric relationships is similar in concept to the way people go about many common activities. Food preparation, for example, offers an appropriate comparison. A recipe for making eight pancakes calls for 1 cup pancake mix, 3 4 cup milk, and one egg. The “equation” representing the preparation of pancakes per this recipe is

1 cup mix + 3 4 cup milk + 1 egg 8 pancakes

If two dozen pancakes are needed for a big family breakfast, the ingredient amounts must be increased proportionally according to the amounts given in the recipe. For example, the number of eggs required to make 24 pancakes is

24 pancakes × 1 egg 8 pancakes = 3 eggs

Balanced chemical equations are used in much the same fashion to determine the amount of one reactant required to react with a given amount of another reactant, or to yield a given amount of product, and so forth. The coefficients in the balanced equation are used to derive stoichiometric factors that permit computation of the desired quantity. To illustrate this idea, consider the production of ammonia by reaction of hydrogen and nitrogen:

N 2 ( g ) + 3 H 2 ( g ) 2 NH 3 ( g )

This equation shows ammonia molecules are produced from hydrogen molecules in a 2:3 ratio, and stoichiometric factors may be derived using any amount (number) unit:

2 NH 3 molecules 3 H 2 molecules or 2 doz NH 3 molecules 3 doz H 2 molecules or 2 mol NH 3 molecules 3 mol H 2 molecules

These stoichiometric factors can be used to compute the number of ammonia molecules produced from a given number of hydrogen molecules, or the number of hydrogen molecules required to produce a given number of ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation.

Moles of reactant required in a reaction

How many moles of I 2 are required to react with 0.429 mol of Al according to the following equation (see [link] )?

2 Al + 3 I 2 2 AlI 3
This figure shows three photos with an arrow leading from one to the next. The first photo shows a small pile of iodine and aluminum on a white surface. The second photo shows a small amount of purple smoke coming from the pile. The third photo shows a large amount of purple and gray smoke coming from the pile.
Aluminum and iodine react to produce aluminum iodide. The heat of the reaction vaporizes some of the solid iodine as a purple vapor. (credit: modification of work by Mark Ott)

Solution

Referring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest is 3 mol I 2 2 mol Al . The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor:

This figure shows two pink rectangles. The first is labeled, “Moles of A l.” This rectangle is followed by an arrow pointing right to a second rectangle labeled, “Moles of I subscript 2.”
mol I 2 = 0.429 mol Al × 3 mol I 2 2 mol Al = 0.644 mol I 2

Check your learning

How many moles of Ca(OH) 2 are required to react with 1.36 mol of H 3 PO 4 to produce Ca 3 (PO 4 ) 2 according to the equation 3 Ca ( OH ) 2 + 2 H 3 PO 4 Ca 3 ( PO 4 ) 2 + 6 H 2 O?

Answer:

2.04 mol

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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