Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 97

A 3.05 g sample of \(\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{s})\) is introduced into an evacuated 2.18 L flask and then heated to \(250^{\circ} \mathrm{C}\).What is the total gas pressure, in atmospheres, in the flask at \(250^{\circ} \mathrm{C}\) when the \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) has completely decomposed? $$\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{s}) \longrightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$

Short Answer

Expert verified
The total gas pressure, in atmospheres, in the flask at \(250^{\circ} \mathrm{C}\) when the \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) has completely decomposed is \(1.25 atm\).

Step by step solution

01

Finding the Number of Moles of \(\mathrm{NH}_{4} \(\mathrm{NO}_{3}\)

Begin by calculating the number of moles of the compound \(\mathrm{NH}_{4} \(\mathrm{NO}_{3}\). Molar mass of \(\mathrm{NH}_{4} \(\mathrm{NO}_{3}\) is approximately 80.052 g/mol. So, the number of moles can be calculated using the formula \(\text{Number of moles = Mass / Molar Mass}\). Upon substituting the given values, we get \(\text{Number of moles = }3.05 g /\ 80.052 g/mol = 0.0381 mol\).
02

Applying the Stoichiometry

Next, apply the stoichiometry of the reaction. The balanced chemical equation shows that 1 mole of the reactant produces 1 mole of \(\mathrm{N}_{2}\mathrm{O}(\mathrm{g})\) and 2 moles of \(\mathrm{H}_{2}\mathrm{O}(\mathrm{g})\) upon decomposition. Therefore, the total moles of gas produced = 0.0381 mol \( + 2 * 0.0381 mol = 0.1143 mol\).
03

Ideal Gas Law Calculation

Finally, use the ideal gas law equation \(P V = nRT\) to solve for the pressure. In this equation, P represents the pressure we want to determine, V is the volume which is 2.18 L, n is the number of moles calculated in the previous step, R is the gas constant which is about 0.0821 L.atm/mol.K, and T is the temperature in Kelvin, which is 250^0 C + 273 = 523 K. Substituting these values, we get \(P = nRT / V\). Thus, \(P = 0.1143 mol * 0.0821 L.atm/(mol.K) * 523 K / 2.18 L\), which gives the pressure = 1.25 atm.

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.

Start your free trial

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

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

You purchase a bag of potato chips at an ocean beach to take on a picnic in the mountains. At the picnic, you notice that the bag has become inflated, almost to the point of bursting. Use your knowledge of gas behavior to explain this phenomenon.

What is the pressure, in pascals, exerted by \(1242 \mathrm{g}\) CO(g) when confined at \(-25^{\circ} \mathrm{C}\) to a cylindrical tank \(25.0 \mathrm{cm}\) in diameter and \(1.75 \mathrm{m}\) high?

A 12.8 L cylinder contains \(35.8 \mathrm{g} \mathrm{O}_{2}\) at \(46^{\circ} \mathrm{C}\). What is the pressure of this gas, in atmospheres?

In your own words, define or explain each term or symbol. (a) atm; (b) STP; (c) \(R ;\) (d) partial pressure; (e) \(u_{\mathrm{rms}}\).

What is the partial pressure of \(\mathrm{Cl}_{2}(\mathrm{g}),\) in millimeters of mercury, at \(0.00^{\circ} \mathrm{C}\) and 1.00 atm in a gaseous mixture that consists of \(46.5 \% \mathrm{N}_{2}, 12.7 \% \mathrm{Ne},\) and \(40.8 \%\) \(\mathrm{Cl}_{2},\) by mass?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation
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