Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 209

Air is approximately \(21 \%\) by volume oxygen, \(\mathrm{O}_{2}\). How many grams of oxygen are present in \(200.0 \mathrm{~L}\) of air at \(1.00 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) ? (Hint: Use the ideal gas equation from Chapter \(11 .\) )

Short Answer

Expert verified
The mass of oxygen present in 200.0 L of air at 1.00 atm and 25°C is approximately 54.72 grams.

Step by step solution

01

Find the number of moles of air

Use the Ideal Gas Equation to find the number of moles of air, n_air: PV = nRT Where: - P is the pressure (1.00 atm) - V is the volume (200.0 L) - n is the number of moles - R is the ideal gas constant (0.0821 L⋅atm/mol⋅K) - T is the temperature (25°C = 298.15 K, after converting to Kelvin) Rearrange the equation to solve for n_air: n_air = PV / RT
02

Calculate the number of moles of air

Plug the values into the equation and calculate n_air: n_air = (1.00 atm × 200.0 L) / (0.0821 L⋅atm/mol⋅K × 298.15 K) n_air ≈ 8.143 mol
03

Find the number of moles of oxygen

Given that the air is approximately 21% by volume oxygen, we can find the number of moles of oxygen (n_oxygen) by multiplying n_air by the volume percentage: n_oxygen = n_air × 0.21 n_oxygen ≈ 8.143 mol × 0.21 ≈ 1.71 mol
04

Calculate the mass of oxygen

Now, we need to convert the number of moles of oxygen to grams. To do this, we will use the molecular weight of oxygen, which is 32 g/mol: mass_oxygen = n_oxygen × (32 g/mol) mass_oxygen ≈ 1.71 mol × 32 g/mol ≈ 54.72 g
05

Present the final answer

The mass of oxygen present in 200.0 L of air at 1.00 atm and 25°C is approximately 54.72 grams.

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

Which one of the following steps in the dissolving process must have a negative value for \(\Delta E\) ? Explain your answer. (a) The physical separation of solute particles (b) The formation of solvent-solute interactions (c) The physical separation of solvent particles (d) None of the above

A student plans to divide the molarity of his solution by its volume to determine the number of moles of solvent in it. He is making two mistakes here. Identify the mistakes and correct both of them.

Combustion analysis reveals vitamin \(\mathrm{C}\) to be \(40.9 \%\) by mass \(\mathrm{C}\) and \(4.58 \%\) by mass \(\mathrm{H}\). The only other element present is oxygen. A solution of \(19.40 \mathrm{~g}\) of vitamin \(\mathrm{C}\) in \(100.0 \mathrm{~g}\) of water freezes at \(22.05^{\circ} \mathrm{C}\). What is the molecular formula of vitamin \(C\) ?

Why do most solids become more soluble in water with increasing temperature? (Hint: Think about what happens to the water molecules as temperature increases.)

In a titration of \(\mathrm{HCl}\) with \(\mathrm{NaOH}\), you read the buret to see how much base it took to neutralize the acid (turn the indicator pink). You do this titration to determine the molar concentration of the acid. Why is it important to accurately know the amount (volume) of acid that you originally put into the flask?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

Inorganic 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