Question

5.117 Liquid oxygen was first prepared by heating potassium chlorate, \(\mathrm{KClO}_{3},\) in a closed vessel to obtain oxygen at high pressure. The oxygen was cooled until it liquefied. $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) $$ If \(171 \mathrm{~g}\) of potassium chlorate reacts in a \(2.70-\mathrm{L}\) vessel, which was initially evacuated, what pressure of oxygen will be attained when the temperature is finally cooled to \(25^{\circ} \mathrm{C} ?\) Use the preceding chemical equation and ignore the volume of solid product.

Short answer

The pressure of oxygen at 25°C will be approximately 18.9 atm.

Step-by-step solution

Calculate Moles of KClO3

First, determine the molar mass of potassium chlorate, \(\mathrm{KClO_3}\), which is calculated as the sum of the atomic masses: \(\text{K} = 39.10\,\text{g/mol}, \text{Cl} = 35.45\,\text{g/mol}, \text{O} = 16.00\,\text{g/mol}\). Thus, \(\text{Molar Mass of } \mathrm{KClO_3} = 39.10 + 35.45 + (3 \times 16.00) = 122.55\,\text{g/mol}\). Now, divide the mass of \(\mathrm{KClO_3}\) by its molar mass to get moles: \(\text{Moles of } \mathrm{KClO_3} = \frac{171}{122.55} \approx 1.395\,\text{mol}\).

Determine Moles of Produced O2

Using stoichiometry, the balanced equation \(2 \mathrm{KClO_3} \rightarrow 2 \mathrm{KCl} + 3 \mathrm{O_2}\) shows that 2 moles of \(\mathrm{KClO_3}\) produce 3 moles of \(\mathrm{O_2}\). Thus, 1 mole of \(\mathrm{KClO_3}\) produces \(\frac{3}{2}\) moles of \(\mathrm{O_2}\). Calculate the moles of \(\mathrm{O_2}\) produced: \(\text{Moles of } \mathrm{O_2} = \frac{3}{2} \times 1.395 \approx 2.0925\,\text{mol}\).

Calculate Pressure using Ideal Gas Law

Use the ideal gas law \(PV = nRT\) to find the pressure of \(\mathrm{O_2}\). Here, \(n = 2.0925\,\text{mol}\), \(R = 0.0821\,\text{L atm/mol K}\), and \(T = 25^{\circ}C = 298\,\text{K}\). The volume \(V\) is given as \(2.70\,\text{L}\). Substitute these into the equation: \(P \cdot 2.70 = 2.0925 \cdot 0.0821 \cdot 298\). Solve for \(P\): \(P = \frac{2.0925 \cdot 0.0821 \cdot 298}{2.70}\ \approx 18.9\,\text{atm}\).

Key concepts

Stoichiometry

When dealing with chemical equations, stoichiometry helps us understand the quantitative relationships between the reactants and the products. It's crucial because it allows us to anticipate how much product we can get from a specific amount of reactant. In our case with the decomposition of potassium chlorate (\( \text{KClO}_3 \)), stoichiometry tells us the ratio of \( \text{KClO}_3 \) to oxygen gas \( \text{O}_2 \) produced.

Based on the balanced equation:

• 2 moles of \( \text{KClO}_3 \) yields 3 moles of \( \text{O}_2 \).

This ratio means that for every 2 moles of \( \text{KClO}_3 \) decomposed, we are able to produce 3 moles of \( \text{O}_2 \). When calculating the actual amount of \( \text{O}_2 \), we started by finding moles of \( \text{KClO}_3 \). For the given 171 grams, we calculated approximately 1.395 moles of \( \text{KClO}_3 \), which then allows us to find that 2.0925 moles of \( \text{O}_2 \) are produced using the stoichiometric ratio.

Pressure Calculation

Finding the pressure of a gas through the Ideal Gas Law is a vital application in chemistry, particularly after a chemical reaction has produced a gas and you're interested in its properties, like pressure.

The Ideal Gas Law equation is: \[ PV = nRT \]

• \( P \) is pressure,
• \( V \) is volume,
• \( n \) is the number of moles,
• \( R \) is the universal gas constant (0.0821 \( \text{L} \cdot \text{atm/mol} \cdot \text{K} \)),
• and \( T \) is temperature in Kelvin.

In our exercise, the number of moles (\( n \)) is 2.0925, the temperature is given as 25°C, which is 298 Kelvin, and the volume of the vessel is 2.70 L. By substituting these values into the Ideal Gas Law, we solve for \( P \), leading us to the final pressure of approximately 18.9 atm. Understanding how to manipulate this formula helps predict how gases behave under different conditions following a reaction.

Chemical Reactions

Understanding chemical reactions is a fundamental part of chemistry. These reactions involve reorganizing atoms to transform reactants into products. In the reaction presented, potassium chlorate \( \text{KClO}_3 \) decomposes when heated, producing potassium chloride \( \text{KCl} \) and oxygen gas \( \text{O}_2 \).

Here’s a closer look at what's happening:

• The reactant, \( \text{KClO}_3 \), breaks down into \( \text{KCl} \) and \( \text{O}_2 \).
• We observe that the products formed have different physical states; \( \text{KCl} \) remains as a solid, whereas \( \text{O}_2 \) is released as a gas.

This type of reaction, known as a decomposition reaction, involves a single compound splitting into two or more simpler substances. Understanding this is crucial for predicting the outcomes of reactions and for further calculations such as determining pressures, the volume of gases, etc. By mastering these concepts, students gain valuable insights into how substances interact and change during chemical processes.