Question

A chemist weighed out 5.14 g of a mixture containing unknown amounts of \(\mathrm{BaO}(s)\) and \(\mathrm{CaO}(s)\) and placed the sample in a \(1.50-\mathrm{L}\) flask containing \(\mathrm{CO}_{2}(g)\) at \(30.0^{\circ} \mathrm{C}\) and 750 . torr. After the reaction to form BaCO_ \(_{3}(s)\) and \(\mathrm{CaCO}_{3}(s)\) was completed, the pressure of \(\mathrm{CO}_{2}(g)\) remaining was 230 . torr. Calculate the mass percentages of \(\mathrm{CaO}(s)\) and \(\mathrm{BaO}(s)\) in the mixture.

Short answer

The mass percentages of \(\mathrm{CaO}(s)\) and \(\mathrm{BaO}(s)\) in the mixture are approximately 87.38% and 12.62%, respectively.

Step-by-step solution

Calculate Moles of CO2

Firstly, we need to find the moles of \(\mathrm{CO}_{2}(g)\) that have reacted. We can use the ideal gas law to do this, which states: \(PV = nRT\) where: P: pressure (atm) V: volume (L) n: moles of gas molecules R: gas constant = 0.0821 L atm/mol K T: temperature (K) First, we need to convert the pressure from torr to atm, and temperature from Celsius to Kelvin. \(P_{1} = (750\,\text{torr})\left(\frac{1\,\text{atm}}{760\,\text{torr}}\right) = 0.9868 \,\text{atm}\) \(T_{1} = 30.0 + 273.15 = 303.15\,\mathrm{K}\) Now, we can determine the moles of \(\mathrm{CO}_{2}(g)\) initially: \(n_{1} = \frac{P_{1}V}{RT_{1}} = \frac{(0.9868\,\text{atm})(1.50\,\text{L})}{(0.0821\,\text{L atm/mol K})(303.15\,\text{K})} = 0.0603\,\text{mol}\) We do the same for the final pressure: \(P_{2} = (230\,\text{torr})\left(\frac{1\,\text{atm}}{760\,\text{torr}}\right) = 0.3026\,\text{atm}\) And the moles of \(\mathrm{CO}_{2}(g)\) after the reaction: \(n_{2} = \frac{P_{2}V}{RT_{1}} = \frac{(0.3026\,\text{atm})(1.50\,\text{L})}{(0.0821\,\text{L atm/mol K})(303.15\,\text{K})} = 0.0187\,\text{mol}\) We can now find the moles of \(\mathrm{CO}_{2}(g)\) that have reacted: \(\Delta n_{\mathrm{CO}_2} = n_{1} - n_{2} = 0.0603\,\mathrm{mol} - 0.0187\,\mathrm{mol} = 0.0416\,\mathrm{mol}\)

Using Stoichiometry

Now that we have the moles of \(\mathrm{CO}_{2}(g)\) that reacted, we can set up stoichiometry equations to solve for the moles of \(\mathrm{BaO}(s)\) and \(\mathrm{CaO}(s)\). Let x = moles of \(\mathrm{BaO}(s)\) and y = moles of \(\mathrm{CaO}(s)\). The balanced chemical equations for the reactions are as follows: \(\mathrm{BaO(s)}+\mathrm{CO}_{2}(g) \rightarrow \mathrm{BaCO}_{3}(s)\) \(\mathrm{CaO(s)}+\mathrm{CO}_{2}(g) \rightarrow \mathrm{CaCO}_{3}(s)\) Both \(\mathrm{BaO}(s)\) and \(\mathrm{CaO}(s)\) react with \(\mathrm{CO}_{2}(g)\) in a 1:1 stoichiometric ratio. Therefore, we can create the equations: \(x + y = \Delta n_{\mathrm{CO}_2} = 0.0416\,\mathrm{mol}\) \[137.33x + 56.08y = 5.14\,\mathrm{g} \] Where 137.33 g/mol is the molar mass of \(\mathrm{BaO}(s)\) and 56.08 g/mol is the molar mass of \(\mathrm{CaO}(s)\).

Solving for moles of BaO and CaO

Now, we can solve for x and y using Substitution or Elimination Method. For this example, we will use the Substitution Method. From equation 1, we can isolate x: \(x = 0.0416 - y\) Now, we substitute this expression of x into equation 2: \(\begin{aligned} 137.33(0.0416 - y) + 56.08y = 5.14\,\mathrm{g} \end{aligned}\) Solve for y: \(y \approx 0.0369\,\text{mol}\) Now, substitute the value of y back into the expression for x: \(x \approx 0.0416 - 0.0369 \approx 0.0047\,\text{mol}\)

Finding Mass Percentages

Now that we have the moles of \(\mathrm{BaO}(s)\) and \(\mathrm{CaO}(s)\), we can determine the mass percentages in the mixture: \(\text{Mass Percent of BaO} = \frac{\text{mass of BaO}}{\text{total mass}} \times 100\) \(\text{Mass Percent of BaO} = \frac{(0.0047\,\text{mol})(137.33\,\text{g/mol})}{5.14\,\text{g}} \times 100 \approx 12.62\%\) \(\text{Mass Percent of CaO} = \frac{\text{mass of CaO}}{\text{total mass}} \times 100\) \(\text{Mass Percent of CaO} = \frac{(0.0369\,\text{mol})(56.08\,\text{g/mol})}{5.14\,\text{g}} \times 100 \approx 87.38\%\) So, the mass percentages of \(\mathrm{CaO}(s)\) and \(\mathrm{BaO}(s)\) in the mixture are approximately 87.38% and 12.62%, respectively.

Key concepts

Chemical Reactions

Chemical reactions are the processes in which substances, also known as reactants, transform into new substances, called products. In this exercise, we focus on the reactions involving barium oxide (\(\mathrm{BaO}\)) and calcium oxide (\(\mathrm{CaO}\)) reacting with carbon dioxide (\(\mathrm{CO}_2\)) to form their respective carbonates.

• For barium oxide, the reaction is: \(\mathrm{BaO(s)} + \mathrm{CO}_{2}(g) \rightarrow \mathrm{BaCO}_{3}(s)\)
• For calcium oxide, the reaction is: \(\mathrm{CaO(s)} + \mathrm{CO}_{2}(g) \rightarrow \mathrm{CaCO}_{3}(s)\)

These reactions highlight the stoichiometric relationship, where each mole of metal oxide reacts with one mole of carbon dioxide. Understanding the stoichiometry is crucial as it determines how much of each reactant is needed to completely react, as well as the amount of products formed. This 1:1 ratio is a basic concept in stoichiometry that dictates how chemical equations are balanced and predicts the outcome of reactions, which is instrumental when calculating quantities in a chemical reaction.

Ideal Gas Law

The ideal gas law is a fundamental principle used to relate the pressure, volume, temperature, and moles of gas in a system. It is expressed with the equation \(PV = nRT\), where:

• \(P\) is the pressure of the gas (in atm)
• \(V\) is the volume of the gas (in liters)
• \(n\) is the number of moles
• \(R\) is the universal gas constant, which equals 0.0821 L atm/mol K
• \(T\) is the temperature (in Kelvin)

To apply the ideal gas law, the conditions of the gas need to be converted into compatible units. For example, pressure often needs conversion from torr to atm, and temperature from Celsius to Kelvin, as seen in the exercise. By understanding the initial and final pressures and temperatures, the change in the moles of gas can be calculated. This is crucial in exercises involving gaseous reactions where we need to figure out how many moles of a reaction gas have been consumed or produced by applying the ideal gas law.

Mass Percentage Calculations

Mass percentage is a way to express the concentration of a component in a mixture. It is calculated using the formula:\[\text{Mass Percent} = \frac{\text{mass of component}}{\text{total mass of mixture}} \times 100\]In this exercise, knowing the mass percentage of \(\mathrm{BaO}\) and \(\mathrm{CaO}\) in the mixture is essential to understanding the composition of the original sample.

• After finding the moles of \(\mathrm{BaO}\) and \(\mathrm{CaO}\), convert these moles to grams using their respective molar masses: 137.33 g/mol for \(\mathrm{BaO}\) and 56.08 g/mol for \(\mathrm{CaO}\).
• The mass percentages are calculated by dividing the mass of each oxide by the total mass of the mixture and then multiplying by 100.

This provides insight into the proportion of each compound in the sample, aiding in understanding its chemical properties and potential reactions. Mass percentage is an important concept in chemistry, making it easier to communicate the concentration of elements in compounds.