Reaction Stoichiometry (7.3)

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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 to 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 the 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, [latex]\frac{3}{4}[/latex]  cup milk, and one egg. The “equation” representing the preparation of pancakes per this recipe is

1cup mix+34cup milk+1egg⟶8pancakes1cup mix+34cup milk+1egg⟶8pancakes

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

24pancakes×1egg8pancakes=3eggs24pancakes×1egg8pancakes=3eggs

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 the computation of the desired quantity. To illustrate this idea, consider the production of ammonia by reaction of hydrogen and nitrogen:

N2(g)+3H2(g)⟶2NH3(g)N2(g)+3H2(g)⟶2NH3(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:

2NH3molecules3H2moleculesor2 dozNH3molecules3 dozH2moleculesor2 molNH3molecules3 molH2molecules2NH3molecules3H2moleculesor2 dozNH3molecules3 dozH2moleculesor2 molNH3molecules3 molH2moleculesThese 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.

EXAMPLE 7.8

Moles of Reactant Required in a Reaction

How many moles of I2 are required to react with 0.429 mol of Al according to the following equation (see Figure 7.10)?

2Al+3I2⟶2AlI3

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.
Figure 7.10 Aluminum and iodine react to produce aluminum iodide. The heat of the reaction vaporizes some of the solid iodine as a purple vapor. By: Rice University Openstax CC BY NC SA

Solution

Referring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest is [latex]\frac{3\text{ mol }I_{2}}{2\text{ mol Al }}[/latex]. 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.”
Figure 7.11 By: Rice University OpenstaxCC BY NC SA

mol I2=0.429mol Al×3 molI22mol Al=0.644 molI2mol I2=0.429mol Al×3 molI22mol Al=0.644 molI2

Check Your Learning

How many moles of Ca(OH)2 are required to react with 1.36 mol of H3PO4 to produce Ca3(PO4)2 according to the equation:

3Ca(OH)2+2H3PO4⟶Ca3(PO4)2+6H2O?3Ca(OH)2+2H3PO4⟶Ca3(PO4)2+6H2O? 

Answer: 2.04 mol

EXAMPLE 7.9

Number of Product Molecules Generated by a Reaction

How many carbon dioxide molecules are produced when 0.75 mol of propane is combusted according to this equation?

C3H8 + 5O2 ⟶ 3CO2 + 4H2O

Solution

The approach here is the same as for Example 7.8, though the absolute number of molecules is requested, not the number of moles of molecules. This will simply require the use of the moles-to-numbers conversion factor, Avogadro’s number.

The balanced equation shows that carbon dioxide is produced from propane in a 3:1 ratio:

[latex]\frac{3\text{ mol }CO_{2}}{1\text{ mol }C_{3}H_{8}}[/latex]

Using this stoichiometric factor, the provided molar amount of propane, and Avogadro’s number,

This figure shows two pink rectangles. The first is labeled, “Moles of C subscript 3 H subscript 8.” This rectangle is followed by an arrow pointing right to a second rectangle labeled, “Moles of C O subscript 2.”
Figure 7.12 By: Rice University OpenstaxCC BY NC SA

 

0.76 mol C3H8 x [latex]\frac{3\cancel{\text{ mol }CO_{2}}}{1\cancel{\text{ mol }C_{3}H_{8}}} \times \frac{6.022 \times 10^{23} CO_{2} \text{molecules}}{\cancel{\text{ mol }CO_{2}}}[/latex] = 1.4 x 1024 CO2 molecules

Check Your Learning

How many NH3 molecules are produced by the reaction of 4.0 mol of Ca(OH)2 according to the following equation:
(NH4)2SO4 + Ca(OH)2 ⟶ 2NH3 + CaSO4 + 2H2O

Answer: 4.8 x 1024 NH3 molecules

 

These examples illustrate the ease with which the amounts of substances involved in a chemical reaction of known stoichiometry may be related. Directly measuring numbers of atoms and molecules is, however, not an easy task, and the practical application of stoichiometry requires that we use the more readily measured property of mass.

EXAMPLE 7.10

Relating Masses of Reactants and Products

What mass of sodium hydroxide, NaOH, would be required to produce 16 g of the antacid milk of magnesia [magnesium hydroxide, Mg(OH)2] by the following reaction?

 

MgCl2(aq) + 2NaOH (aq) ⟶ Mg(OH)2(s) + 2NaCl (aq)

Solution

The approach used previously in Example 7.8 and Example 7.9 is likewise used here; that is, we must derive an appropriate stoichiometric factor from the balanced chemical equation and use it to relate the amounts of the two substances of interest. In this case, however, masses (not molar amounts) are provided and requested, so additional steps of the sort learned in the previous chapter are required. The calculations required are outlined in this flowchart:

This figure shows four rectangles. The first is shaded yellow and is labeled, “Mass of M g ( O H ) subscript 2.” This rectangle is followed by an arrow pointing right to a second rectangle which is shaded pink and is labeled, “Moles of M g ( O H ) subscript 2.” This rectangle is followed by an arrow pointing right to a third rectangle which is shaded pink and is labeled, “Moles of N a O H.” This rectangle is followed by an arrow pointing right to a fourth rectangle which is shaded yellow and is labeled, “Mass of N a O H.”
Figure 7.13 By: Rice University OpenstaxCC BY NC SA
16 g Mg(OH)2  x  [latex]\frac{1\cancel{\text{mol Mg} (OH)_{2}}}{58.3 \cancel{\text{g Mg}(OH)_{2}}} \times \frac{2\cancel{\text{mol NaOH}}}{1\cancel{\text{mol Mg}(OH)_{2}}} \times \frac{40.0 g NaOH}{1 \cancel{\text{mol NaOH}}}[/latex] = 22 g NaOH

Check Your Learning

What mass of gallium oxide, Ga2O3, can be prepared from 29.0 g of gallium metal? The equation for the reaction is 

Answer: 39.0 g

 

EXAMPLE 7.11

Relating Masses of Reactants

What mass of oxygen gas, O2, from the air is consumed in the combustion of 702 g of octane, C8H18, one of the principal components of gasoline?

2C8H18 + 25O2 ⟶ 16CO2 + 18H2O

Solution

The approach required here is the same as for the approach in Example 7.10, differing only in that the provided and requested masses are both for reactant species.

 

This figure shows four rectangles. The first is shaded yellow and is labeled, “Mass of C subscript 8 H subscript 18.” This rectangle is followed by an arrow pointing right to a second rectangle which is shaded pink and is labeled, “Moles of C subscript 8 H subscript 18.” This rectangle is followed by an arrow pointing right to a third rectangle which is shaded pink and is labeled, “Moles of O subscript 2.” This rectangle is followed by an arrow pointing right to a fourth rectangle which is shaded yellow and is labeled, “Mass of O subscript 2.”
Figure 7.14 By: Rice University OpenstaxCC BY NC SA

702 g C8H18  x  [latex]\frac{1\cancel{\text{ mol }C_{8}H_{18}}}{114.23\cancel{g C_{8}H_{18}}} \times \frac{25\cancel{\text{ mol } O_{2}}}{2\cancel{\text{ mol }C_{8}H_{18}}} \times \frac{32.00 g O_{2}}{\cancel{\text{ mol} O_{2}}}[/latex] = 2.46 x 103 g O2

Check Your Learning

What mass of CO is required to react with 25.13 g OF Fe2O3 according to the equation:

Fe2O3 + 3CO ⟶ 2Fe +3CO2

Answer: 13.22 g

 

These examples illustrate just a few instances of reaction stoichiometry calculations. Numerous variations on the beginning and ending computational steps are possible depending upon what particular quantities are provided and sought (volumes, solution concentrations, and so forth). Regardless of the details, all these calculations share a common essential component: the use of stoichiometric factors derived from balanced chemical equations. Figure 7.15

  • provides a general outline of the various computational steps associated with many reaction stoichiometry calculations.
This flowchart shows 10 rectangles connected by double headed arrows. To the upper left, a rectangle is shaded lavender and is labeled, “Volume of pure substance A.” This rectangle is followed by a horizontal double headed arrow labeled, “Density.” It connects to a second rectangle which is shaded yellow and is labeled, “Mass of A.” This rectangle is followed by a double headed arrow which is labeled, “Molar Mass,” that connects to a third rectangle which is shaded pink and is labeled, “Moles of A.” To the left of this rectangle is a horizontal double headed arrow labeled, “Molarity,” which connects to a lavender rectangle which is labeled, “Volume of solution A.” The pink, “Moles of A,” rectangle is also connected with a double headed arrow below and to the left. This arrow is labeled “Avogadro’s number.” It connects to a green shaded rectangle that is labeled, “Number of particles of A.” To the right of the pink “Moles of A,” rectangle is a horizontal double headed arrow which is labeled, “Stoichiometric factor.” It connects to a second pink rectangle which is labeled, “Moles of B.” A double headed arrow which is labeled, “Molar mass,” extends from the top of this rectangle above and to the right to a yellow shaded rectangle labeled, “Mass of B.” A horizontal double headed arrow which is labeled, “Density” links to a lavender rectangle labeled, “Volume of substance B,” to the right. A horizontal double headed arrow labeled, “Molarity,” extends right to the of the pink “Moles of B” rectangle. This arrow connects to a lavender rectangle that is labeled, “Volume of substance B.” Another double headed arrow extends below and to the right of the pink “Moles of B” rectangle. This arrow is labeled “Avogadro’s number,” and it extends to a green rectangle which is labeled, “Number of particles of B.”
Figure 7.15 The flowchart depicts the various computational steps involved in most reaction stoichiometry calculations. By: Rice University OpenstaxCC BY NC SA

CHEMISTRY IN EVERYDAY LIFE

Airbags

Airbags (Figure 7.16) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, NaN3. When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of NaN3 to initiate its decomposition:

2NaN3(s)⟶3N2(g)+2Na(s) 

This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second (~0.03–0.1 s). Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the airbag and ensure its proper function. For example, a small mass (~100 g) of NaN3 will generate approximately 50 L of N2.

This photograph shows the inside of an automobile from the driver’s side area. The image shows inflated airbags positioned just in front of the driver’s and passenger’s seats and along the length of the passenger side over the windows. A large, round airbag covers the steering wheel.
Figure 7.16 Airbags deploy upon impact to minimize serious injuries to passengers. By: Rice University OpenstaxCC BY NC SA

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