KEY2CHEM

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. $$7.5\text{ g}$$

Solution

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. $$0.272\text{ mol}$$

Solution

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. $$12.2\text{ g}$$

Solution

B. $$24.4\text{ g}$$

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