Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

At \(0^{\circ} \mathrm{C}\) a \(1.0\) - \(\mathrm{L}\) flask contains \(5.0 \times 10^{-2}\) mole of \(\mathrm{N}_{2}, 1.5 \times 10^{2}\) \(\mathrm{mg} \mathrm{O}_{2}\), and \(5.0 \times 10^{21}\) molecules of \(\mathrm{NH}_{3} .\) What is the partial pressure of each gas, and what is the total pressure in the flask?

Short Answer

Expert verified
The partial pressures of N2, O2, and NH3 in the flask are \(108.4 \, \text{kPa}\), \(12.7 \, \text{kPa}\), and \(22.5 \, \text{kPa}\) respectively, and the total pressure in the flask is \(141.3 \, \text{kPa}\).

Step by step solution

01

Calculate the moles of each gas

First, we need to calculate the moles of each gas present in the flask. We're given the moles of N2 directly and the mass of O2. For NH3, we must convert the number of molecules to moles. 1. Moles of N2: \(5.0 \times 10^{-2}\) moles 2. Moles of O2: We need to convert the given mass to moles. To do this, we can use the molar mass of O2 (32 g/mol): Moles of O2 = \( \frac{1.5 \times 10^2 \, \text{mg}}{32 \, \text{g/mol}} \) = \( \frac{0.15 \, \text{g}}{32 \, \text{g/mol}}\) = \(4.69 \times 10^{-3}\) moles 3. Moles of NH3: To convert the given number of molecules to moles, we can use Avogadro's number (6.022 x 10^23 molecules/mol): Moles of NH3 = \( \frac{5.0 \times 10^{21}\, \text{molecules}}{6.022 \times 10^{23} \, \text{molecules/mol}} \) = \(8.30 \times 10^{-3}\) moles
02

Calculate the mole fractions for each gas

Now, we need to find the mole fractions of each gas in the mixture. To do this, we'll divide the number of moles of each gas by the total moles in the flask. 1. Mole fraction of N2: \( \frac{\text{moles of N2}}{\text{total moles}} = \frac{5.0 \times 10^{-2}\, \text{moles}}{(5.0 \times 10^{-2} + 4.69 \times 10^{-3} + 8.30 \times 10^{-3})\, \text{moles}} \) = 0.767 2. Mole fraction of O2: \( \frac{\text{moles of O2}}{\text{total moles}} = \frac{4.69 \times 10^{-3}\, \text{moles}}{(5.0 \times 10^{-2} + 4.69 \times 10^{-3} + 8.30 \times 10^{-3})\, \text{moles}} \) = 0.090 3. Mole fraction of NH3: \( \frac{\text{moles of NH3}}{\text{total moles}} = \frac{8.30 \times 10^{-3}\, \text{moles}}{(5.0 \times 10^{-2} + 4.69 \times 10^{-3} + 8.30 \times 10^{-3})\, \text{moles}} \) = 0.159
03

Calculate the total pressure using the Ideal Gas Law

Now, we'll use the Ideal Gas Law (PV = nRT) to find the total pressure in the flask. We're given the volume (1.0 L), and we can use the total number of moles calculated previously. The temperature is given in Celsius, so we must first convert it to Kelvin (0°C = 273.15 K). 1. Total moles = \(5.0 \times 10^{-2} + 4.69 \times 10^{-3} + 8.30 \times 10^{-3}\) = 0.065 moles 2. Convert the temperature = 273.15 K 3. Using the Ideal Gas Law: Total pressure P = \( \frac{nRT}{V} = \frac{(0.065\, \text{moles})(8.314\, \text{J/mol K})(273.15\, \text{K})}{1.0\, \text{L}} \) = 141.3 kPa (R is given in J/mol K, so the pressure will be in kPa, and 1 L = 0.001 m^3)
04

Calculate the partial pressures for each gas

Using the mole fractions calculated earlier, we can determine the partial pressure of each gas by multiplying the mole fraction by the total pressure: 1. Partial pressure of N2: (0.767)(141.3 kPa) = 108.4 kPa 2. Partial pressure of O2: (0.090)(141.3 kPa) = 12.7 kPa 3. Partial pressure of NH3: (0.159)(141.3 kPa) = 22.5 kPa Therefore, the partial pressures of N2, O2, and NH3 are 108.4 kPa, 12.7 kPa, and 22.5 kPa, respectively, and the total pressure in the flask is 141.3 kPa.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is crucial for understanding how gases behave under different conditions. It is represented by the equation PV = nRT, where P is the pressure, V is the volume, n is the amount of substance in moles, R is the universal gas constant, and T is the temperature in Kelvin.

In the context of partial pressure calculations, this law allows us to compute the total pressure exerted by a mixture of gases in a container when the temperature, volume, and total number of moles are known. It is important to remember to always use Kelvin for the temperature and ensure consistency in the units used throughout the calculation. Additionally, R may vary depending on the pressure units in use, so it's crucial to match R to the units you want for pressure.
Mole Fraction
The mole fraction is a way of expressing the concentration of a particular component in a mixture of gases. It is the ratio of the number of moles of that component to the total number of moles of all components in the mixture.

To calculate the mole fraction, represented by the symbol \( x \), for a particular gas, you can use the formula \( x = \frac{n_{\text{component}}}{n_{\text{total}}} \), where \( n_{\text{component}} \) is the number of moles of the gas in question, and \( n_{\text{total}} \) is the sum of moles of all gases in the mixture. It is a dimensionless quantity and can be used to determine the partial pressure of a gas in a mixture by multiplying the mole fraction by the total pressure, as shown previously in the step by step solution.
Avogadro's Number
Avogadro's number, approximately 6.022 x 10\(^{23}\), is the number of particles, often atoms or molecules, in one mole of a substance. It serves as a bridge between the macroscopic scale that we can observe and measure (such as grams or liters) and the microscopic scale of individual atoms and molecules.

When you have a specific number of particles and need to find the number of moles, you divide the number of particles by Avogadro's number. Conversely, if you start with moles and need the number of particles, you multiply the number of moles by Avogadro's number. This is used in gas calculations to convert between moles and molecules of a gas, allowing further calculations such as finding the mole fraction or partial pressures.
Molar Mass
Molar mass is the mass of one mole of a particular substance and is expressed in grams per mole (g/mol). It is a physical property unique to each substance, usually found on the periodic table by adding up the atomic masses of the elements present in a compound.

For gas mixture problems, the molar mass allows for the conversion from the given mass of a gas to the number of moles, which is essential for calculating mole fractions and partial pressures. To convert mass to moles, you divide the given mass by the molar mass of the substance. For instance, if you are given a mass in milligrams, make sure to convert it to grams before using the molar mass for the conversion.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the following samples of gases at the same temperature. Arrange each of these samples in order from lowest to highest: a. pressure b. average kinetic energy c. density d. root mean square velocity Note: Some samples of gases may have equal values for these attributes. Assume the larger containers have a volume twice the volume of the smaller containers, and assume the mass of an argon atom is twice the mass of a neon atom.

Consider separate \(1.0\) -L gaseous samples of \(\mathrm{H}_{2}, \mathrm{Xe}, \mathrm{Cl}_{2}\), and \(\mathrm{O}_{2}\) all at STP. a. Rank the gases in order of increasing average kinetic energy. b. Rank the gases in order of increasing average velocity. c. How can separate \(1.0\) -L samples of \(\mathrm{O}_{2}\) and \(\mathrm{H}_{2}\) each have the same average velocity?

A \(15.0\) -L rigid container was charged with \(0.500\) atm of krypton gas and \(1.50\) atm of chlorine gas at \(350 .{ }^{\circ} \mathrm{C}\). The krypton and chlorine react to form krypton tetrachloride. What mass of krypton tetrachloride can be produced assuming \(100 \%\) yield?

A compound contains only \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{N}\). It is \(58.51 \% \mathrm{C}\) and \(7.37 \% \mathrm{H}\) by mass. Helium effuses through a porous frit \(3.20\) times as fast as the compound does. Determine the empirical and molecular formulas of this compound.

Calculate the root mean square velocities of \(\mathrm{CH}_{4}(\mathrm{~g})\) and \(\mathrm{N}_{2}(\mathrm{~g})\) molecules at \(273 \mathrm{~K}\) and \(546 \mathrm{~K}\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free