Stoichiometry of Synthesis and Decomposition Reactions

Quantitative calculations can be performed for synthesis and decomposition reactions to determine the quantities of reactants or products in a specific process.

Example 1.

Silver metal can be obtained by heating silver oxide (\(\require{mhchem}\ce{Ag2O}\)) strongly in a decomposition reaction. If \(5.4\text{ g } \require{mhchem}\ce{Ag2O}\) are heated and the reaction goes to completion, how many grams of \(\require{mhchem}\ce{Ag}\) can be obtained?

                  \(\require{mhchem}\ce{2 Ag2O(s) -> 4 Ag(s) + O2(g)}\)


A. \(5.0\text{ g}\)

B. \(2.5\text{ g}\)

C. \(7.5\text{ g}\)









A. \(5.0\text{ g}\)

Based on the law of conservation of matter, all silver in \(\require{mhchem}\ce{Ag2O}\) must be converted to \(\require{mhchem}\ce{Ag(s)}\).

\(5.4\text{ g }\require{mhchem}\ce{Ag2O} \times \frac{1\text{ mol }\ce{Ag2O} }{231.74\text{ g }\ce{Ag2O} } \times \frac{4\text{ mol Ag}}{2 \text{ mol }\ce{Ag2O} } \times \frac{107.87\text{ g Ag}}{1\text{ mol Ag}} = 5.0\text{ g Ag} \)


Example 2.

If \(50.0\text{ g}\) of \(\require{mhchem}\ce{KClO3}\) are heated in the following decomposition reaction, how many moles of \(\require{mhchem}\ce{O2}\) can be obtained? 

\(\require{mhchem}\ce{2 KClO3(s) -> 2 KCl(s) + 3 O2(g)}\)


A. \(0.612\text{ mol}\)

B. \(0.408\text{ mol}\)   

C. \(0.272\text{ mol}\)






A. \(0.612\text{ mol}\)

\(50.0\text{ g }\require{mhchem}\ce{KClO3} \times \frac{1\text{ mol }\ce{KClO3} }{122.55\text{ g }\ce{KClO3} } \times \frac{3\text{ mol O}_2}{2 \text{ mol }\ce{KClO3} } = 0.612\text{ mol O}_2\)

Example 3.

If \(12.9\text{ g Al}\) react with excess \(\require{mhchem}\ce{O2(g)}\) by the following synthesis reaction, how many grams of \(\require{mhchem}\ce{Al2O3(s)}\) can be obtained?

 \(\require{mhchem}\ce{4 Al(s) + 3 O2(g)-> 2 AlO3(s) }\)



A. \(48.8\text{ g}\)

B. \(24.4\text{ g}\)

C. \(12.2\text{ g}\)





B. \(24.4\text{ g}\)

\(12.9\text{ g }\require{mhchem}\ce{Al} \times \frac{1\text{ mol }\ce{Al} }{26.98\text{ g }\ce{Al} } \times \frac{2\text{ mol }\ce{Al2O3}}{4 \text{ mol }\ce{Al} } \times \frac{101.96\text{ g }\ce{Al2O3}}{1\text{ mol }\ce{Al2O3}} = 24.4\text{ g }\ce{Al2O3}\)