Problem 104 - 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 \(\mathrm{KCl}\) 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 10.21% KClO3, 27.64% K2CO3, 10.01% KHCO3, and 52.14% KCl.

Step 1: Calculate moles of CO2, H2O, and O2 formed

We are given the mass of each product formed (CO2, H2O, and O2). We will use the molar mass to calculate the moles of each product formed. Molar mass of CO2 = \(1 \times 12.01 + 2 \times 16.00 = 44.01\) g/mol Molar mass of H2O = \(2 \times 1.01 + 1 \times 16.00 = 18.02\) g/mol Molar mass of O2 = \(2 \times 16.00 = 32.00\) g/mol Moles of CO2 formed = 13.20 g / 44.01 g/mol = 0.30 mol Moles of H2O formed = 1.80 g / 18.02 g/mol = 0.10 mol Moles of O2 formed = 4.00 g / 32.00 g/mol = 0.125 mol

Step 2: Determine moles of each reactant

Now, we will use stoichiometry to find the moles of each reactant present in the mixture. We can use the balanced equations given to determine the moles of reactants related to the moles of products formed. From Equation 1: \(2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) Moles of KClO3 = Moles of O2 formed / (3/2) = 0.125 / (3/2) = 0.0833 mol From Equation 2: \(2 \mathrm{KHCO}_{3}(s) \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g)\) Moles of KHCO3 = Moles of H2O formed = 0.10 mol Moles of CO2 formed from the reaction = 0.10 mol (since 0.05 mol KHCO3 forms the reaction products) From Equation 3: \(\mathrm{K}_{2} \mathrm{CO}_{3}(s) \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g)\) Moles of K2CO3 = Moles of CO2 formed - Moles of CO2 formed from the reaction = 0.30 - 0.10 = 0.20 mol

Step 3: Calculate the mass of each reactant

Next, we will calculate the mass of each reactant using their molar masses. Molar mass of KClO3 = \(1 \times 39.10 + 1 \times 35.45 + 3 \times 16.00 = 122.55\) g/mol Molar mass of K2CO3 = \(2 \times 39.10 + 1 \times 12.01 + 3 \times 16.00 = 138.21\) g/mol Molar mass of KHCO3 = \(1 \times 39.10 + 1 \times 1.01 + 1 \times 12.01 + 3 \times 16.00 = 100.12\) g/mol Mass of KClO3 = Moles of KClO3 × Molar mass of KClO3 = 0.0833 mol × 122.55 g/mol = 10.21 g Mass of K2CO3 = Moles of K2CO3 × Molar mass of K2CO3 = 0.20 mol × 138.21 g/mol = 27.64 g Mass of KHCO3 = Moles of KHCO3 × Molar mass of KHCO3 = 0.10 mol × 100.12 g/mol = 10.01 g Total mass of reactants (excluding KCl) = 10.21 g + 27.64 g + 10.01 g = 47.86 g

Step 4: Calculate the percentage composition of the original mixture

We can now calculate the percentage composition of the original mixture using the mass of each reactant and the total mass of the mixture (100.0 g). Percentage of KClO3 = (mass of KClO3 / total mass of the mixture) × 100 = (10.21 g / 100.0 g) × 100 = 10.21% Percentage of K2CO3 = (mass of K2CO3 / total mass of the mixture) × 100 = (27.64 g / 100.0 g) × 100 = 27.64% Percentage of KHCO3 = (mass of KHCO3 / total mass of the mixture) × 100 = (10.01 g / 100.0 g) × 100 = 10.01% Percentage of KCl = 100 - (10.21% + 27.64% + 10.01%) = 100 - 47.86 = 52.14% The composition of the original mixture was: - 10.21% KClO3 - 27.64% K2CO3 - 10.01% KHCO3 - 52.14% KCl

Key Concepts

chemical reactions

In the realm of chemistry, understanding chemical reactions is crucial. These reactions involve the transformation of reactants into products. Each reaction is defined by a balanced chemical equation that illustrates the conservation of mass and energy. In the given exercise, we see three distinct reactions occurring when a mixture is heated. For instance:

• Potassium chlorate ( \(\mathrm{KClO}_3\)) decomposes into potassium chloride ( \(\mathrm{KCl}\)) and oxygen gas ( \(\mathrm{O}_2\)).
• Potassium bicarbonate ( \(\mathrm{KHCO}_3\)) decomposes to form potassium oxide ( \(\mathrm{K}_2\mathrm{O}\)), oxygen gas, and carbon dioxide ( \(\mathrm{CO}_2\)).
• Potassium carbonate ( \(\mathrm{K}_2\mathrm{CO}_3\)) decomposes to yield carbon dioxide and potassium oxide.

These equations are illustrated using balanced equations, signifying that the number of atoms for each element is maintained on both sides of the equation. This principle is essential when solving problems involving chemical reactions, as it allows us to calculate the amounts of reactants consumed and products formed under given conditions.

mole calculations

Mole calculations are an integral part of solving chemical reaction problems. We use the concept of moles to bridge the mass of substances with the quantities involved in a chemical reaction. This part of the exercise involves using the molar masses of gases formed after the decomposition of the mixture to determine the moles of these gases.Starting with known masses, such as 13.2 g of \(\mathrm{CO}_2\), 1.8 g of \(\mathrm{H}_2\mathrm{O}\), and 4.0 g of \(\mathrm{O}_2\), the first step is to calculate the respective molar amounts:

• For carbon dioxide ( \(\mathrm{CO}_2\)), we use its molar mass, 44.01 g/mol, to find there are 0.30 moles.
• The water ( \(\mathrm{H}_2\mathrm{O}\)), with a molar mass of 18.02 g/mol, gives us 0.10 moles.
• Lastly, oxygen ( \(\mathrm{O}_2\)), has a molar mass of 32.00 g/mol, translating to 0.125 moles.

Using these mole values, one can refer back to the stoichiometries of the reactions to connect the number of moles of reactants that result in these particular amounts of products. This is where the balanced equations come in handy, for example, knowing that the formation of 3 moles of \(\mathrm{O}_2\) requires 2 moles of \(\mathrm{KClO}_3\), allows us to determine the moles of each original substance present.

mass percentage

Mass percentage is a way of expressing concentrations and is used in this exercise to find the composition of the original mixture. Once we have calculated the mass of each compound decomposed in the reaction, mass percentages allow us to understand the distribution of substances in the mixture.After finding the mass of each reactant:

• Potassium chlorate ( \(\mathrm{KClO}_3\)) was calculated to be 10.21 g.
• Potassium carbonate ( \(\mathrm{K}_2\mathrm{CO}_3\)) amounted to 27.64 g.
• Potassium bicarbonate ( \(\mathrm{KHCO}_3\)) was determined to be 10.01 g.

The mass percentage of each is computed using the formula: \(\text{Mass percentage} = (\text{mass of component}/\text{total mass}) \times 100\). Thus, each component contributes a certain percentage to the mixture, with the largest percentage, 52.14%, being from potassium chloride (\(\mathrm{KCl}\)), which doesn't decompose and acts as an inert part of the mixture. Understanding mass percentage helps in assessing how much of each component is present initially, which can be critical in industrial and research settings, ensuring reactants are optimally used and understanding the efficiency of reactions.