Problem 121 - Chemistry: The Central Science in SI Units, Global Edition

Question

A \(4.00-\mathrm{g}\) 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 \(97.33 \mathrm{kPa}\) 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 \(20.0 \mathrm{kPa}\). (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 number of moles of CO2 that have reacted is ∆n = n_initial - n_final. The mass percentage of CaO in the mixture is calculated as (mass of CaO / 4.00 g) × 100, where mass of CaO is obtained using the moles of CaO (x) found in step 5.

Step 1: Find the initial moles of CO2

We can find the initial moles of CO2 gas using the Ideal Gas Law, which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. The initial pressure of CO2 is 97.33 kPa, and the volume is 1.00 L. The temperature, given in Celsius, should be converted to Kelvin: T = 25 + 273.15 = 298.15 K. Using the pressure in atm (1 kPa = 0.00986923 atm), we have, P = 97.33 kPa × 0.00986923 atm/kPa ≈ 0.9606 atm Now we can find the initial moles of CO2: \(n_{initial} = \frac{PV}{RT}\)

Step 2: Find the final moles of CO2

Similar to step 1, we will use the Ideal Gas Law to find the final moles of CO2 after reaction. The final pressure is given as 20.0 kPa. P = 20.0 kPa × 0.00986923 atm/kPa ≈ 0.19738 atm Again, using the Ideal Gas Law, \(n_{final} = \frac{PV}{RT}\)

Step 3: Calculate the moles of CO2 that have reacted

The difference between the initial and final moles of CO2 will give us the moles of CO2 that have reacted: ∆n = n_initial - n_final

Step 4: Determine the stoichiometry of the reaction

Since one mole of CO2 reacts with one mole of CaO and one mole of BaO, we can find the moles of CaO and BaO that reacted with CO2: moles of CaO = x moles of BaO = ∆n - x We also know that the mass of the mixture is 4.00 g, so: mass of CaO + mass of BaO = 4.00 g Using the molar masses of CaO (56.08 g/mol) and BaO (153.33 g/mol), we can write: 56.08x + 153.33(∆n - x) = 4.00

Step 5: Solve for x

Rearrange the equation from step 4 to solve for x, the moles of CaO: x = (4.00 - 153.33∆n) / (56.08 - 153.33) Now substitute the value of ∆n obtained in step 3 and find the moles of CaO, x.

Step 6: Calculate the mass percentage of CaO

Finally, we can find the mass percentage of CaO in the mixture using the mass of CaO and the total mass of the mixture: mass percentage of CaO = (mass of CaO / 4.00 g) × 100 Substitute the value of x found in step 5, and find the mass percentage of CaO.

Key Concepts

Chemical Reaction Stoichiometry

Stoichiometry is a fascinating topic in chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. Imagine baking cookies. If your recipe needs 2 cups of flour for every cup of sugar, you can scale the ingredients up or down depending on how many cookies you want. Similarly, in a chemical reaction such as our exercise involving \[\text{CaO + CO}_2 \rightarrow \text{CaCO}_3\] and \[\text{BaO + CO}_2 \rightarrow \text{BaCO}_3\], stoichiometry tells us that one mole of \(\text{CO}_2\) will react with one mole of either \(\text{CaO}\) or \(\text{BaO}\).
This means that for every mole of \({\text{CO}_2}\) used, one mole of one of these oxides is consumed to form a mole of the corresponding carbonate. The concept of reaction stoichiometry helps in determining how much of each reactant is needed or how much of a product is formed in a reaction.
Understanding stoichiometry allows us to predict the outcome of chemical reactions and calculate the quantities needed, which is crucial for both academic and practical applications.

Molar Mass

Molar mass is the mass of one mole, or \(6.022 \times 10^{23}\) particles, of a substance, usually expressed in grams per mole \(\text{g/mol}\). It's like weighing a dozen of eggs instead of one. For our scenario, consider \(\text{CaO}\) and \(\text{BaO}\). The molar mass of \(\text{CaO}\) is approximately 56.08 \(\text{g/mol}\), and for \(\text{BaO}\), it is roughly 153.33 \(\text{g/mol}\).

• To calculate molar mass, you add up the atomic masses of all the atoms in a molecule. For \(\text{CaO}\), this is the sum of calcium \((\approx 40.08 \text{g/mol})\) and oxygen \((\approx 16.00 \text{g/mol})\).
• Similarly, for \(\text{BaO}\), add the atomic masses of barium \((\approx 137.33 \text{g/mol})\) and oxygen.

Knowing the molar mass is essential to convert between grams and moles, which is often required when dealing with real-world chemical reactions, as it allows us to relate how much reactant or product is involved.

Mass Percentage Calculation

Mass percentage calculation is a common way to express the concentration of a component in a mixture. In simpler terms, it's like asking how much of your chocolate chip cookie recipe is actually chocolate chips. For the problem at hand, we want to find the mass percentage of \(\text{CaO}\) in our original 4.00 g mixture.
To calculate mass percentage, you use the formula:\[\text{Mass Percentage} = \left(\frac{\text{mass of component}}{\text{total mass of mixture}}\right) \times 100\%\]In our case,

• Calculate the mass of \(\text{CaO}\) involved by multiplying its moles by its molar mass.
• Then take this mass and divide it by the total mass of the sample, which is 4.00 g in this scenario.
• Finally, multiply by 100 to convert this fraction to a percentage.

This calculation provides insight into the composition of a mixture, allowing us to determine the proportion of \(\text{CaO}\) and compare it with other substances present.

Gas Pressure Conversion

Gas pressure conversion is an essential skill in chemistry and physics, especially when working with gases, since pressures can be reported in different units depending on the context. The Ideal Gas Law often uses standard units such as atm, but our exercise initially uses kPa.

• To convert kPa to atm, you use the conversion factor where 1 kPa equals approximately 0.00986923 atm.
• This is important because calculations using the Ideal Gas Law \(PV = nRT\) (where \(n\) is moles, \(R\) is the gas constant, and \(T\) is temperature) require consistent units.
• By standardizing units, errors in calculations are minimized, especially when dealing with reactions that depend on precise measurements, like in our example.

Understanding and correctly applying these conversions ensures accuracy when predicting and analyzing the behavior of gases under various conditions, making it a crucial aspect of chemical experimentation and industrial processes.