Problem 121 - Chemistry The Central Science In Si Units

Question

A \(4.00-g\) sample of a mixture of \(C a O\) and \(B a O\) is placed in a 1.00-L vessel containing \(\mathrm{CO}_{2}\) gas at a pressure of \(97.33 \mathrm{kPa}\) and a temperature of \(25^{\circ} \mathrm{C}\). The \(\mathrm{CO}_{2}\) reacts with the 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 \(20.0 \mathrm{kPa}\). (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 moles of CO2 reacted were approximately 0.0312. The mass percentage of CaO in the mixture is approximately 11.37%.

Step 1: Determine Initial Moles of CO2

Use the ideal gas law to find the initial moles of \(\mathrm{CO}_2\). At a pressure of \(97.33 \text{ kPa}\), volume of \(1.00 \text{ L}\), and temperature \(25^{\circ} \text{C}\) (which is \(298 \text{ K}\)), the initial moles can be calculated using \(PV = nRT\). Assuming \(R = 8.314 \text{ J/mol}\cdot\text{K}\), convert the pressure from kPa to Pa: \(97.33 \times 1000 \text{ Pa}\). Solve for \(n\): \[ n = \frac{PV}{RT} = \frac{97,330 \times 1.00}{8.314 \times 298} = \frac{97,330}{2475.172} \approx 0.0393 \text{ moles}\]

Step 2: Determine Final Moles of CO2

Repeat the ideal gas law calculation for the final pressure of \(20.0 \text{ kPa}\). Convert \(20.0 \text{ kPa}\) to \(20,000 \text{ Pa}\). Using the same equation, \[ n_{\text{final}} = \frac{20,000 \times 1.00}{8.314 \times 298} = \frac{20,000}{2475.172} \approx 0.00808 \text{ moles}\]

Step 3: Calculate Reacted Moles of CO2

Subtract the final moles of \(\mathrm{CO}_2\) from the initial moles to find the moles that have reacted: \[ n_{\text{reacted}} = 0.0393 - 0.00808 \approx 0.0312 \text{ moles}\]

Step 4: Setup Equations for Masses

The reaction stoichiometry shows \(1 \text{ mole of CO}_2\) reacts with either \(1 \text{ mole of CaO}\) or \(1 \text{ mole of BaO}\). Suppose \(x\) is the moles of \(\text{CaO}\) and \(y\) is the moles of \(\text{BaO}\), then:\[x + y = 0.0312\]Also, calculate the mass of the sample assuming it is entirely \(\text{CaO}\) and \(\text{BaO}\): \[56.08x + 153.33y = 4.00\]

Step 5: Solve the System of Equations

You have two equations:\[1. x + y = 0.0312\]\[2. 56.08x + 153.33y = 4.00\]Solving these equations, substitute for \(y = 0.0312 - x\) in the second equation:\[56.08x + 153.33(0.0312 - x) = 4.00\]\[56.08x + 4.7878 - 153.33x = 4.00\]Simplifying, solve for \(x\):\[-97.25x = -0.7878\]\[x \approx 0.0081 \text{ moles of CaO}\]

Step 6: Calculate Mass and Percentage of CaO

Convert \(x = 0.0081 \text{ moles}\) of \(\text{CaO}\) to grams using its molar mass (56.08 g/mol):\[\text{mass of } \text{CaO} = 0.0081 \times 56.08 \approx 0.4549 \text{ g}\]The percentage of \(\text{CaO}\) in the mixture is:\[\frac{0.4549}{4.00} \times 100\% \approx 11.37\%\]

Key Concepts

Ideal Gas Law

The Ideal Gas Law is fundamental in understanding the behavior of gases. It relates the pressure, volume, temperature, and number of moles of a gas using the equation: \( PV = nRT \). Here, \( P \) is the pressure in pascals, \( V \) is the volume in liters, \( n \) is the number of moles, \( R \) is the universal gas constant (8.314 \( ext{J/mol} \cdot ext{K} \)), and \( T \) is the temperature in Kelvin.

To solve problems involving the Ideal Gas Law, remember to:

• Convert all units to their respective SI units - pressure in pascals, volume in liters, and temperature in Kelvin.
• Use the formula \( n = \frac{PV}{RT} \) to find the number of moles when pressure, volume, and temperature are known. This calculation is crucial in determining the amount of gas participating in a reaction.

For example, in the exercise, to find the initial moles of \( ext{CO}_2 \), you would use \( P = 97,330 \) Pa, \( V = 1.00 \) L, and \( T = 298 \) K. Inserting these into the equation yields the initial moles of \( ext{CO}_2 \) before the reaction begins.

Chemical Reactions

Chemical reactions describe the process where substances, known as reactants, transform into different substances, called products. In the exercise, carbon dioxide \( ext{CO}_2 \) reacts with calcium oxide \( ext{CaO} \) and barium oxide \( ext{BaO} \) to form calcium carbonate \( ext{CaCO}_3 \) and barium carbonate \( ext{BaCO}_3 \).

Key aspects to consider during a chemical reaction, particularly for stoichiometric calculations, include:

• The balanced chemical equation, which ensures the same number of each type of atom on both sides of the equation.
• Mole-to-mole conversions, which help in translating moles of one substance into moles of another using the coefficients from the balanced equation.
• Conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction, allowing you to compute total reactants' mass equals total products' mass.

In the exercise, the stoichiometry tells us that each mole of \( ext{CO}_2 \) can react with one mole of either \( ext{CaO} \) or \( ext{BaO} \), allowing for calculations of reacted moles to determine how much of each compound is present.

Mole Calculations

Mole calculations are a core part of stoichiometry, allowing for the conversion between mass, moles, and numbers of particles. Understanding and applying mole concepts helps in quantifying reactants and products in chemical reactions.

Here's a brief on important mole calculations:

• To convert from moles to grams, use the formula: \( ext{mass} = ext{moles} imes ext{molar mass} \).
• To find the percentage composition of a compound in a mixture, use: \( rac{ ext{mass of component}}{ ext{total mass of mixture}} imes 100 \% \).
• Use known quantities (like total mass or total moles) to deduce unknowns through system of equations, especially useful in mixtures.

In the given problem, after determining moles of \( ext{CaO} \), you convert it to grams using its molar mass 56.08 g/mol, thereby allowing the computation of the mass percentage of \( ext{CaO} \) in the mixture. This step helps link the chemical calculations to real-world quantities, enabling practical applications of theoretical chemistry knowledge.