Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 52

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) is used to disinfect contact lenses. How many milliliters of \(\mathrm{O}_{2}(\mathrm{g})\) at \(22^{\circ} \mathrm{C}\) and \(752 \mathrm{mmHg}\) can be liberated from \(10.0 \mathrm{mL}\) of an aqueous solution containing \(3.00 \% \mathrm{H}_{2} \mathrm{O}_{2}\) by mass? The density of the aqueous solution of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(1.01 \mathrm{g} / \mathrm{mL}\) $$2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(1)+\mathrm{O}_{2}(\mathrm{g})$$

Short Answer

Expert verified
Therefore, 111 mL of \(O_{2}\) gas can be liberated from 10.0 mL of an aqueous solution containing 3.00% \(H_{2}O_{2}\) by mass.

Step by step solution

01

Calculate the mass of \(H_{2}O_{2}\)

First, find the mass fraction of \(H_{2}O_{2}\) in the solution. The solution is made up of 3.00% \(H_{2}O_{2}\) by mass. Therefore, in 10.0 mL of the solution, the mass of \(H_{2}O_{2}\) is \(10.0 \, mL * 1.01 \, g/mL * 0.03 = 0.303 \, g\).
02

Convert mass of \(H_{2}O_{2}\) to moles

The molar mass of \(H_{2}O_{2}\) is \(16*2+1*2 = 34 \, g/mol\). So, we have \(0.303 \, g / 34 \, g/mol = 0.00892 \, mol\) of \(H_{2}O_{2}\).
03

Determine the amount of \(O_{2}\) liberated

From the balanced equation, 2 moles of \(H_{2}O_{2}\) gives 1 mole of \(O_{2}\). Therefore, \(0.00892 \, mol * (1/2) = 0.00446 \, mol\) of \(O_{2}\) is produced.
04

Apply the Ideal Gas Law

Finally, use the ideal gas law, \(PV=nRT\), where \(P=752 \, mmHg = 752/760 \, atm = 0.990 \, atm\), \(R=0.0821 \, atm.L/mol.K\), and \(T=22°C=22+273=295 \, K\). Solving for \(V\), we have \(V=nRT/P = 0.00446 \, mol * 0.0821 \, L/mol.K * 295 \, K / 0.990 \, atm = 0.111 \, L = 111 \, mL\).

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

We want to change the volume of a fixed amount of gas from \(725 \mathrm{mL}\) to 2.25 L while holding the temperature constant. To what value must we change the pressure if the initial pressure is \(105 \mathrm{kPa} ?\)

The density of phosphorus vapor is \(2.64 \mathrm{g} / \mathrm{L}\) at \(310^{\circ} \mathrm{C}\) and \(775 \mathrm{mm}\) Hg. What is the molecular formula of the phosphorus under these conditions?

For a fixed amount of gas at a fixed pressure, changing the temperature from \(100.0^{\circ} \mathrm{C}\) to \(200 \mathrm{K}\) causes the gas volume to (a) double; (b) increase, but not to twice its original value; (c) decrease; (d) stay the same.

A 2.650 g sample of a gaseous compound occupies \(428 \mathrm{mL}\) at \(24.3^{\circ} \mathrm{C}\) and \(742 \mathrm{mmHg} .\) The compound consists of \(15.5 \%\) C \(, 23.0 \%\) Cl, and \(61.5 \%\) F, by mass. What is its molecular formula?

What is the volume, in liters, occupied by a mixture of 15.2 \(\mathrm{g} \mathrm{Ne}(\mathrm{g})\) and \(34.8 \mathrm{g} \mathrm{Ar}(\mathrm{g})\) at 7.15 atm pressure and \(26.7^{\circ} \mathrm{C} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

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