Question

A chemist weighed out \(5.14 \mathrm{~g}\) 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 \(\mathrm{BaCO}_{3}(s)\) 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 BaO and CaO in the mixture are approximately 84.53% and 15.47%, respectively.

Step-by-step solution

Calculate moles of CO2 reacted

Subtract the final pressure of CO2 from the initial pressure to get the total pressure change of CO2 due to the reaction. - Initial CO2 pressure: 750 torr - Final CO2 pressure: 230 torr Total pressure change of CO2: 750 torr - 230 torr = 520 torr Using the ideal gas law (\(PV = nRT\)), we can find the moles of CO2 that reacted. Convert the pressure change to atmospheres, and use R = 0.0821 L atm/(mol K). - Pressure change in atm: 520 torr * (1 atm / 760 torr) = 0.684 atm - Volume: 1.50 L - Temperature: 30.0°C + 273.15 = 303.15 K \[n = \frac{P \cdot V}{R \cdot T} = \frac{0.684 \, atm \cdot 1.50 \, L}{0.0821 \frac{L \, atm}{mol \, K} \cdot 303.15 \, K} = 0.04254 \, mol\]

Use stoichiometry to find moles of BaO and CaO

Since BaO and CaO both react with CO2 in a 1:1 ratio, we can say that: - moles of BaO reacted + moles of CaO reacted = moles of CO2 reacted - Let x be the moles of BaO reacted, so (0.04254 - x) represents the moles of CaO reacted.

Calculate the mass of BaO and CaO

Use the molar mass of BaO and CaO to find their masses. The sum of their masses is equal to the given mixture mass (5.14 g). - Molar mass of BaO: 137.3 g/mol (Ba) + 16.0 g/mol (O) = 153.3 g/mol - Molar mass of CaO: 40.1 g/mol (Ca) + 16.0 g/mol (O) = 56.1 g/mol - Mass of BaO = x * molar mass of BaO = 153.3x g - Mass of CaO = (0.04254 - x) * molar mass of CaO = 56.1(0.04254 - x) g Sum of masses = 5.14 g, therefore: 153.3x + 56.1(0.04254 - x) = 5.14

Solve for x

Now, we'll solve the equation 153.3x + 56.1(0.04254 - x) = 5.14 for x: 153.3x + 56.1(0.04254 - x) = 5.14 153.3x + 2.383 - 56.1x = 5.14 97.2x = 2.757 x = 0.02835

Calculate mass percentages

Now that we have x, which represents the moles of BaO that reacted, we can find the mass percentages of BaO and CaO: - Mass of BaO = 153.3 * 0.02835 = 4.345 g - Mass of CaO = 5.14 g - 4.345 g = 0.795 g Mass percentage of BaO: \(\frac{4.345 \, g}{5.14 \, g} \times 100\% = 84.53\%\) Mass percentage of CaO: \(\frac{0.795 \, g}{5.14 \, g} \times 100\% = 15.47\%\) The mass percentages of BaO and CaO in the mixture are approximately 84.53% and 15.47%, respectively.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that describes the behavior of an ideal gas. It is expressed as \(PV = nRT\), where:

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

To solve problems using the Ideal Gas Law, always ensure that units are consistent. This means typically converting pressure into atmospheres, volume into liters, and temperature into Kelvin.
In the exercise, the pressure change of \(\text{CO}_2\) due to reaction is key to determining the moles of gas reacted. Calculations involve converting the pressure from torr to atmospheres and using the known values of other variables in the equation.

Molar Mass

Molar Mass is an essential concept when dealing with stoichiometry and reactions. It is the mass of a given substance (chemical element or chemical compound) divided by its amount of substance (mole).
For individual elements, the molar mass is taken from the periodic table, given in grams per mole (g/mol).
For compounds, the molar mass is the sum of the molar masses of its constituent elements. For example, in the reaction:

• The molar mass of \(\text{BaO}\) is determined by summing the atomic masses of barium (Ba) and oxygen (O), which are 137.3 g/mol and 16.0 g/mol, respectively, resulting in a total of 153.3 g/mol.
• Similarly, the molar mass of \(\text{CaO}\) is the sum of calcium (Ca) at 40.1 g/mol and oxygen (O) at 16.0 g/mol, which adds up to 56.1 g/mol.

These values are crucial for determining the masses of each compound that react and are formed.

Mass Percentage Calculation

Mass Percentage Calculation is a common way to express the composition of a mixture or compound. It refers to the percentage by mass of a component in a mixture, calculated using the formula:\[\text{Mass Percentage} = \left(\frac{\text{Mass of component}}{\text{Total mass of mixture}}\right) \times 100\%.\]For this example, after solving for the individual masses of \(\text{BaO}\) and \(\text{CaO}\), we calculate their percentages in the original mixture:

• The mass of \(\text{BaO}\) was found to be 4.345 g, and when divided by the total mass of the sample, 5.14 g, it gives a mass percentage of 84.53%.
• The mass of \(\text{CaO}\), 0.795 g, when subjected to the same calculation, results in a mass percentage of 15.47%.

These percentages tell us the composition of the initial mixture, showing the prevalence of each component.

Gas Reaction Stoichiometry

Gas Reaction Stoichiometry involves using balanced chemical equations to determine the relationships between reactants and products in gas reactions. This stoichiometric approach helps predict the amount of products produced or reactants needed based on the mole ratios in the balanced equation.
In reactions involving gases, you often rely on the Ideal Gas Law to find the moles of gas. This requires knowing the changes in gas quantities (like pressure differences) due to the reaction.
In the given exercise:

• The reaction of \(\text{BaO}\) and \(\text{CaO}\) with \(\text{CO}_2\) to form carbonates is a 1:1 stoichiometry each, which means for every mole of \(\text{BaO}\) or \(\text{CaO}\) that reacts, an equal mole of \(\text{CO}_2\) is used.
• This stoichiometric ratio was vital in determining the initial unknown moles of \(\text{BaO}\) and \(\text{CaO}\) that reacted, which was then used to find their masses and eventually their mass percentages.

This application of stoichiometry helps in translating observed chemical changes to quantifiable masses useful in lab and industry scenarios.