Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 98

A mixture contains only \(\mathrm{NaCl}\) and \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\). A 1.45 -g sample of the mixture is dissolved in water and an excess of NaOH is added, producing a precipitate of \(\mathrm{Al}(\mathrm{OH})_{3}\). The precipitate is filtered, dried, and weighed. The mass of the precipitate is \(0.107 \mathrm{g} .\) What is the mass percent of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in the sample?

Short Answer

Expert verified
The mass percent of $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ in the 1.45-g sample is approximately 16.14%.

Step by step solution

01

Calculate the moles of Al(OH)3 formed

First, we need to find the moles of Al(OH)3 formed in the reaction. We can do this by using the molar mass of Al(OH)3 and the given mass of the precipitate. The molar mass of Al(OH)3 is: Al: 26.98 g/mol O: 16.00 g/mol H: 1.01 g/mol Molar mass of Al(OH)3 = Al + 3(O + 3H) = 26.98 + 3(16.00 + 3(1.01)) ≈ 78.00 g/mol Now we can calculate the moles of Al(OH)3 formed: Moles of Al(OH)3 = mass of precipitate / molar mass of Al(OH)3 Moles of Al(OH)3 = 0.107 g / 78.00 g/mol ≈ 0.00137 mol
02

Calculate the moles of Al2(SO4)3 in the sample

Since there is a 1:1 ratio between moles of Al(OH)3 and moles of Al2(SO4)3, we can simply use the moles of Al(OH)3 to find the moles of Al2(SO4)3 in the sample. Moles of Al2(SO4)3 = Moles of Al(OH)3 / 2 = 0.00137 mol / 2 ≈ 0.000685 mol
03

Calculate the mass of Al2(SO4)3 in the sample

Now, we can find the mass of Al2(SO4)3 in the sample by multiplying the moles of Al2(SO4)3 by the molar mass of Al2(SO4)3. The molar mass of Al2(SO4)3 is: Al: 26.98 g/mol S: 32.07 g/mol O: 16.00 g/mol Molar mass of Al2(SO4)3 = 2(Al) + 3(2S + 4O) = 2(26.98) + 3(2(32.07) + 4(16.00)) ≈ 342.15 g/mol Mass of Al2(SO4)3 = moles of Al2(SO4)3 * molar mass of Al2(SO4)3 Mass of Al2(SO4)3 = 0.000685 mol * 342.15 g/mol ≈ 0.234 g
04

Calculate the mass percent of Al2(SO4)3 in the sample

Finally, we can find the mass percent of Al2(SO4)3 in the sample by dividing the mass of Al2(SO4)3 by the total mass of the sample and multiplying by 100. Mass percent of Al2(SO4)3 = (Mass of Al2(SO4)3 / Mass of the sample) * 100 Mass percent of Al2(SO4)3 = (0.234 g / 1.45 g) * 100 ≈ 16.14 % Thus, the mass percent of Al2(SO4)3 in the sample is approximately 16.14%.

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

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CuSO}_{4}(a q)\) b. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CaCl}_{2}(a q)\) c. \(\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+\mathrm{MgI}_{2}(a q)\) d. \(\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q)+\mathrm{AlBr}_{3}(a q)\)

In the spectroscopic analysis of many substances, a series of standard solutions of known concentration are measured to generate a calibration curve. How would you prepare standard solutions containing \(10.0,25.0,50.0,75.0,\) and \(100 .\) ppm of copper from a commercially produced 1000.0 -ppm solution? Assume each solution has a final volume of \(100.0 \mathrm{mL}\). (See Exercise 123 for definitions.)

Calculate the concentration of all ions present when \(0.160 \mathrm{g}\) of \(\mathrm{MgCl}_{2}\) is dissolved in \(100.0 \mathrm{mL}\) of solution.

Which of the following solutions of strong electrolytes contains the largest number of ions: \(100.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\), \(50.0 \mathrm{mL}\) of \(0.200 \mathrm{M} \mathrm{BaCl}_{2},\) or \(75.0 \mathrm{mL}\) of \(0.150 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4} ?\)

The concentration of a certain sodium hydroxide solution was determined by using the solution to titrate a sample of potassium hydrogen phthalate (abbreviated as KHP). KHP is an acid with one acidic hydrogen and a molar mass of \(204.22 \mathrm{g} / \mathrm{mol}\). In the titration, \(34.67 \mathrm{mL}\) of the sodium hydroxide solution was required to react with 0.1082 g KHP. Calculate the molarity of the sodium hydroxide.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Making Measurements

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