Problem 98 - Chemistry: The Central Science

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

The composition of the original mixture was 5.11 g of KClO\(_3\), 20.02 g of KHCO\(_3\), 13.82 g of K\(_2\)CO\(_3\), and 61.05 g of KCl.

Step 1: Find the moles of products

Let's convert the mass of each product to moles: - For H2O: \[\text{moles of H}_{2}\text{O} = \frac{\text{mass of H}_{2}\text{O}}{\text{molar mass of H}_{2}\text{O}}=\frac{1.80\text{ g}}{18.02\text{ g/mol}} = 0.1\text{ mol}\] - For CO2: \[\text{moles of CO}_{2} = \frac{\text{mass of CO}_{2}}{\text{molar mass of CO}_{2}}=\frac{13.20\text{ g}}{44.01\text{ g/mol}}=0.3\text{ mol}\] - For O2: \[\text{moles of O}_{2} = \frac{\text{mass of O}_{2}}{\text{molar mass of O}_{2}}=\frac{4.00\text{ g}}{32.00\text{ g/mol}}=0.125\text{ mol}\]

Step 2: Calculate the moles of each reactant

Use the equations and the moles of products found in the previous step to determine the moles of each reactant: - For KClO3: \[2\text{ KClO}_{3} \longrightarrow 3\text{ O}_{2}\] \[\text{moles of KClO}_{3} = \frac{0.125 \text{ mol O}_{2}}{3} \times 2 = 0.0417\text{ mol}\] - For KHCO3: \[2\text{ KHCO}_{3} \longrightarrow \text{H}_{2}\text{O} + 2\text{ CO}_{2}\] \[\text{moles of KHCO}_{3} = \frac{0.1 \text{ mol H}_{2}\text{O}}{1} \times 2 = 0.2\text{ mol}\] - For K2CO3: \[\text{K}_{2}\text{CO}_{3} \longrightarrow \text{CO}_{2}\] \[\text{moles of K}_{2}\text{CO}_{3} = 0.3 \text{ mol CO}_{2} - 0.2\text{ mol KHCO}_{3} = 0.1\text{ mol}\]

Step 3: Calculate the masses of the reactants

Now that we have the moles of the reactants, we can convert each to mass and determine the composition of the original mixture: - For KClO3: \[\text{mass of KClO}_{3} = 0.0417\text{ mol} \times 122.55 \text{ g/mol}=5.11 \text{ g}\] - For KHCO3: \[\text{mass of KHCO}_{3} = 0.2\text{ mol} \times 100.12 \text{ g/mol} =20.02 \text{ g}\] - For K2CO3: \[\text{mass of K}_{2}\text{CO}_{3} = 0.1\text{ mol} \times 138.21\text{ g/mol} =13.82\text{ g}\] - For KCl: Since KCl does not react, its mass can be found as follows: \[\text{mass of KCl} = 100.0\text{ g} - 5.11 \text{ g} - 20.02 \text{ g}- 13.82 \text{ g} = 61.05\text{ g}\]

Step 4: Present the composition of the original mixture

The composition of the original mixture was: - 5.11 g of KClO3 - 20.02 g of KHCO3 - 13.82 g of K2CO3 - 61.05 g of KCl

Key Concepts

Chemical Reactions

Chemical reactions are the processes where substances, known as reactants, are transformed into different substances, called products. In these reactions, chemical bonds are broken and new bonds form, leading to the production of new molecules.
Each chemical reaction follows a specific balanced equation, which indicates the quantities of reactants and products. Balancing these equations is crucial because it ensures the conservation of mass; no atoms are lost or gained in a chemical reaction.

• The reaction of potassium chlorate (\(\mathrm{KClO}_3\)) decomposing into potassium chloride (\(\mathrm{KCl}\)) and oxygen gas (\(\mathrm{O}_2\)) can be represented by a balanced equation: \[2\mathrm{KClO}_3(s) \longrightarrow 2\mathrm{KCl}(s) + 3\mathrm{O}_2(g)\]
• Similarly, potassium bicarbonate (\(\mathrm{KHCO}_3\)) decomposes into various gases and potassium carbonate (\(\mathrm{K}_2\mathrm{CO}_3\)) produces carbon dioxide.

Understanding chemical reactions helps predict the amounts of products that can be formed, which is essential for stoichiometry calculations.

Mass Calculations

Mass calculations in stoichiometry involve determining the mass of reactants and products based on their moles and molar masses. By using the balanced chemical equation, we can find out how much of each substance is involved in the reaction.
To perform mass calculations, follow these steps:

• Determine the molar mass of each substance involved in the reaction. Molar mass is obtained from the periodic table by summing the atomic masses of each element in a compound.
• Convert the given masses of products into moles using their molar masses. For example, to find the moles of \(\mathrm{H}_2\mathrm{O}\): \[\text{moles of H}_2\text{O} = \frac{1.80\ \text{g}}{18.02\ \text{g/mol}} = 0.1\ \text{mol}\]
• Use the moles obtained to calculate the masses of the reactants based on the stoichiometry of the reaction.

This approach is highly effective for determining how much original substance was used or created in a chemical process.

Moles to Grams Conversion

Converting moles to grams is a key step in stoichiometry, essential for linking chemical equations to real-world quantities. This conversion is straightforward when you know the molar mass of the chemical involved. The formula is:
\[\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}\]
Here's how you use it:

• Identify the number of moles given in the problem. For instance, if you have 0.0417 moles of \(\mathrm{KClO}_3\), write down this value.
• Find the molar mass from the periodic table. The molar mass of \(\mathrm{KClO}_3\) is approximately 122.55 g/mol.
• Multiply the moles by the molar mass to get the mass: \[\text{mass of KClO}_3 = 0.0417 \times 122.55 = 5.11 \text{ g}\]

This method will help you find the mass for any substance involved, ensuring accurate representations of reactants and products across chemical reactions.