Problem 104 - Chemistry: The Central Science in SI Units, Global Edition

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 \(100.0\mathrm{~g}\) mixture is \(10.21~g\, \mathrm{KClO}_{3}\), \(20.02~g\, \mathrm{KHCO}_{3}\), \(13.82~g\, \mathrm{K}_{2} \mathrm{CO}_{3}\), and \(55.95~g\, \mathrm{KCl}\).

Step 1: Identify the corresponding coefficients in the balanced equations

In the balanced chemical equations, the coefficients represent the stoichiometric ratio of each reactant and product. We will use these coefficients to relate the amount of reactants in the original mixture to the amount of products formed. For \(\mathrm{KClO}_{3}(s)\): 2 moles of \(\mathrm{KClO}_{3}\) produce 3 moles of \(\mathrm{O}_{2}\): $$2 \mathrm{KClO}_{3} \longrightarrow 3 \mathrm{O}_{2}$$ For \(\mathrm{KHCO}_{3}(s)\): 2 moles of \(\mathrm{KHCO}_{3}\) produce 1 mole of \(\mathrm{H}_{2} \mathrm{O}\) and 2 moles of \(\mathrm{CO}_{2}\): $$2 \mathrm{KHCO}_{3} \longrightarrow \mathrm{H}_{2} \mathrm{O} + 2 \mathrm{CO}_{2}$$ For \(\mathrm{K}_{2} \mathrm{CO}_{3}(s)\): 1 mole of \(\mathrm{K}_{2} \mathrm{CO}_{3}\) produces 1 mole of \(\mathrm{CO}_{2}\): $$\mathrm{K}_{2} \mathrm{CO}_{3} \longrightarrow \mathrm{CO}_{2}$$

Step 2: Obtain the moles of each gas produced in the given mixture

First, we need to find the moles of \(\mathrm{CO}_{2}, \mathrm{O}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\) produced from the given mass of each gas: - Moles of \(\mathrm{CO}_{2}\): $$\dfrac{13.20~g}{44.01\frac{g}{mol}} = 0.3~mol\,\, \mathrm{CO}_{2}$$ - Moles of \(\mathrm{O}_{2}\): $$\dfrac{4.00~g}{32.00\frac{g}{mol}} = 0.125~mol\,\, \mathrm{O}_{2}$$ - Moles of \(\mathrm{H}_2 \mathrm{O}\): $$\dfrac{1.80~g}{18.02\frac{g}{mol}} = 0.1~mol\,\, \mathrm{H}_2 \mathrm{O}$$

Step 3: Determine the amount of reacting compounds in the original mixture

Use the stoichiometric ratios of the balanced chemical equations to determine the moles of \(\mathrm{KClO}_{3}, \mathrm{KHCO}_{3}\), and \(\mathrm{K}_{2} \mathrm{CO}_{3}\) in the original \(100.0 \mathrm{~g}\) mixture: - From \(\mathrm{KClO}_{3}\): $$\dfrac{0.125~mol\,\, \mathrm{O}_{2}}{3/2} = 0.0833~mol\, \mathrm{KClO}_{3}$$ - From \(\mathrm{KHCO}_{3}\): $$\dfrac{0.1~mol \,\mathrm{H}_{2} \mathrm{O}}{1/2} = 0.2~mol\, \mathrm{KHCO}_{3}$$ - From \(\mathrm{K}_{2} \mathrm{CO}_{3}\): $$\dfrac{0.3~mol\,\, \mathrm{CO}_{2}}{2 + 1} = 0.1~mol\, \mathrm{K}_{2} \mathrm{CO}_{3}$$

Step 4: Calculate the mass of each compound in the original mixture

Now, determine the mass of each compound in the original mixture using their molar masses: - Mass of \(\mathrm{KClO}_{3}\): $$0.0833~mol \times 122.55\frac{g}{mol} = 10.21~g\, \mathrm{KClO}_{3}$$ - Mass of \(\mathrm{KHCO}_{3}\): $$0.2~mol \times 100.12\frac{g}{mol} = 20.02~g\, \mathrm{KHCO}_{3}$$ - Mass of \(\mathrm{K}_{2} \mathrm{CO}_{3}\): $$0.1~mol \times 138.21\frac{g}{mol} = 13.82~g\, \mathrm{K}_{2} \mathrm{CO}_{3}$$

Step 5: Calculate the mass of \(\mathrm{KCl}\) in the original mixture

Subtract the total mass of the other compounds from \(100.0 \mathrm{~g}\) to find the quantity of \(\mathrm{KCl}\): Mass of \(\mathrm{KCl}\): $$100.0~g - 10.21~g - 20.02~g - 13.82~g = 55.95~g\, \mathrm{KCl}$$

Step 6: Write the final composition of the original mixture

The composition of the original \(100.0 \mathrm{~g}\) mixture is: - \(10.21~g\, \mathrm{KClO}_{3}\) - \(20.02~g\, \mathrm{KHCO}_{3}\) - \(13.82~g\, \mathrm{K}_{2} \mathrm{CO}_{3}\) - \(55.95~g\, \mathrm{KCl}\)

Key Concepts

Stoichiometry

Stoichiometry is the part of chemistry concerned with the relative quantities of reactants and products in chemical reactions. It allows us to predict how much of each product can be formed from a given amount of reactants. To do this, we use the balanced chemical equation, which provides the ratios of reactants to products in a reaction.

In a balanced equation, like those in the exercise above, each chemical species is represented in a specific ratio represented by coefficients in front of each molecule. For instance, in the decomposition of potassium chlorate (\(\mathrm{KClO}_3\)), the equation shows that 2 moles of \(\mathrm{KClO}_3\) produce 3 moles of \(\mathrm{O}_2\). This ratio is crucial in stoichiometry as it helps determine the exact amount of reactants needed to produce a specific amount of product.

Here’s how you can use this ratio:

• Identify the balanced equation.
• Use the coefficients to set up a proportion between the reactants and products.
• Assume complete reaction of the compound, meaning no leftovers unless specified.

Gas Production in Chemical Reactions

In decomposition reactions, gases are often produced as products. These gases can be measured to help understand the reaction better, especially in a lab setting. When a compound like \(\mathrm{KClO}_3\) decomposes, it forms \(\mathrm{O}_2\) as one of the gaseous products. Similarly, \(\mathrm{KHCO}_3\) decomposes to produce both \(\mathrm{H}_2\mathrm{O}\) and \(\mathrm{CO}_2\) gases.

Understanding the production of gases:

• The amounts of gases produced can tell us about the composition of the original reactants.
• By measuring the mass of the gases, we can backtrack to find the reacting quantities of the compounds involved.
• It’s crucial to have a thorough understanding of the balanced equations to interpret the data correctly.

These gas measurements can then be converted into moles, using their respective molar masses, to further aid in calculations.

Molar Calculations

Molar calculations are essential in chemistry for converting between mass, moles, and molecules. The mole is a fundamental unit that represents a specific number of particles, usually atoms or molecules, which is Avogadro's number. It allows chemists to convert mass into moles to understand and calculate the amount of substances involved in a reaction more easily.

Here’s how you can perform molar calculations in chemical reactions:

• First, find the molar mass of the compound, which is the weight of one mole of that compound in grams.
• To find moles from mass, use the formula: \(\text{moles} = \frac{\text{mass of substance}}{\text{molar mass}}\).
• In the given problem, you can calculate the moles of \(\mathrm{CO}_2\), \(\mathrm{O}_2\), and \(\mathrm{H}_2\mathrm{O}\) using their respective molar masses.

By converting the masses of gases produced into moles, you can utilize stoichiometric coefficients effectively to deduce the quantity of initial reactants required in the reaction.