Question

A mixture containing \(\mathrm{KClO}_{3}, \mathrm{~K}_{2} \mathrm{CO}_{3}, \mathrm{KHCO}_{3},\) and \(\mathrm{KCl}\) was heated, producing \(\mathrm{CO}_{2}, \mathrm{O}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\) gases according to the following equations: $$ \begin{aligned} 2 \mathrm{KClO}_{3}(s) & \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \\\ 2 \mathrm{KHCO}_{3}(s) & \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g) \\ \mathrm{K}_{2} \mathrm{CO}_{3}(s) & \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g) \end{aligned} $$ The KCl does not react under the conditions of the reaction. If \(100.0 \mathrm{~g}\) of the mixture produces \(1.80 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}, 13.20 \mathrm{~g}\) of \(\mathrm{CO}_{2}\), and \(4.00 \mathrm{~g}\) of \(\mathrm{O}_{2}\), what was the composition of the original mixture? (Assume complete decomposition of the mixture.)

Short answer

The composition of the original 100.0 g mixture is: - \(10.2\,\text{g}\) of \({\mathrm{KClO}_3}\) - \(13.8\,\text{g}\) of \({\mathrm{K}_{2}\mathrm{CO}_{3}}\) - \(20.0\,\text{g}\) of \({\mathrm{KHCO}_3}\) - \(56.0\,\text{g}\) of \({\mathrm{KCl}}\)

Step-by-step solution

Find Moles of Produced Gases

From the masses of the produced gases, we will find their respective moles: - For \(\mathrm{H}_{2}\mathrm{O}\): Molar mass of \(\mathrm{H}_{2}\mathrm{O} = 2 \times 1.01 + 16.00 = 18.02\,\text{g/mol}\) Moles of \(\mathrm{H}_{2}\mathrm{O} = \frac{1.80\,\text{g}}{18.02\,\text{g/mol}} = 0.10\,\text{mol}\) - For \(\mathrm{CO}_{2}\): Molar mass of \(\mathrm{CO}_{2} = 12.01 + 2 \times 16.00 = 44.01\,\text{g/mol}\) Moles of \(\mathrm{CO}_{2} = \frac{13.20\,\text{g}}{44.01\,\text{g/mol}} = 0.30\,\text{mol}\) - For \(\mathrm{O}_{2}\): Molar mass of \(\mathrm{O}_{2} = 2 \times 16.00 = 32.00\,\text{g/mol}\) Moles of \(\mathrm{O}_{2} = \frac{4.00\,\text{g}}{32.00\,\text{g/mol}} = 0.125\,\text{mol}\)

Find Moles of Original Components

Now, we will use stoichiometry to find the moles of original components based on the moles of the produced gases. a) From the first equation, the \({\mathrm{KClO}_3}\) mole-to-\({\mathrm{O}_2}\) mole ratio is 2:3. Thus, moles of \({\mathrm{KClO}_3}\) = \(\frac{2}{3} \times\)moles of \({\mathrm{O}_2} = \frac{2}{3} \times 0.125 = 0.0833\,\text{mol}\) b) From the second equation, the \({\mathrm{KHCO}_3}\) mole-to-\({\mathrm{H}_{2}\mathrm{O}}\) mole ratio is 2:1. Thus, moles of \({\mathrm{KHCO}_3} = 2 \times\)moles of \({\mathrm{H}_{2}\mathrm{O}} = 2 \times 0.10 = 0.20\,\text{mol}\) c) From the second equation, the \({\mathrm{KHCO}_3}\) mole-to-\({\mathrm{CO}_{2}}\) mole ratio is also 2:2. This means that 0.20 moles of \({\mathrm{CO}_{2}}\) are produced by the \({\mathrm{KHCO}_3}\). d) From the third equation, the \({\mathrm{K}_{2}\mathrm{CO}_{3}}\) mole-to-\({\mathrm{CO}_{2}}\) mole ratio is 1:1. Considering that a total of 0.30 moles of \({\mathrm{CO}_{2}}\) were produced, 0.10 moles of \({\mathrm{K}_{2}\mathrm{CO}_{3}}\) are left.

Calculate the Masses of the Original Components

Now that we have the moles of the original components, we can calculate their masses. - For \({\mathrm{KClO}_3}\): Molar mass of \({\mathrm{KClO}_3} = 39.10 + 35.45 + 3 \times 16 = 122.55\,\text{g/mol}\) Mass of \({\mathrm{KClO}_3} = 0.0833\,\text{mol} \times 122.55\,\text{g/mol} = 10.2\,\text{g}\) - For \({\mathrm{K}_{2}\mathrm{CO}_{3}}\): Molar mass of \({\mathrm{K}_{2}\mathrm{CO}_{3}} = 2 \times 39.10 + 12.01 + 3 \times 16 = 138.21\,\text{g/mol}\) Mass of \({\mathrm{K}_{2}\mathrm{CO}_{3}} = 0.10\,\text{mol} \times 138.21\,\text{g/mol} = 13.8\,\text{g}\) - For \({\mathrm{KHCO}_3}\): Molar mass of \({\mathrm{KHCO}_3} = 39.10 + 1.01 + 12.01 + 3 \times 16 = 100.12\,\text{g/mol}\) Mass of \({\mathrm{KHCO}_3} = 0.20\,\text{mol} \times 100.12\,\text{g/mol} = 20.0\,\text{g}\) Since the KCl does not react and the total mass of the components is 100.0 g, Mass of \({\mathrm{KCl}}\) = \(100.0 - 10.2 - 13.8 - 20.0 = 56.0\,\text{g}\) Therefore, the composition of the original mixture is as follows: - \(10.2\,\text{g}\) of \({\mathrm{KClO}_3}\) - \(13.8\,\text{g}\) of \({\mathrm{K}_{2}\mathrm{CO}_{3}}\) - \(20.0\,\text{g}\) of \({\mathrm{KHCO}_3}\) - \(56.0\,\text{g}\) of \({\mathrm{KCl}}\)

Key concepts

Understanding Chemical Reaction Equations

Chemical reaction equations are the backbone of stoichiometry and give a concise representation of the reactants transforming into products during a chemical reaction. They show the substances involved and their quantitative relationships. In the exercise provided, we see a mixture of potassium compounds decomposing to form gases such as carbon dioxide (CO2), oxygen (O2), and water vapor (H2O). The written equations translate this process, indicating the decomposition of solid potassium chlorate (KClO3) into potassium chloride (KCl) and oxygen gas, potassium bicarbonate (KHCO3) breaking down into potassium oxide (K2O), water vapor, and carbon dioxide, and potassium carbonate (K2CO3) yielding potassium oxide and carbon dioxide.

Encoded within these equations are the stoichiometric coefficients, which tell us exactly how many molecules of each substance participate in the reaction. These coefficients are key to stoichiometric calculations, as they help us determine the precise amounts of reactants needed to produce a given amount of products. For example, the coefficient '2' in front of KClO3 and O2 means that two moles of potassium chlorate produce three moles of oxygen gas. Understanding and balancing these chemical equations are essential skills for anyone dealing with chemical reactions.

Molar Mass Calculation

Calculating molar mass is crucial in converting between the mass of a substance and the amount of substance in moles. It is the weight of one mole of a chemical compound or element, typically expressed in grams per mole (g/mol). The molar mass of a compound is calculated by adding up the atomic masses of all atoms present in the formula.

In our exercise, you can see how the molar mass is determined for each gas produced. The molar mass for water (H2O) is found by adding together the atomic mass of hydrogen (about 1.01 g/mol) and oxygen (about 16.00 g/mol), summing up to a molar mass of 18.02 g/mol. This step is repeated for each gas, and these values allow us to convert the mass of each substance produced (which is given in grams) into moles. These conversion are vital because they provide a common measurement (moles) that enables us to use the stoichiometric ratios from the chemical equations to calculate the number of moles of the original compounds present in the mixture.

Gas Evolution from Reactions

Gas evolution is an important phenomenon to understand in chemical reactions as it often indicates the progression or completion of a reaction. In the exercise, we observe the production of CO2, O2, and H2O gases as the potassium compounds decompose upon heating. Gas evolution can be detected through bubbling, changes in mass, or even energy changes (heat, light, etc.).

To ascertain the composition of the original mixture, the exercise requires an understanding of the quantitative relationship between the solid reactants and the gases produced. By tracking the amount of gases evolved (in grams) and calculating their moles, we derive the proportion of reactants that have undergone decomposition. When a reaction releases a gas, the stoichiometry - which in this case uses the mole ratios derived from the balanced chemical equations - allows us to backtrack and determine the moles of the original substances. This process showcases the practical application of stoichiometry in solving real-world problems involving gas evolution in chemical reactions.