Question

On heating, potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\) decomposes to yield potassium chloride and oxygen gas. In one experiment, a student heated \(20.4 \mathrm{~g}\) of \(\mathrm{KClO}_{3}\) until the decomposition was complete. (a) Write a balanced equation for the reaction. (b) Calculate the volume of oxygen (in liters) if it was collected at 0.962 atm and \(18.3^{\circ} \mathrm{C}\)

Short answer

The balanced equation is \(2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2\). Volume of \(\text{O}_2\) is \(6.16\,\text{L}\).

Step-by-step solution

Write the Balanced Equation

The decomposition of potassium chlorate (\( \text{KClO}_3 \)) produces potassium chloride (\( \text{KCl} \)) and oxygen gas (\( \text{O}_2 \)). The balanced chemical equation is:\[ 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \] This equation shows that 2 moles of \( \text{KClO}_3 \) decompose to produce 2 moles of \( \text{KCl} \) and 3 moles of \( \text{O}_2 \).

Calculate Molar Mass of KClO3

The molar mass of \( \text{KClO}_3 \) is calculated as follows:- Potassium (K): 39.10 g/mol- Chlorine (Cl): 35.45 g/mol- Oxygen (O): 16.00 g/mol (3 atoms)Thus, the molar mass of \( \text{KClO}_3 \) is:\[ 39.10 + 35.45 + 3 \times 16.00 = 122.55 \text{ g/mol} \]

Calculate Moles of KClO3

Given 20.4 g of \( \text{KClO}_3 \), we calculate the moles:\[\text{Moles of } \text{KClO}_3 = \frac{20.4 \text{ g}}{122.55 \text{ g/mol}} \approx 0.166 \text{ moles} \]

Use Stoichiometry to Find Moles of O2

According to the balanced equation, 2 moles of \( \text{KClO}_3 \) produce 3 moles of \( \text{O}_2 \). Hence, 0.166 moles of \( \text{KClO}_3 \) will produce:\[\text{Moles of } \text{O}_2 = 0.166 \times \frac{3}{2} \approx 0.249 \text{ moles}\]

Use PV=nRT to Calculate Volume of O2

To find the volume of \( \text{O}_2 \), use the ideal gas law \( PV = nRT \). Since pressure \( P \) is in atm, volume \( V \) is in liters, temperature \( T \) is in Kelvin, and \( R = 0.0821 \text{ L atm/mol K} \), we first convert the temperature:\[T = 18.3 + 273.15 = 291.45 \text{ K}\]Using the values, \( n = 0.249 \), \( P = 0.962 \text{ atm} \):\[V = \frac{nRT}{P} = \frac{0.249\times 0.0821 \times 291.45}{0.962} \approx 6.155 \text{ L}\]

Conclusion: Report Volume of O2

Therefore, the volume of oxygen gas collected is approximately \( 6.16 \text{ L} \).

Key concepts

Balancing Chemical Equations

Balancing a chemical equation is crucial to ensure that the law of conservation of mass is obeyed. This means that the number of atoms of each element must be the same on both sides of the equation. In the decomposition of potassium chlorate (\( \text{KClO}_3 \)), for example, we need to ensure that the number of potassium, chlorine, and oxygen atoms in the reactants equals the number in the products. For \( \text{KClO}_3 \), the balanced equation is:\[ 2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2 \] This equation shows that 2 moles of \( \text{KClO}_3 \) decompose into 2 moles of potassium chloride \( \text{KCl} \) and 3 moles of oxygen gas \( \text{O}_2 \).

• Start by counting the number of each type of atom in the reactants and products.
• Adjust the coefficients (the numbers in front of compounds) to balance the atoms. For instance, you need 3 \( \text{O}_2 \) molecules to balance the 6 oxygen atoms in 2 \( \text{KClO}_3 \).

Balancing helps in calculating the quantities of reactants and products during a chemical reaction, which is essential for stoichiometric calculations.

Ideal Gas Law Calculations

The Ideal Gas Law is a useful equation to relate the pressure, volume, temperature, and number of moles of a gas. The formula is expressed as \( PV = nRT \), where:

• \( P \) stands for pressure in atmospheres (atm).
• \( V \) stands for volume in liters.
• \( n \) is the number of moles of gas.
• \( R \) is the ideal gas constant (0.0821 L atm/mol K).
• \( T \) is the temperature in Kelvin.

For example, in the decomposition exercise, we calculated the volume of oxygen gas produced using the Ideal Gas Law. First, the temperature in Celsius was converted to Kelvin: \[ T = 18.3 + 273.15 = 291.45 \text{ K} \] Next, we plugged the known values \( n = 0.249 \) moles (from stoichiometry), \( P = 0.962 \) atm, and \( R = 0.0821 \) L atm/mol K into the formula:\[ V = \frac{0.249 \times 0.0821 \times 291.45}{0.962} \approx 6.155 \text{ L} \]This produces the volume of oxygen gas at the given conditions, showing how crucial the Ideal Gas Law can be.

Molar Mass Calculation

Calculating the molar mass of a compound helps in determining the number of moles of that compound from a given mass, which is pivotal for stoichiometric calculations. To find the molar mass of \( \text{KClO}_3 \):

• Potassium (K): \( 39.10 \) g/mol
• Chlorine (Cl): \( 35.45 \) g/mol
• Oxygen (O): \( 16.00 \) g/mol for each of the 3 atoms

Adding these gives:\[ 39.10 + 35.45 + 3 \times 16.00 = 122.55 \text{ g/mol} \]Knowing this value allows the conversion from grams to moles, which is calculated as follows: \[\text{Moles of } \text{KClO}_3 = \frac{20.4 \text{ g}}{122.55 \text{ g/mol}} \approx 0.166 \text{ moles}\]Understanding molar mass is fundamental when engaging with reactions, as it facilitates the calculation of how much of each substance is involved in the chemical change.

Gas Volume Calculation

Gas volume calculations are critical in scenarios where gases are collected or produced in a chemical reaction. In the given exercise, after determining the moles of oxygen gas \(( \text{O}_2 )\) formed using stoichiometry, the volume was then calculated using the Ideal Gas Law.To calculate the volume of gas, consider the conditions given:

• Pressure \( P \): 0.962 atm
• Temperature in Celsius: 18.3
• Converted Temperature in Kelvin: \( 18.3 + 273.15 = 291.45 \text{ K} \)
• Moles of gas \( n \): \( 0.249 \text{ moles} \)

Substituting these values into the Ideal Gas Law equation \( PV = nRT \), we find:\[ V = \frac{0.249 \times 0.0821 \times 291.45}{0.962} \approx 6.155 \text{ L} \]The calculation provides the volume of \( \text{O}_2 \) gas produced under the specified conditions. Understanding how to perform these calculations is valuable when dealing with gases, as it helps predict how the gas will behave under different temperature and pressure conditions.