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(a) What are the mole fractions of each component in a mixture of 15.08 \(\mathrm{g}\) of \(\mathrm{O}_{2}, 8.17 \mathrm{g}\) of \(\mathrm{N}_{2},\) and 2.64 \(\mathrm{g}\) of \(\mathrm{H}_{2}\) ? (b) What is the partial pressure in atm of each component of this mixture if it is held in a \(15.50-\mathrm{L}\) vessel at \(15^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
(a) The mole fractions of Oâ‚‚, Nâ‚‚, and Hâ‚‚ are approximately 0.226, 0.140, and 0.634, respectively. (b) The partial pressures of Oâ‚‚, Nâ‚‚, and Hâ‚‚ in the mixture are approximately 0.576 atm, 0.357 atm, and 1.617 atm, respectively.

Step by step solution

01

Calculate moles of each component

First, we need to convert grams to moles for each component using their molar masses: - O₂: Molar mass = 32 g/mol, so moles = 15.08 g / 32 g/mol ≈ 0.471 moles - N₂: Molar mass = 28 g/mol, so moles = 8.17 g / 28 g/mol ≈ 0.292 moles - H₂: Molar mass = 2 g/mol, so moles = 2.64 g / 2 g/mol = 1.32 moles
02

Calculate total moles

Add the moles of each component to find the total molesia: Total moles = moles of O₂ + moles of N₂ + moles of H₂ Total moles = 0.471 + 0.292 + 1.32 ≈ 2.083 moles
03

Determine the mole fractions

Now, we will calculate the mole fraction of each component by dividing their individual moles by the total moles: - Mole Fraction of O₂ = moles of O₂ / Total moles = 0.471 / 2.083 ≈ 0.226 - Mole Fraction of N₂ = moles of N₂ / Total moles = 0.292 / 2.083 ≈ 0.140 - Mole Fraction of H₂ = moles of H₂ / Total moles = 1.32 / 2.083 ≈ 0.634
04

Calculate the total pressure

First, we must convert the temperature from Celsius to Kelvin: Temperature = 15°C + 273.15 = 288.15 K Now, we'll use the Ideal Gas Law (PV = nRT) to find the total pressure. In this case, we have: - n (number of moles) = 2.083 mol - R (gas constant) = 0.0821 L atm/mol K - T (temperature) = 288.15 K - V (volume) = 15.50 L Solving for P (pressure): P = (nRT) / V P = (2.083 mol * 0.0821 L atm/mol K * 288.15 K) / 15.5 L P ≈ 2.55 atm
05

Calculate partial pressures

Finally, we'll find the partial pressures of each component by multiplying their mole fractions by the total pressure: - Partial Pressure of O₂ = Mole Fraction of O₂ * Total Pressure = 0.226 * 2.55 atm ≈ 0.576 atm - Partial Pressure of N₂ = Mole Fraction of N₂ * Total Pressure = 0.140 * 2.55 atm ≈ 0.357 atm - Partial Pressure of H₂ = Mole Fraction of H₂ * Total Pressure = 0.634 * 2.55 atm ≈ 1.617 atm So, the partial pressures of O₂, N₂, and H₂ in the mixture are 0.576 atm, 0.357 atm, and 1.617 atm, respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
The Ideal Gas Law is a key principle in chemistry that relates the state variables of a gas to each other. It combines several gas laws into one formula, which is \( PV = nRT \). Here, \( P \) represents the pressure of the gas, \( V \) represents its volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) stands for temperature in Kelvin.

This law makes it incredibly easy to predict how a gas will respond to changes in volume, temperature, or pressure, provided we know its other properties. For example, if the temperature and amount of gas are held constant, reducing the volume of a gas will increase its pressure.

The Ideal Gas Law is most accurate for "ideal gases," which are hypothetical gases that perfectly adhere to the assumptions of the law, namely, gas particles with no volume and no intermolecular forces. Real gases approximate this behavior under high temperatures and low pressures.

When solving problems using the Ideal Gas Law, it is crucial to convert all measurements to the appropriate units, like using Kelvin for temperature and liters for volume. This ensures that the values are compatible with the gas constant \( R = 0.0821 \) L atm/mol K.
Partial Pressures
In a gaseous mixture, each individual gas exerts its own pressure as if it occupied the entire volume alone. This is known as the partial pressure of the gas. Understanding partial pressures is important when dealing with mixed gases because it helps us understand how each gas behaves in the mixture.

The partial pressure \( P_i \) of a component in a mixture is determined by multiplying its mole fraction \( X_i \) by the total pressure \( P \) of the gas mixture: \[ P_i = X_i \times P \].
  • Mole fraction \( X_i \) is the ratio of moles of a particular gas to the total moles in the mixture.
  • Total pressure is the sum of all partial pressures in the mixture.
Each component of a gas mixture contributes to the total pressure proportionally to its mole fraction. This connection is known as Dalton's Law of Partial Pressures, which is incredibly useful for calculating individual gases’ contributions in a mixture, given the total pressure.
Molar Mass
Molar mass, a fundamental concept in chemistry, is the mass of one mole of a substance. It can be thought of as the "weight" of a mole, linking the macroscopic scale we see, in grams, with the microscopic scale of atoms and molecules.

To find the molar mass of a compound, you sum the atomic masses of all atoms present in a molecule of the compound. For example, the molar mass of \( \text{O}_2 \) is 32 g/mol because each oxygen atom has an atomic mass of approximately 16 g/mol.
  • Molar mass allows us to convert between grams and moles, an essential step in stoichiometry.
  • Understanding molar mass enables the calculation of a substance's mole fraction in any given mixture, as demonstrated in the initial problem.
For gases, knowing the molar mass is particularly important because it directly affects the number of moles of gas present in a given mass, influencing variables like partial pressures or total pressure when using the ideal gas law. Overcoming this bridge between mass and moles is a fundamental skill in interpreting and predicting chemical behavior.

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Most popular questions from this chapter

\((\mathbf{a})\) Amonton's law expresses the relationship between pressure and temperature. Use Charles's law and Boyle's law to derive the proportionality relationship between \(P\) and \(T\) . \((\mathbf{b})\) If a car tire is filled to a pressure of 32.0 \(\mathrm{lb} / \mathrm{in.}^{2}\) (psi) measured at \(75^{\circ} \mathrm{F},\) what will be the tire pressure if the tires heat up to \(120^{\circ} \mathrm{F}\) during driving?

Which of the following statements is false? \begin{equation}\begin{array}{l}{\text { (a) Gases are far less dense than liquids. }} \\ {\text { (b) Gases are far more compressible than liquids. }} \\\ {\text { (c) Because liquid water and liquid carbon tetrachloride do }} \\\ {\text { not mix, neither do their vapors. }} \\ {\text { (d) The volume occupied by a gas is determined by the volume }} \\ {\text { of its container. }}\end{array}\end{equation}

Consider a mixture of two gases, \(A\) and \(B,\) confined in a closed vessel. A quantity of a third gas, \(C,\) is added to the same vessel at the same temperature. How does the addition of gas C affect the following: (a) the partial pressure of gas A, (b) the total pressure in the vessel, (c) the mole fraction of gas B?

On a single plot, qualitatively sketch the distribution of molecular speeds for (a) \(\operatorname{Kr}(g)\) at \(-50^{\circ} \mathrm{C},(\mathbf{b}) \mathrm{Kr}(g)\) at \(0^{\circ} \mathrm{C},\) (c) \(\operatorname{Ar}(g)\) at \(0^{\circ} \mathrm{C} .[\) Section 10.7\(]\)

In the Dumas-bulb technique for determining the molar mass of an unknown liquid, you vaporize the sample of a liquid that boils below \(100^{\circ} \mathrm{C}\) in a boiling-water bath and determine the mass of vapor required to fill the bulb. From the following data, calculate the molar mass of the unknown liquid: mass of unknown vapor, 1.012 g; volume of bulb, \(354 \mathrm{cm}^{3} ;\) pressure, 742 torr; temperature, \(99^{\circ} \mathrm{C}\) .

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