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Reaction Stoichiometry Examples

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 image below)?

\(\mathrm{2Al} + \mathrm{3I_2} \longrightarrow \mathrm{2AlI_3}\)


Aluminum and iodine react to produce aluminum iodide. The heat of the reaction vaporizes some of the solid iodine as a purple vapor. Image credit: modification of work by Mark Ott


Referring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest is \(\frac{3\text{ mol }\mathrm{I_2}}{2\text{ mol Al}}.\) The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor:


Image credit: OpenStax, Chemistry

\mathrm{mol \; I_2} & = & \mathrm{0.429 \; \enclose{horizontalstrike}{mol \; Al}} × \frac{3\text{ mol }\mathrm{I_2}}{2\text{ mol Al}} \\
& = & \mathrm{0.644 \; mol \; I_2} \\

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?

\(\mathrm{C_3H_8} + \mathrm{5O_2} \longrightarrow \mathrm{3CO_2} + \mathrm{4H_2O}\)


The approach here is the same as for the first example above, though the absolute number of molecules is requested, not the number of moles of molecules. This will simply require 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:

\(\cfrac{\mathrm{3 \; mol \; CO_2}}{\mathrm{1 \; mol \; C_3H_8}}\)

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


Image credit: OpenStax, Chemistry

\(\require{cancel}\mathrm{0.75 \; \cancel{mol \; C_3H_8}} × \cfrac{\mathrm{3 \; \cancel{mol \; CO_2}}}{\mathrm{1 \; \cancel{mol \; C_3H_8}}} × \cfrac{6.022 × 10^{23} \mathrm{\; CO_2 \; molecules}}{\mathrm{\cancel{mol \; CO_2}}} = 1.4 × 10^{24} \mathrm{\; CO_2 \; molecules}\)

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.

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?

\(\mathrm{MgCl_2}(aq) + \mathrm{2NaOH}(aq) \longrightarrow \mathrm{Mg(OH)_2}(s) + \mathrm{2NaCl}(aq)\)


The approach used previously in the examples above 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 last set of lessons in Chemistry 101 are required. The calculations required are outlined in this flowchart:


Image credit: OpenStax, Chemistry

\(\require{cancel}\mathrm{16 \; \cancel{g \; Mg(OH)_2}} × \cfrac{\mathrm{1 \; \cancel{mol \; Mg(OH)_2}}}{\mathrm{58.3 \; \cancel{g \; Mg(OH)_2}}} × \cfrac{\mathrm{2 \; \cancel{mol \; NaOH}}}{\mathrm{1 \; \cancel{mol \; Mg(OH)_2}}} × \cfrac{\mathrm{40.0 \; \cancel{g \; NaOH}}}{\mathrm{\cancel{mol \; NaOH}}} = \mathrm{22 \; g \; NaOH}\)

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?

\(\mathrm{2C_8H_{18}} + \mathrm{25O_2} \longrightarrow \mathrm{16CO_2} + \mathrm{18H_2O}\)


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


\(\require{cancel}\mathrm{702 \; \cancel{g \; C_8H_{18}}} × \cfrac{\mathrm{1 \; \cancel{mol \; C_8H_{18}}}}{\mathrm{114.23 \; \cancel{g \; C_8H_{18}}}} × \cfrac{\mathrm{25 \; \cancel{mol \; O_2}}}{\mathrm{2 \; \cancel{mol \; C_8H_{18}}}} × \cfrac{\mathrm{32.00 \; \cancel{g \; O_2}}}{\mathrm{\cancel{mol \; O_2}}} = 2.46 × 10^3 \mathrm{\; g \; O_2}\)

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. The figure provides a general outline of the various computational steps associated with many reaction stoichiometry calculations.


The flowchart depicts the various computational steps involved in most reaction stoichiometry calculations. Image credit: OpenStax, Chemistry

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