Question

The partial pressure of \(\mathrm{CH}_{4}(g)\) is 0.175 atm and that of \(\mathrm{O}_{2}(g)\) is 0.250 atm in a mixture of the two gases. a. What is the mole fraction of each gas in the mixture? b. If the mixture occupies a volume of \(10.5 \mathrm{L}\) at \(65^{\circ} \mathrm{C}\), calculate the total number of moles of gas in the mixture. c. Calculate the number of grams of each gas in the mixture.

Short answer

The mole fractions of CH4 and O2 are approximately 0.412 and 0.588, respectively. The total number of moles in the mixture is approximately 0.137 mol, and the mass of each gas in the mixture is approximately 0.904 g for CH4 and 2.58 g for O2.

Step-by-step solution

Calculate the mole fraction of each gas

To find the mole fraction of each gas, simply divide the individual partial pressure of the gas by the total pressure of the mixture. Let's denote the mole fraction of methane (CH4) as X_CH4 and the mole fraction of oxygen (O2) as X_O2. Find the total pressure of the mixture by adding the partial pressures of CH4 and O2: Total Pressure (P_total) = P_CH4 + P_O2 P_total = 0.175 atm (CH4) + 0.250 atm (O2) = 0.425 atm Now, calculate the mole fraction of each gas: X_CH4 = P_CH4 / P_total X_O2 = P_O2 / P_total

Calculate the mole fractions

Calculate the mole fractions using the formula from step 1: X_CH4 = 0.175 atm / 0.425 atm ≈ 0.412 X_O2 = 0.250 atm / 0.425 atm ≈ 0.588

Calculate the total number of moles in the mixture

We can use the Ideal Gas Law to find the total number of moles in the mixture. The Ideal Gas Law is given by PV = nRT, where P is the total pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. First, we need to convert the temperature from Celsius to Kelvin: Temperature (T) = 65℃ + 273.15 = 338.15 K Now plug the values into the Ideal Gas Law formula and solve for n: (0.425 atm) * (10.5 L) = n * (0.0821 L atm/mol K) * (338.15 K) Divide both sides by the product of the gas constant and the temperature: n = (0.425 atm * 10.5 L) / (0.0821 L atm/mol K * 338.15 K)

Find the total number of moles

Calculate the total number of moles using the formula from step 3: n ≈ 0.137 mol

Calculate the number of moles of individual gases in the mixture

Now that we know the total number of moles in the mixture, we can find the number of moles of each individual gas by multiplying the total number of moles by the mole fraction of each gas: Moles of CH4 = n * X_CH4 Moles of O2 = n * X_O2

Find the number of moles of individual gases

Calculate the number of moles of CH4 and O2 using the formula from step 5: Moles of CH4 ≈ 0.137 mol * 0.412 ≈ 0.0564 mol Moles of O2 ≈ 0.137 mol * 0.588 ≈ 0.0806 mol

Calculate the mass of individual gases in the mixture

Now we can calculate the mass of each individual gas in the mixture by multiplying the number of moles of each gas by its molar mass: Mass of CH4 = moles of CH4 * molar mass of CH4 Mass of O2 = moles of O2 * molar mass of O2

Find the mass of individual gases

Calculate the mass of each gas using the formula from step 7: Mass of CH4 ≈ 0.0564 mol * (16.04 g/mol) ≈ 0.904 g Mass of O2 ≈ 0.0806 mol * (32.00 g/mol) ≈ 2.58 g To sum up, the mole fraction of CH4 is ≈ 0.412 and O2 is ≈ 0.588, the total number of moles in the mixture is ≈ 0.137 mol, and the mass of each gas in the mixture is ≈ 0.904 g for CH4 and ≈ 2.58 g for O2.

Key concepts

Mole Fraction

The mole fraction is an important concept in chemistry, especially when dealing with mixtures of gases. It is defined as the ratio of the number of moles of a particular component to the total number of moles in the mixture. It provides a way to express the concentration of a component within a mixture. To calculate the mole fraction of a gas in a mixture, you can use its partial pressure divided by the total pressure of the gas mixture. This method is used because, according to Dalton’s Law of Partial Pressures, the pressure exerted by a gas mixture is equal to the sum of the partial pressures of each individual gas. For example, in this exercise:

• The total pressure is the sum of the partial pressures of CH extsubscript{4} and O extsubscript{2}.
• Mole fraction of CH extsubscript{4} is: \(X_{CH_4} = \frac{P_{CH_4}}{P_{total}}\)
• Mole fraction of O extsubscript{2} is: \(X_{O_2} = \frac{P_{O_2}}{P_{total}}\)

This gives us an easy way to express how much of each type of gas is present in a gaseous mixture. Using the mole fraction, one can easily calculate the contributions of each gas towards the mixture's properties.

Partial Pressure

Partial pressure is the pressure that a single gas in a mixture would exert if it were the only gas present in the volume occupied by the entire gas mixture. In multi-gas systems, like the atmosphere or an enclosed gas mixture, each component gas exerts its pressure independently of the others.This concept is grounded in Dalton's Law of Partial Pressures which states that the total pressure exerted by a gaseous mixture is the sum total of the partial pressures of the individual gases.In practice:

• If you know the total pressure of a gas mixture, the partial pressure of one gas can be found using its mole fraction:
• For example, \(P_{gas} = X_{gas} \times P_{total}\)

Partial pressures play a critical role in various applications, such as determining the concentration of gases for chemical reactions, calculations involving gas solubility in liquids, and understanding behaviors of gases under different conditions.

Molar Mass

Molar mass, also known as molecular weight, is the mass of a given substance (chemical element or compound) divided by the amount of substance. It is typically expressed in units of grams per mole (g/mol). For an element, the molar mass is numerically equal to the atomic weight of the element. For a compound, it's the sum of the atomic masses of all the atoms in a molecule of that compound. This value helps in converting moles of a substance to grams, which is essential in various chemical calculations. For instance, when calculating the mass of gases in a mixture:

• Determine the number of moles of the gas.
• Multiply by the molar mass to find the mass:
• For CH extsubscript{4}, using the molar mass of 16.04 g/mol.
• For O extsubscript{2}, using the molar mass of 32.00 g/mol.

Hence, knowing the molar mass helps in bridging the gap between the microscopic world of atoms/molecules and the macroscopic world of grams/liters used in laboratory settings.