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A mixture of gases contains \(0.75 \mathrm{~mol} \mathrm{~N}_{2}, 0.30 \mathrm{~mol} \mathrm{O}_{2}\) and \(0.15 \mathrm{~mol} \mathrm{CO}_{2}\). If the total pressure of the mixture is \(1.56 \mathrm{~atm}\), what is the partial pressure of each component?

Short Answer

Expert verified
The partial pressures of the components in the mixture are: N2: \(0.975 \: \mathrm{atm}\) O2: \(0.39 \: \mathrm{atm}\) CO2: \(0.195 \: \mathrm{atm}\)

Step by step solution

01

Calculate the total moles of the mixture.

As given, we have 0.75 mol of N2, 0.30 mol of O2, and 0.15 mol of CO2. To find the total moles in the mixture, we simply add up the moles of each gas: Total moles = 0.75 mol N2 + 0.30 mol O2 + 0.15 mol CO2 = 1.2 mol
02

Determine the mole fraction of each component.

The mole fraction of each gas is the ratio of the moles of that gas to the total moles in the mixture. Let's calculate the mole fractions for N2, O2, and CO2. Mole fraction of N2 (X_N2) = (moles of N2) / (total moles) = (0.75 mol) / (1.2 mol) = 0.625 Mole fraction of O2 (X_O2) = (moles of O2) / (total moles) = (0.30 mol) / (1.2 mol) = 0.25 Mole fraction of CO2 (X_CO2) = (moles of CO2) / (total moles) = (0.15 mol) / (1.2 mol) = 0.125
03

Calculate the partial pressure of each component.

Now, we can use Dalton's Law of Partial Pressures and the mole fractions to calculate the partial pressure of each gas. The partial pressure of each component (P_i) can be calculated as: P_i = X_i × P_total Partial pressure of N2 (P_N2) = X_N2 × P_total = 0.625 × 1.56 atm = 0.975 atm Partial pressure of O2 (P_O2) = X_O2 × P_total = 0.25 × 1.56 atm = 0.39 atm Partial pressure of CO2 (P_CO2) = X_CO2 × P_total = 0.125 × 1.56 atm = 0.195 atm In conclusion, the partial pressures of the components in the mixture are: N2: \(0.975 \: \mathrm{atm}\) O2: \(0.39 \: \mathrm{atm}\) CO2: \(0.195 \: \mathrm{atm}\)

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Key Concepts

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

Dalton's Law of Partial Pressures
Dalton's Law of Partial Pressures is a crucial concept in understanding the behavior of gas mixtures. This law states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of individual gases. Consider a scenario where we have different gases in a container; Dalton's law allows us to pinpoint the pressure contribution of each gas in the mixture.

The application of Dalton’s law is beautifully straightforward when calculating the partial pressures in a sample mixture. For instance, if we know the total pressure of a gas mixture and the mole fraction of each gas (more on that later), we can calculate each gas's partial pressure by simply multiplying its mole fraction by the total pressure. This is exactly what we saw in our exercise example, where partial pressures of nitrogen, oxygen, and carbon dioxide were computed based on their mole fractions.
Mole Fraction
The mole fraction is another fundamental concept in chemistry that quantifies the proportion of a substance within a mixture. It is defined as the number of moles of a component divided by the total number of moles of all components in the mixture. Symbolized by the Greek letter Chi \(\chi\), the mole fraction is a dimensionless number that plays a vital role in the stoichiometry of gases and the application of Dalton’s Law.

Understanding how to calculate the mole fraction is essential for determining the particular behavior of each gas in a mixture. For example, in the given solution, the mole fractions of nitrogen (N2), oxygen (O2), and carbon dioxide (CO2) were crucial in calculating their respective partial pressures. The mole fraction represents the “share” of the gas in the overall mixture, thereby determining its impact on properties like partial pressure.
Gas Laws in Chemistry
Gas laws in chemistry relate to a series of foundational principles that explain how gases interact and behave under different conditions. These laws include Boyle’s Law, Charles’s Law, Gay-Lussac's Law, and Avogadro’s Law. Each law focuses on a specific property of gases, such as pressure, volume, temperature, or the amount of gas.

Understanding these gas laws helps us predict and quantify how changing one condition will affect others. However, when dealing with mixtures of gases, as in the exercise example, Dalton’s Law of Partial Pressures becomes significantly useful. It complements the other gas laws by focusing on mixtures and establishing the relationship between the individual gases’ partial pressures and the total pressure.
Stoichiometry of Gases
The stoichiometry of gases involves the quantitative relationship between reactants and products in chemical reactions involving gases. It incorporates the principles of various gas laws to help determine the volume, pressure, and temperature conditions of gases involved in a chemical reaction.

To apply stoichiometry to gases effectively, one must often convert between moles, volume, and pressure using the ideal gas law. These conversions are necessary when solving real-world problems, like finding the amounts of reactants needed for a given reaction or estimating the yield of a product. The exercise example showcased the stoichiometry of gas mixtures, pertaining to the way we calculated mole fractions and then used these fractions along with Dalton’s Law to find the partial pressures of each gas in the mixture.

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Most popular questions from this chapter

A sample of \(3.00 \mathrm{~g}\) of \(\mathrm{SO}_{2}(g)\) originally in a 5.00-L vessel at \(21^{\circ} \mathrm{C}\) is transferred to a \(10.0\) - \(\mathrm{L}\) vessel at \(26^{\circ} \mathrm{C}\). A sample of \(2.35 \mathrm{~g} \mathrm{~N}_{2}(g)\) originally in a 2.50-L vessel at \(20^{\circ} \mathrm{C}\) is transferred to this same \(10.0\) - \(\mathrm{L}\) vessel. (a) What is the partial pressure of \(\mathrm{SO}_{2}(g)\) in the larger container? (b) What is the partial pressure of \(\mathrm{N}_{2}(\mathrm{~g})\) in this vessel? (c) What is the total pressure in the vessel?

An herbicide is found to contain only \(C, H, N\), and \(C 1\) The complete combustion of a \(100.0-\mathrm{mg}\) sample of the herbicide in excess oxygen produces \(83.16 \mathrm{~mL}\) of \(\mathrm{CO}_{2}\) and \(73.30 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) vapor at STP. A separate analysis shows that the sample also contains \(16.44 \mathrm{mg}\) of \(\mathrm{Cl}\). (a) Determine the percent composition of the substance. (b) Calculate its empirical formula.

Suppose you have two 1-L flasks, one containing \(\mathrm{N}_{2}\) at STP, the other containing \(\mathrm{CH}_{4}\) at STP. How do these systems compare with respect to (a) number of molecules, (b) density, (c) average kinetic energy of the molecules, (d) rate of effusion through a pinhole leak?

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 (see drawing). From the following data, calculate the molar mass of the unknown liquid: mass of unknown vapor, \(1.012 \mathrm{~g}\); volume of bulb, \(354 \mathrm{~cm}^{3}\); pressure, 742 torr; temperature, \(99^{\circ} \mathrm{C}\).

Magnesium can be used as a "getter" in evacuated enclosures, to react with the last traces of oxygen. (The magnesium is usually heated by passing an electric current through a wire or ribbon of the metal.) If an enclosure of \(0.382 \mathrm{~L}\) has a partial pressure of \(\mathrm{O}_{2}\) of \(3.5 \times 10^{-6}\) torr at \(27^{\circ} \mathrm{C}\), what mass of magnesium will react according to the following equation? $$ 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s) $$

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