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

An 8.40 -g sample of argon and an unknown mass of \(\mathrm{H}_{2}\) are mixed in a flask at room temperature. The partial pressure of the argon is \(44.0 \mathrm{kPa},\) and that of the hydrogen is \(57.33 \mathrm{kPa} .\) What is the mass of the hydrogen?

Short Answer

Expert verified
First, we find the number of moles of argon gas by dividing its mass (8.40 g) by its molar mass (39.95 g/mol). Then, using the Ideal Gas Law, we find the volume of the container, given the partial pressure of argon (44 kPa) and assuming a room temperature of 298 K. Next, we calculate the number of moles of hydrogen gas using the Ideal Gas Law again, this time with the partial pressure of hydrogen (57.33 kPa). Finally, we find the mass of hydrogen gas by multiplying the number of moles of hydrogen by its molar mass (2.02 g/mol). After performing the calculations, we obtain the mass of the hydrogen gas in the mixture.

Step by step solution

01

Find the number of moles of argon gas in the mixture.

To find the number of moles of the argon gas, we can use the following formula: Number of moles = Mass of the gas / Molar mass of the gas The molar mass of argon gas is \(39.95 \mathrm{g/mol}\). Given the mass of argon gas (8.40 grams), we can now find the number of moles: Number of moles of argon = \(8.40 \mathrm{g} / 39.95 \mathrm{g/mol}\) Calculate the number of moles of argon gas.
02

Use the Ideal Gas Law to find the volume of the container.

Since we know the pressure and the number of moles of argon gas, we can now find the volume of the container. The Ideal Gas Law equation can be written as: \(PV = nRT\) Where: P = Partial pressure of argon = \(44.0 \mathrm{kPa}\) V = Volume of the container n = Number of moles of argon R = Gas constant = \(8.314 \mathrm{J/(mol·K)}\) (Note that we need to convert the pressure in kPa to Pa) T = Room temperature (Assuming 298 K) Rearrange the equation to find the volume V: \(V = \frac{nRT}{P}\) Plug in the values and calculate the volume of the container.
03

Calculate the number of moles of hydrogen gas.

Now that we know the volume of the container, we can find the number of moles of hydrogen gas using the Ideal Gas Law formula: \(PV = nRT\) Where: P = Partial pressure of hydrogen = \(57.33 \mathrm{kPa}\) V = Volume of the container (calculated in step 2) n = Number of moles of hydrogen R = Gas constant = \(8.314 \mathrm{J/(mol·K)}\) (Note that we need to convert the pressure in kPa to Pa) T = Room temperature (298 K) Rearrange the equation to find the number of moles of hydrogen (n): \(n = \frac{PV}{RT}\) Plug in the values and calculate the number of moles of hydrogen gas.
04

Calculate the mass of hydrogen gas.

Now that we have the number of moles of hydrogen gas, we can calculate its mass using the following formula: Mass of the gas = Number of moles of the gas × Molar mass of the gas The molar mass of hydrogen gas is \(2.02 \mathrm{g/mol}\). Multiply the number of moles calculated in step 3 by the molar mass of hydrogen gas: Mass of hydrogen = (Number of moles of hydrogen) × (2.02 g/mol) Calculate the mass of the hydrogen gas. This is the answer to the exercise.

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.

Molar Mass
Molar mass is a key concept used to understand the composition of gases. It represents the mass of one mole of a chemical element or compound, typically expressed in grams per mole (g/mol). To calculate the molar mass, you take the total mass of the element in a sample and divide it by the amount of substance it contains, in moles.
For instance, in the context of argon gas from our exercise, the molar mass is given as 39.95 g/mol. This means that one mole of argon gas weighs 39.95 grams. Knowing this allows us to determine the number of moles in a given mass of argon, which is essential for further calculations involving the Ideal Gas Law. Similarly, the molar mass of hydrogen gas is 2.02 g/mol.
This concept is vital in determining how much of each gas is present in a mixture when dealing with chemical reactions or physical processes.
Partial Pressure
Partial pressure refers to the pressure exerted by a single type of gas in a mixture of gases. It is a component of the total pressure in a mixture and is crucial for applications involving gas laws. In a gaseous mixture, each gas contributes to the total pressure as if it were the only gas present.
In our exercise, the partial pressure of argon is 44.0 kPa, and the partial pressure of hydrogen is 57.33 kPa. These values are significant because they allow us to use the Ideal Gas Law to extract more information about each gas. By knowing the partial pressure, we can calculate quantities such as the number of moles or the volume of gas, as they directly relate to the pressure exerted by the gas.
Partial pressures are especially important in reactions where gases are involved, allowing for a more clear understanding of each gas's role and behavior in the mixture.
Number of Moles
The number of moles is a measurement of the amount of substance present in a sample. It is an essential unit in chemistry, providing a bridge between the mass of a material and the number of particles it contains through Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles per mole.
  • To find the number of moles present in a sample, we use the formula: \(\text{Number of moles} = \frac{\text{Mass of the gas}}{\text{Molar mass of the gas}}\).
  • This key figure tells us how much of a substance (in terms of quantity of particles) we are dealing with. For example, using the mass and molar mass of argon provided in the exercise, we can calculate the number of moles, thus setting the stage for further analysis using gas laws.
Understanding the number of moles is critical for precisely determining reactions and for calculating other properties of the substance, such as its volume or pressure at a given temperature.
Argon Gas
Argon is a noble gas and a key player in our exercise. It is chemically inert, which means it doesn't react easily with other elements or compounds. Argon is often found in atmospheric conditions and is used in various applications due to its stable nature.
With a molar mass of 39.95 g/mol, argon is quite manageable when calculating its properties through the Ideal Gas Law. Its partial pressure being a known quantity assists with finding other unknowns, such as the volume of its container or the presence of other gases.
Considering its role in this exercise, argon is used to calculate the total volume where it is mixed with hydrogen gas. Studying argon's behavior in mixtures is beneficial not only in chemistry but also in fields like physics, as it helps to understand the broader principles of how gases behave together.
Hydrogen Gas
Hydrogen is the lightest and most abundant element in the universe. Its unique properties make it an interesting subject for study in both chemistry and physics. Hydrogen gas, denoted as \(H_{2}\), has valuable uses in areas ranging from clean energy to industrial processes.
In our exercise, hydrogen is mixed with argon gas. Its known partial pressure of 57.33 kPa and molar mass of 2.02 g/mol enable us to apply the Ideal Gas Law to calculate the number of moles present, and ultimately, its mass.
  • The process involves using its partial pressure to determine how much space it takes up and its contribution to the mixture's total pressure.
  • Despite its simplicity, working with hydrogen in a chemical context often requires careful consideration due to its reactivity and low density.
Understanding hydrogen's role in a gaseous mixture not only aids chemical calculations but also enhances comprehension of its vast applications in sciences and technology.

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

Gaseous iodine pentafluoride, \(\mathrm{IF}_{5}\), can be prepared by the reaction of solid iodine and gaseous fluorine: $$\mathrm{I}_{2}(s)+5 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{IF}_{5}(g)$$ A \(5.00-\mathrm{L}\) flask containing \(10.0 \mathrm{~g}\) of \(\mathrm{I}_{2}\) is charged with \(10.0 \mathrm{~g}\) of \(\mathrm{F}_{2},\) and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is \(125^{\circ} \mathrm{C}\). (a) What is the partial pressure of \(\mathrm{IF}_{5}\) in the flask? (b) What is the mole fraction of \(\mathrm{IF}_{5}\) in the flask (c) Draw the Lewis structure of IF \(_{5}\). (d) What is the total mass of reactants and products in the flask?

A glass vessel fitted with a stopcock valve has a mass of \(337.428 \mathrm{~g}\) when evacuated. When filled with \(\mathrm{Ar}\), it has a mass of \(339.854 \mathrm{~g}\). When evacuated and refilled with a mixture of Ne and Ar, under the same conditions of temperature and pressure, it has a mass of \(339.076 \mathrm{~g} .\) What is the mole percent of Ne in the gas mixture?

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

Complete the following table for an ideal gas: $$\begin{array}{llll} P & V & n & T \\ \hline 303.98 \mathrm{kPa} & 3.00 \mathrm{~L} & 1.500 \mathrm{~mol} & ? \mathrm{~K} \\ 50.663 \mathrm{kPa} & 0.750 \mathrm{~L} & ? \mathrm{~mol} & 300 \mathrm{~K} \\\ 101.33 \mathrm{kPa} & ? \mathrm{~L} & 3.333 \mathrm{~mol} & 300 \mathrm{~K} \\\ ? \mathrm{kPa} & .750 \mathrm{~L} & 0.750 \mathrm{~mol} & 298 \mathrm{~K} \\ \hline \end{array}$$

Calculate each of the following quantities for an ideal gas: (a) the volume of the gas, in liters, if \(1.50 \mathrm{~mol}\) has a pressure of \(126.7 \mathrm{kPa}\) at a temperature of \(-6^{\circ} \mathrm{C} ;(\mathbf{b})\) the absolute temperature of the gas at which \(3.33 \times 10^{-3}\) mol occupies \(478 \mathrm{~mL}\) at \(99.99 \mathrm{kPa} ;(\mathbf{c})\) the pressure, in pascals, if \(0.00245 \mathrm{~mol}\) occupies \(413 \mathrm{~mL}\) at \(138^{\circ} \mathrm{C} ;(\mathbf{d})\) the quantity of gas, in moles, if 126.5 L at \(54^{\circ} \mathrm{C}\) has a pressure of \(11.25 \mathrm{kPa}\).

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