Problem 115 - Chemistry Central Science

Question

A 4.00-g sample of a mixture of \(\mathrm{CaO}\) and \(\mathrm{BaO}\) is placed in a 1.00-L vessel containing \(\mathrm{CO}_{2}\) gas at a pressure of 730 torr and a temperature of \(25^{\circ} \mathrm{C}\). The \(\mathrm{CO}_{2}\) reacts with the \(\mathrm{CaO}\) and \(\mathrm{BaO}\), forming \(\mathrm{CaCO}_{3}\) and \(\mathrm{BaCO}_{3} .\) When the reaction is complete, the pressure of the remaining \(\mathrm{CO}_{2}\) is 150 torr. (a) Calculate the number of moles of \(\mathrm{CO}_{2}\) that have reacted. (b) Calculate the mass percentage of \(\mathrm{CaO}\) in the mixture.

Step-by-Step Solution

Answer

The number of moles of CO₂ that have reacted is 0.03150 mol, and the mass percentage of CaO in the mixture is approximately 24.51%.

Step 1: Convert pressure of CO₂ to moles

We'll first calculate the initial moles of CO₂ using the ideal gas equation: PV = nRT. Rearrange it for n: n = PV / RT We have P = 730 torr (initial pressure), V = 1.00 L (volume of the vessel), T = 25°C (temperature), and R = 62.364 L⋅torr / (K⋅mol) (gas constant). We will convert temperature to Kelvin: T(K) = T(°C) + 273.15 T(K) = 25 + 273.15 T(K) = 298.15 K Now, calculate the initial moles of CO₂: n = PV / RT n = (730 * 1.00) / (62.364 * 298.15) n = 730 / (62.364 * 298.15) n = 0.03966 mol Similarly, calculate the final moles of CO₂ (after the reaction) using the final pressure of 150 torr: n = (150 * 1.00) / (62.364 * 298.15) n = 150 / (62.364 * 298.15) n = 0.00816 mol

Step 2: Calculate the number of moles of CO₂ that have reacted

To find the number of moles of CO₂ that have reacted, subtract the final moles of CO₂ from the initial moles of CO₂: Δn = n_initial - n_final Δn = 0.03966 - 0.00816 Δn = 0.03150 mol

Step 3: Calculate the moles of CaO and BaO

By stoichiometry, each mole of CO₂ reacts with one mole of either CaO or BaO. Thus, the moles of CaO + BaO = moles of CO₂ that have reacted: moles_combined_CaO_and_BaO = 0.03150 mol Let x be the moles of CaO and (0.03150 - x) be the moles of BaO. We know the total mass of the CaO and BaO mixture is 4.00 g. Using the molar mass of CaO (56.08 g/mol) and BaO (153.33 g/mol), we can write: mass_CaO + mass_BaO = 4.00 g 56.08 * x + 153.33 * (0.03150 - x) = 4.00

Step 4: Solve the equation to find the moles of CaO

We now have to solve the equation for x (moles of CaO): 56.08 * x + 153.33 * (0.03150 - x) = 4.00 After solving for x, we have: x = 0.01748 mol (moles of CaO)

Step 5: Calculate the mass percentage of CaO

Now that we have the moles of CaO, we can find the mass of CaO and calculate its mass percentage in the mixture. mass_CaO = moles_CaO * molar_mass_CaO mass_CaO = 0.01748 * 56.08 mass_CaO = 0.9804 g Finally, the mass percentage of CaO is: mass_percentage_CaO = (mass_CaO / total_mass_mixture) * 100 mass_percentage_CaO = (0.9804 / 4.00) * 100 mass_percentage_CaO ≈ 24.51% The mass percentage of CaO in the mixture is approximately 24.51%.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a key concept to grasp in gas stoichiometry. It is expressed by the equation \( PV = nRT \), where:

• \( P \) stands for pressure
• \( V \) is the volume
• \( n \) represents the number of moles
• \( R \) is the gas constant
• \( T \) is the temperature in Kelvin

Understanding how to apply this equation is crucial for determining the number of moles of a gas under certain conditions. In the provided exercise, we used the Ideal Gas Law to compute the initial and final moles of \( \mathrm{CO}_{2} \) gas. By converting the ambient temperature (from Celsius to Kelvin) and using the given pressure in torr, we accurately determined the moles present before and after the chemical reaction. This allows for further calculations, such as determining the amount of gas that reacted.

Molar Mass Calculation

Molar mass calculation is important when dealing with chemical reactions involving gaseous substances. The molar mass of a compound is the sum of the atomic masses of its constituent elements. For example, in the reaction given, we deal with \( \mathrm{CaO} \) and \( \mathrm{BaO} \). Knowing their molar masses (56.08 g/mol for \( \mathrm{CaO} \) and 153.33 g/mol for \( \mathrm{BaO} \)) is essential for further calculations. By multiplying the number of moles by the molar mass, the mass of each compound can be found. This forms the basis for other calculations, like determining how much of each compound was present after the reaction. Always ensure you have the correct molar masses to avoid errors in the calculations that follow.

Chemical Reactions

Chemical reactions are processes where substances are transformed into different substances. In this exercise, the chemical reaction occurs between \( \mathrm{CaO} \) and \( \mathrm{BaO} \) with \( \mathrm{CO}_{2} \), resulting in the formation of \( \mathrm{CaCO}_{3} \) and \( \mathrm{BaCO}_{3} \). Each mole of \( \mathrm{CO}_{2} \) reacts with one mole of either \( \mathrm{CaO} \) or \( \mathrm{BaO} \), which is a stoichiometric relationship. Understanding stoichiometry helps determine how many moles of reactants are needed and what will be produced. In the problem, we used the stoichiometric relationship to find out how much \( \mathrm{CO}_{2} \) had reacted and the totals for \( \mathrm{CaO} \) and \( \mathrm{BaO} \). This understanding is vital for solving many chemical equation problems.

Mass Percentage Calculation

Mass percentage calculation is used to determine the proportion of a specific component in a mixture. It is expressed as a percentage and is calculated by dividing the mass of the component by the total mass of the mixture and then multiplying by 100.In the exercise, we calculated the mass percentage of \( \mathrm{CaO} \) in a mixture of \( \mathrm{CaO} \) and \( \mathrm{BaO} \). We determined the mass of \( \mathrm{CaO} \) from the number of moles multiplied by its molar mass. Then, using the total mass of the mixture, we calculated its mass percentage. This step involves repeatedly verifying calculations to prevent inaccuracies. Understanding mass percentage is useful for analyzing mixtures and understanding their components.