4.2: Reaction Stoichiometry

  Learning Objectives

• 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
• Explain the concepts of theoretical yield, atom economy, and limiting reactants/reagents.
• Derive the theoretical yield for a reaction under specified conditions.
• Calculate the percent yield for a reaction.
• Calculate the atom economy for a reaction.

The relative amounts of reactants and products represented in a balanced chemical equation are often referred to as stoichiometric amounts. All the exercises of the preceding module involved stoichiometric amounts of reactants. For example, when calculating the amount of product generated from a given amount of reactant, it was assumed that any other reactants required were available in stoichiometric amounts (or greater). In this module, more realistic situations are considered, in which reactants are not present in stoichiometric amounts.

Stoichiometry

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. Cooking, for example, offers an appropriate comparison. Suppose a recipe for making eight pancakes calls for 1 cup pancake mix, \(\dfrac{3}{4}\) cup milk, and one egg. The “equation” representing the preparation of pancakes per this recipe is

\[\mathrm{1\:cup\: mix+\dfrac{3}{4}\:cup\: milk+1\: egg \rightarrow 8\: pancakes} \label{4.4.1} \]

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

\[\mathrm{24\: \cancel{pancakes} \times \dfrac{1\: egg}{8\: \cancel{pancakes}}=3\: eggs} \label{4.4.2} \]

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:

\[\ce{N2}(g)+\ce{3H2}(g)\rightarrow \ce{2NH3}(g) \label{4.4.3} \]

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

\[\ce{\dfrac{2NH3 \: molecules}{3H2 \: molecules}\: or \: \dfrac{2 \: doz \: NH3\: molecules}{3\: doz\:H2 \:molecules} \: or \: \dfrac{2\: mol\: NH3\: molecules}{3\: mol\: H2\: molecules}} \label{4.4.4} \]

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.

 

Example \(\PageIndex{1}\): 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 \(\PageIndex{1}\))?

\[\ce{2Al + 3I2 \rightarrow 2AlI3} \label{4.4.5} \]

Figure \(\PageIndex{1}\): 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 \(\ce{\dfrac{3\: mol\: I2}{2\: mol\: Al}}\). The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor:

\[\begin{align*} \mathrm{mol\: I_2} &=\mathrm{0.429\: \cancel{mol\: Al}\times \dfrac{3\: mol\: I_2}{2\:\cancel{mol\: Al}}} \\[4pt] &=\mathrm{0.644\: mol\: I_2} \end{align*} \nonumber \]

Exercise \(\PageIndex{1}\)

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 \(\ce{3Ca(OH)2 + 2H3PO4 \rightarrow Ca3(PO4)2 + 6H2O}\) ?

Answer

2.04 mol

 

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 \(\PageIndex{2}\): 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?

\(\ce{MgCl2}(aq)+\ce{2NaOH}(aq)\rightarrow \ce{Mg(OH)2}(s)+\ce{2NaCl}(aq)\)

 

Solution

The approach used previously in Examples \(\PageIndex{1}\) and \(\PageIndex{2}\) 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:

\[\mathrm{16\:\cancel{g\: Mg(OH)_2} \times \dfrac{1\:\cancel{mol\: Mg(OH)_2}}{58.3\:\cancel{g\: Mg(OH)_2}}\times \dfrac{2\:\cancel{mol\: NaOH}}{1\:\cancel{mol\: Mg(OH)_2}}\times \dfrac{40.0\: g\: NaOH}{\cancel{mol\: NaOH}}=22\: g\: NaOH} \nonumber \]

 

Exercise \(\PageIndex{2}\)

What mass of gallium oxide, Ga₂O₃, can be prepared from 29.0 g of gallium metal? The equation for the reaction is \(\ce{4Ga + 3O2 \rightarrow 2Ga2O3}\).

Answer

39.0 g

Example \(\PageIndex{3}\): 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?

\[\ce{2C8H18 + 25O2 \rightarrow 16CO2 + 18H2O} \nonumber \]

Solution

The approach required here is the same as for the Example \(\PageIndex{3}\), differing only in that the provided and requested masses are both for reactant species.

\(\mathrm{702\:\cancel{g\:\ce{C8H18}}\times \dfrac{1\:\cancel{mol\:\ce{C8H18}}}{114.23\:\cancel{g\:\ce{C8H18}}}\times \dfrac{25\:\cancel{mol\:\ce{O2}}}{2\:\cancel{mol\:\ce{C8H18}}}\times \dfrac{32.00\: g\:\ce{O2}}{\cancel{mol\:\ce{O2}}}=2.46\times 10^3\:g\:\ce{O2}}\)

Exercise \(\PageIndex{3}\)

What mass of CO is required to react with 25.13 g of Fe₂O₃ according to the equation \(\ce{Fe2O3 + 3CO \rightarrow 2Fe + 3CO2}\)?

Answer

13.22 g

 

 

Limiting Reactant

Consider another food analogy, making grilled cheese sandwiches (Figure \(\PageIndex{2}\)):

\[\text{1 slice of cheese} + \text{2 slices of bread} \rightarrow \text{1 sandwich} \label{4.5.A} \]

Stoichiometric amounts of sandwich ingredients for this recipe are bread and cheese slices in a 2:1 ratio. Provided with 28 slices of bread and 11 slices of cheese, one may prepare 11 sandwiches per the provided recipe, using all the provided cheese and having six slices of bread left over. In this scenario, the number of sandwiches prepared has been limited by the number of cheese slices, and the bread slices have been provided in excess.

Figure \(\PageIndex{2}\): Sandwich making can illustrate the concepts of limiting and excess reactants.

Consider this concept now with regard to a chemical process, the reaction of hydrogen with chlorine to yield hydrogen chloride:

\[\ce{H2}(g) + \ce{Cl2}(g)\rightarrow \ce{2HCl}(g) \nonumber \]

The balanced equation shows the hydrogen and chlorine react in a 1:1 stoichiometric ratio. If these reactants are provided in any other amounts, one of the reactants are entirely consumed and limit the amount of product generated. This substance is the limiting reactant.  Any substances that are left over are the excess reactant. Identifying the limiting and excess reactants for a given situation requires computing the molar amounts of each reactant provided and comparing them to the stoichiometric amounts represented in the balanced chemical equation.

For example, imagine combining 6 moles of H₂ and 4 moles of Cl₂. Identifying the limiting reactant involves comparing the amount of product expected for the complete reaction of each reactant. Each reactant amount is used to separately calculate the amount of product that would be formed per the reaction’s stoichiometry. The reactant yielding the lesser amount of product is the limiting reactant.

For the example in the previous paragraph, the complete reaction of the hydrogen would yield

\[\mathrm{mol\: HCl\: produced=6\: mol\:H_2\times \dfrac{2\: mol\: HCl}{1\: mol\:H_2}=12\: mol\: HCl} \nonumber \]

The complete reaction of the provided chlorine would produce

\[\mathrm{mol\: HCl\: produced=4\: mol\:Cl_2\times \dfrac{2\: mol\: HCl}{1\: mol\:Cl_2}=8\: mol\: HCl} \nonumber \]

The chlorine will be completely consumed once 8 moles of HCl have been produced. Since enough hydrogen was provided to yield 12 moles of HCl, there will be non-reacted hydrogen remaining once this reaction is complete. Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant (Figure \(\PageIndex{3}\)).  To determine the amount of excess reactant that remains, the amount of hydrogen consumed in the reaction can be subtracted from the starting quantity of hydrogen. 

The amount of hydrogen consumed is

 \[\mathrm{mol\: H_2\: produced=8\: mol\:HCl\times \dfrac{1\: mol\: H_2}{2\: mol\:HCl}=4\: mol\: H_2} \nonumber \]

Subtract the hydrogen consumed from the starting quantity

\[\mathrm{mole\: of\: excess\:H_{2}=6\:mol\:H_{2}\:starting\:-\:4\:mol\:H_{2}\:consumed\:=\:2\:mol\:H_{2}\; excess} \nonumber \] 

Figure \(\PageIndex{3}\) provides a visual representation for the initial and final conditions for this reaction.

 

Figure \(\PageIndex{3}\): When H₂ and Cl₂ are combined in nonstoichiometric amounts, one of these reactants will limit the amount of HCl that can be produced. This illustration shows a reaction in which hydrogen is present in excess and chlorine is the limiting reactant.

A useful way to summarize the calculations required for determining the reaction yield is an Initial, Change, Final (ICF) table.  This format helps organize the problem and shows the connections for solving a wide range of stoichiometry problems.  Set up a table where the first row is the balanced chemical equation, in the second row you list all the initial amounts - in moles, in the third row you list the change that occurs in the reaction, and in the fourth row you list the final amounts remaining when the reaction is complete.  This layout makes it easy to see if you make any mistakes - like picking the wrong limiting reagent.  And clarifies all the interrelationships needed to solve a wide range of problems.  An example is shown in Table \(\PageIndex{1}\) for the reaction of H₂ and Cl₂:

 

Table \(\PageIndex{1}\): ICF table for reaction of H₂ and Cl₂

	H2	Cl2	→	HCl
Initial	6	4		0
Change	4	4		8
Final	2	0		8

Example \(\PageIndex{4}\): Identifying the Limiting Reactant

Silicon nitride is a very hard, high-temperature-resistant ceramic used as a component of turbine blades in jet engines. It is prepared according to the following equation:

\[\ce{3Si}(s)+\ce{2N2}(g)\rightarrow \ce{Si3N4}(s) \nonumber \]

Which is the limiting reactant when 2.00 g of Si and 1.50 g of N₂ react?

Solution

Compute the provided molar amounts of reactants, and then compare these amounts to the balanced equation to identify the limiting reactant.

\[\mathrm{mol\: Si=2.00\:\cancel{g\: Si}\times \dfrac{1\: mol\: Si}{28.09\:\cancel{g\: Si}}=0.0712\: mol\: Si} \nonumber \]

\[\mathrm{mol\:N_2=1.50\:\cancel{g\:N_2}\times \dfrac{1\: mol\:N_2}{28.02\:\cancel{g\:N_2}}=0.0535\: mol\:N_2} \nonumber \]

Compute the amount of product expected for complete reaction of each of the provided reactants. The 0.0712 moles of silicon would yield

\[\mathrm{mol\:Si_3N_4\:produced=0.0712\: mol\: Si\times \dfrac{1\:mol\:Si_3N_4}{3\: mol\: Si}=0.0237\: mol\:Si_3N_4} \nonumber \]

while the 0.0535 moles of nitrogen would produce

\[\mathrm{mol\:Si_3N_4\:produced=0.0535\: mol\:N_2\times \dfrac{1\: mol\:Si_3N_4}{2\: mol\:N_2}=0.0268\: mol\:Si_3N_4} \nonumber \]

Since silicon yields the lesser amount of product, it is the limiting reactant.

This is summarized in the ICF table shown below.  In the Change row, always use the stoichiometric coefficients from the balanced equation to go from one column to another.

	3 Si	2 N2	→	Si3N4
Initial (grams)	2.00 g	1.50 g		0 g
Initial (moles)	0.0712 mol	0.0535 mol		0 mol
Change (moles)	0.0712 mol	0.0475 mol		0.0237 mol
Final (moles)	0 mol	0.0060 mol		0.0237 mol

 

Exercise \(\PageIndex{4}\)

Which is the limiting reactant when 5.00 g of H₂ and 10.0 g of O₂ react and form water?

Answer

O₂

Stoichiometry Calculations - Video

This project was preformed to supply Libretext authors with videos on General Chemistry topics which can be used to enhance their projects.  Also, these videos are meant to act as a learning resource for all General Chemistry students.

 

Video Topics

This video discusses finding the mols and masses of reactants and products of a reaction by using stoichiometric factors (Mol Ratios). A reaction's stoichiometry can be used to create mole ratio conversion factor which link reactants and products. These types of calculation are a simplified version of the Limiting Reactant Problems which will come later.

Link to Video

Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios): https://youtu.be/74mHV0CZcjw

 

Attribution

• Prof. Steven Farmer (Sonoma State University)

Percent Yield

The amount of product that may be produced by a reaction under specified conditions, as calculated per the stoichiometry of an appropriate balanced chemical equation, is called the theoretical yield of the reaction. The amount of product obtained when a reaction is actually carried out is the actual yield.  The actual yield is often less than the theoretical yield. This may happen because a reaction also produces other side reactions and generates other products.  It could also occur because some reactions do not go to completion and some of the limiting reagent remains when the reaction stops.  And finally some products are difficult to collect and so some of the product is lost, which also reduces the actual yield. The extent to which a reaction’s theoretical yield is achieved is commonly expressed as its percent yield:

\[\mathrm{percent\: yield=\dfrac{actual\: yield}{theoretical\: yield}\times 100\%} \nonumber \]

Actual and theoretical yields may be expressed as masses or molar amounts (or any other appropriate property; e.g., volume, if the product is a gas). As long as both yields are expressed using the same units, these units will cancel when percent yield is calculated.

Example \(\PageIndex{5}\): Calculation of Percent Yield

Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained according to the equation:

\[\ce{CuSO4}(aq)+\ce{Zn}(s)\rightarrow \ce{Cu}(s)+\ce{ZnSO4}(aq) \nonumber \]

What is the percent yield?

Solution

The provided information identifies copper sulfate as the limiting reactant, and so the theoretical yield is found by the approach illustrated in the previous module, as shown here:

\[\mathrm{1.274\:\cancel{g\:Cu_SO_4}\times \dfrac{1\:\cancel{mol\:CuSO_4}}{159.62\:\cancel{g\:CuSO_4}}\times \dfrac{1\:\cancel{mol\: Cu}}{1\:\cancel{mol\:CuSO_4}}\times \dfrac{63.55\:g\: Cu}{1\:\cancel{mol\: Cu}}=0.5072\: g\: Cu} \nonumber \]

Using this theoretical yield and the provided value for actual yield, the percent yield is calculated to be

\[\mathrm{percent\: yield=\left(\dfrac{actual\: yield}{theoretical\: yield}\right)\times 100} \nonumber \]

\[\begin{align*}
\mathrm{percent\: yield}&=\mathrm{\left(\dfrac{0.392\: g\: Cu}{0.5072\: g\: Cu}\right)\times 100} \\
&=77.3\%
\end{align*} \nonumber \]

Exercise \(\PageIndex{5}\)

What is the percent yield of a reaction that produces 12.5 g of the gas Freon CF₂Cl₂ from 32.9 g of CCl₄ and excess HF?

\[\ce{CCl4 + 2HF \rightarrow CF2Cl2 + 2HCl} \nonumber \]

Answer

48.3%

 

Atom Economy

The purposeful design of chemical products and processes that minimize the use of environmentally hazardous substances and the generation of waste is known as green chemistry. Green chemistry is a philosophical approach that is being applied to many areas of science and technology, and its practice is summarized by guidelines known as the “Twelve Principles of Green Chemistry”. One of the 12 principles is aimed specifically at maximizing the efficiency of processes for synthesizing chemical products. The atom economy of a process is a measure of this efficiency, defined as the percentage by mass of the desired product of a synthesis relative to the masses of all the reactants used:

\[\mathrm{atom\: economy=\dfrac{mass\: of\: product}{mass\: of\: reactants}\times 100\%} \nonumber \]

Though the definition of atom economy at first glance appears very similar to that for percent yield, be aware that this property represents a difference in the theoretical efficiencies of different chemical processes. The percent yield of a given chemical process, on the other hand, evaluates the efficiency of a process by comparing the yield of product actually obtained to the maximum yield predicted by stoichiometry.

The synthesis of the common nonprescription pain medication, ibuprofen, nicely illustrates the success of a green chemistry approach (Figure \(\PageIndex{3}\)). First marketed in the early 1960s, ibuprofen was produced using a six-step synthesis that required 514 g of reactants to generate each mole (206 g) of ibuprofen, an atom economy of 40%. In the 1990s, an alternative process was developed by the BHC Company (now BASF Corporation) that requires only three steps and has an atom economy of ~80%, nearly twice that of the original process. The BHC process generates significantly less chemical waste; uses less-hazardous and recyclable materials; and provides significant cost-savings to the manufacturer (and, subsequently, the consumer). In recognition of the positive environmental impact of the BHC process, the company received the Environmental Protection Agency’s Greener Synthetic Pathways Award in 1997.

Figure \(\PageIndex{3}\): (a) Ibuprofen is a popular nonprescription pain medication commonly sold as 200 mg tablets. (b) The BHC process for synthesizing ibuprofen requires only three steps and exhibits an impressive atom economy. (credit a: modification of work by Derrick Coetzee)

 

Example \(\PageIndex{6}\): Calculation of Atom Economy

Plants use carbon dioxide and water to produce fructose (C₆H₁₂O₆) and oxygen gas.  Write a balanced chemical equation that describes this reaction and determine the atom economy for the production of fructose.

Solution

\[6 \mathrm{CO_2}(g)+ 6 \mathrm{H_2O}(g)\rightarrow \mathrm{C_6H_{12}O_6}(s)+6 \mathrm{O_2}(aq) \]

 

To determine the atom economy calculate the mass of the desired product using the stoichiometry for the balanced reaction.

This reaction produces 1 mole of fructose with a molar mass of 180.16 g/mol, for a total mass of 180.16 g

This reaction uses 6 moles of carbon dioxide (44.01 g/mol) and 6 moles of water (18.01 g/mol) for a total mass of 372.12 g

The atom economy is:

\[ \frac{180.16 g}{372.12 g} \times 100=48.14% \]

 

Exercise \(\PageIndex{6}\)

Write a balanced chemical equation and determine the atom economy for the production of sucrose C₁₂H₂₂O₁₁ and oxygen gas from carbon dioxide and water.

Answer

47.13%

 

Airbags

Airbags (Figure \(\PageIndex{4}\)) 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:

\[\ce{2NaN3}(s)\rightarrow \ce{3N2}(g)+\ce{2Na}(s) \nonumber \]

 

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 air bag and ensure its proper function. For example, a small mass (~100 g) of NaN₃ will generate approximately 50 L of N₂.

 

Figure \(\PageIndex{4}\): Airbags deploy upon impact to minimize serious injuries to passengers. (credit: Jon Seidman)

Summary

When reactions are carried out using less-than-stoichiometric quantities of reactants, the amount of product generated will be determined by the limiting reactant. The amount of product generated by a chemical reaction is its actual yield. This yield is often less than the amount of product predicted by the stoichiometry of the balanced chemical equation representing the reaction (its theoretical yield). The extent to which a reaction generates the theoretical amount of product is expressed as its percent yield.

A balanced chemical equation may be used to describe a reaction’s stoichiometry (the relationships between amounts of reactants and products). Coefficients from the equation are used to derive stoichiometric factors that subsequently may be used for computations relating reactant and product masses, molar amounts, and other quantitative properties.

 Key Equations

 \(\mathrm{percent\: yield=\left(\dfrac{actual\: yield}{theoretical\: yield}\right)\times 100}\)

\(\mathrm{atom\: economy=\left(\dfrac{mass\: desired\: product}{mass \: all\: reactants}\right)\times 100}\)

Glossary

 

actual yield

amount of product formed in a reaction

excess reactant

reactant present in an amount greater than required by the reaction stoichiometry

limiting reactant

reactant present in an amount lower than required by the reaction stoichiometry, thus limiting the amount of product generated

percent yield

measure of the efficiency of a reaction, expressed as a percentage of the theoretical yield

stoichiometric factor

ratio of coefficients in a balanced chemical equation, used in computations relating amounts of reactants and products

stoichiometry

relationships between the amounts of reactants and products of a chemical reaction

theoretical yield

amount of product that may be produced from a given amount of reactant(s) according to the reaction stoichiometry