Problem 104 - Chemistry The Central Science In Si Units

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.)

Step-by-Step Solution

Answer

Calculated moles of gases are: 0.10 mol of \(\mathrm{H}_{2}\mathrm{O}\), 0.30 mol of \(\mathrm{CO}_2\), and 0.125 mol of \(\mathrm{O}_2\).

Step 1: Calculate moles of gases produced

First, determine the number of moles of each gas produced using the provided mass and their molar masses. The molar mass of water \(\mathrm{H}_{2}\mathrm{O}\) is approximately \(18.02\, \mathrm{g/mol}\). For \(\mathrm{CO}_2\), it is approximately \(44.01\, \mathrm{g/mol}\). For \(\mathrm{O}_2\), it is approximately \(32.00\, \mathrm{g/mol}\):- Moles of \(\mathrm{H}_{2}\mathrm{O}\): \(n_{\mathrm{H}_{2}\mathrm{O}} = \frac{1.80\, \mathrm{g}}{18.02\, \mathrm{g/mol}} = 0.10\, \mathrm{mol}\).- Moles of \(\mathrm{CO}_2\): \(n_{\mathrm{CO}_2} = \frac{13.20\, \mathrm{g}}{44.01\, \mathrm{g/mol}} = 0.30\, \mathrm{mol}\).- Moles of \(\mathrm{O}_2\): \(n_{\mathrm{O}_2} = \frac{4.00\, \mathrm{g}}{32.00\, \mathrm{g/mol}} = 0.125\, \mathrm{mol}\).

Key Concepts

Moles Calculation

Understanding how to calculate moles is crucial in stoichiometry, as it forms the basis for determining the relationships between reactants and products in a chemical reaction. Let's delve into how this works with our given problem.
We began with the mass of each gas produced from the mixture:

• 1.80 g of water (\(\mathrm{H}_2\mathrm{O}\))
• 13.20 g of carbon dioxide (\(\mathrm{CO}_2\))
• 4.00 g of oxygen (\(\mathrm{O}_2\))

Given these masses, we use their respective molar masses to convert each mass to moles. The molar mass is akin to the weight of one mole of a substance, making it easier to use in calculations. Here's a breakdown of each calculation:

• Molar mass of \(\mathrm{H}_2\mathrm{O}\): approximately 18.02 g/mol. \[n_{\mathrm{H}_2\mathrm{O}} = \frac{1.80 \, \text{g}}{18.02 \, \text{g/mol}} = 0.10 \, \text{mol}\]
• Molar mass of \(\mathrm{CO}_2\): approximately 44.01 g/mol. \[n_{\mathrm{CO}_2} = \frac{13.20 \, \text{g}}{44.01 \, \text{g/mol}} = 0.30 \, \text{mol}\]
• Molar mass of \(\mathrm{O}_2\): approximately 32.00 g/mol. \[n_{\mathrm{O}_2} = \frac{4.00 \, \text{g}}{32.00 \, \text{g/mol}} = 0.125 \, \text{mol}\]

This approach allows us to understand the quantities of gases generated, which is vital for assessing the initial composition of the mixture.

Gas Production

Chemical reactions involving the decomposition of compounds often involve the production of gases. In this problem, heating the initial mixture produces \(\mathrm{CO}_2\), \(\mathrm{O}_2\), and \(\mathrm{H}_2\mathrm{O}\) gases.
This is important because observing the amounts of gas produced can provide insight into the reaction progress and completion.

Three Reactions Involved

• \(2 \mathrm{KClO}_3 (s) \rightarrow 2 \mathrm{KCl} (s) + 3 \mathrm{O}_2 (g)\)
• \(2 \mathrm{KHCO}_3 (s) \rightarrow \mathrm{K}_2 \mathrm{O} (s) + \mathrm{H}_2 \mathrm{O} (g) + 2 \mathrm{CO}_2 (g)\)
• \(\mathrm{K}_2 \mathrm{CO}_3 (s) \rightarrow \mathrm{K}_2 \mathrm{O} (s) + \mathrm{CO}_2 (g)\)

Each equation results in the formation of a specific quantity of gas, which aligns with the law of conservation of mass and stoichiometry principles. By understanding the gases produced and their respective moles derived from the molar ratios in the balanced equations, we can determine the composition of the original mixture.
These gas-producing reactions are vital indicators of the occurrence and extent to which the reactions have occurred.

Compound Decomposition

Compound decomposition is a chemical process where a single compound breaks down into simpler compounds or elements. In our scenario, when the original mixture is heated, several decomposition reactions occur.

The Role of Decomposition

The compounds in the mixture:

• Potassium chlorate \( (\mathrm{KClO}_3)\)
• Potassium bicarbonate \((\mathrm{KHCO}_3)\)
• Potassium carbonate \((\mathrm{K}_2\mathrm{CO}_3)\)

undergo thermal decomposition. Each of these decomposes to form different products, including gases:

• Pottassium chlorate yields potassium chloride and oxygen gas.
• Pottassium bicarbonate yields potassium oxide, water vapor, and carbon dioxide gas.
• Pottassium carbonate yields potassium oxide and carbon dioxide gas.

Why Decomposition Matters

Decomposition offers essential insights into the behavior of substances under thermal conditions and helps predict product formation in a controlled environment. This understanding allows chemists to harness decomposition reactions in various applications, from industrial processes to laboratory experiments.By analyzing decomposition reactions, one can decipher the potential of the mixture components and simplify complex mixtures into manageable parts.