Question

A 100.0 -g mixture made up of \(\mathrm{NaClO}_{3}, \mathrm{Na}_{2} \mathrm{CO}_{3}, \mathrm{NaCl}\), and \(\mathrm{NaHCO}_{3}\) is heated, producing \(5.95 \mathrm{~g}\) of oxygen, \(1.67 \mathrm{~g}\) of water, and \(14.5 \mathrm{~g}\) of carbon dioxide. \(\mathrm{NaCl}\) does not react under the conditions of the experiment. The equations for the reactions that take place are: $$ \begin{array}{r} 2 \mathrm{NaClO}_{3}(s) \longrightarrow 2 \mathrm{NaCl}(s)+3 \mathrm{O}_{2}(g) \\\ \mathrm{Na}_{2} \mathrm{CO}_{3}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g) \\ 2 \mathrm{NaHCO}_{3}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{O}(s)+2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} \end{array} $$ Assuming \(100 \%\) decomposition of \(\mathrm{NaClO}_{3}, \mathrm{Na}_{2} \mathrm{CO}_{3},\) and \(\mathrm{NaHCO}_{3},\) what is the composition of the mixture in grams?

Short answer

Based on the given information and reaction equations, the composition of the original 100.0-g mixture of \(\mathrm{NaClO}_{3}, \mathrm{Na}_{2}\mathrm{CO}_{3}, \mathrm{NaCl}\), and \(\mathrm{NaHCO}_{3}\) can be determined using the mass of products produced. After calculating the mass of each reacting component, the mass of non-reacting \(\mathrm{NaCl}\) can be found. Finally, the masses obtained for all four components give the composition of the original mixture in grams.

Step-by-step solution

Determine mass of reactants from products produced

Using the mass of the products produced, we can determine the mass of the reactants that decomposed. From the balanced equations, we know: - From \(\mathrm{NaClO}_{3}\) decomposition, \(2 \mathrm{NaClO}_{3}(s) \longrightarrow 2 \mathrm{NaCl}(s)+3 \mathrm{O}_{2}(g)\). We are given that \(5.95\ \mathrm{g}\) of oxygen is produced. Use the equation to find the mass of \(\mathrm{NaClO}_{3}\) that decomposed: $$\frac{5.95\ \mathrm{g\ O}_{2}}{3\ \mathrm{mol\ O}_{2}} \times \frac{2\ \mathrm{mol\ NaClO}_{3}}{1\ \mathrm{mol\ O}_{2}} \times \frac{106.44\ \mathrm{g\ NaClO}_{3}}{1\ \mathrm{mol\ NaClO}_{3}}$$ - From \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) decomposition, \(\mathrm{Na}_{2}\mathrm{CO}_{3}(s) \longrightarrow \mathrm{Na}_{2}\mathrm{O}(s)+\mathrm{CO}_{2}(g)\). We are given that \(14.5\ \mathrm{g}\) of \(\mathrm{CO}_{2}\) is produced. Use the equation to find the mass of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) that decomposed: $$\frac{14.5\ \mathrm{g\ CO}_{2}}{1\ \mathrm{mol\ CO}_{2}} \times \frac{1\ \mathrm{mol\ Na}_{2}\mathrm{CO}_{3}}{1\ \mathrm{mol\ CO}_{2}} \times \frac{105.99\ \mathrm{g\ Na}_{2}\mathrm{CO}_{3}}{1\ \mathrm{mol\ Na}_{2}\mathrm{CO}_{3}}$$ - From \(\mathrm{NaHCO}_{3}\) decomposition, \(2\ \mathrm{NaHCO}_{3}(s) \longrightarrow \mathrm{Na}_{2}\mathrm{O}(s)+2\mathrm{CO}_{2}(g)+\mathrm{H}_{2}\mathrm{O}\). We are given that \(1.67\ \mathrm{g}\) of water is produced. Use the equation to find the mass of \(\mathrm{NaHCO}_{3}\) that decomposed: $$\frac{1.67\ \mathrm{g\ H}_{2}\mathrm{O}}{1\ \mathrm{mol\ H}_{2}\mathrm{O}} \times \frac{2\ \mathrm{mol\ NaHCO}_{3}}{1\ \mathrm{mol\ H}_{2}\mathrm{O}} \times \frac{84.01\ \mathrm{g\ NaHCO}_{3}}{1\ \mathrm{mol\ NaHCO}_{3}}$$

Calculate the mass of NaCl

We know that the total mass of all four components in the mixture is 100.0 g. Using the values calculated in step 1, find the mass of non-reacting \(\mathrm{NaCl}\): $$\mathrm{Mass \ of \ NaCl} = 100.0\ \mathrm{g} - (\mathrm{Mass \ of \ NaClO}_{3} + \mathrm{Mass \ of \ Na}_{2}\mathrm{CO}_{3} + \mathrm{Mass \ of \ NaHCO}_{3})$$

Calculate the composition of the original mixture

Using the mass values obtained in steps 1 and 2, we now have the composition of the original mixture of \(\mathrm{NaClO}_{3}, \mathrm{Na}_{2}\mathrm{CO}_{3}, \mathrm{NaCl},\) and \(\mathrm{NaHCO}_{3}\) in grams.