Answered: 25.0g AlCl3 was mixed with 30.0g H2SO4.…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Aluminum sulfate can be made by the following reaction:

2AlCl_{3 (aq)} + 3H₂SO₄ _(aq) → Al₂(SO₄)_{3 (aq)} + 6HCl _(aq)

It is quite soluble in water, so to isolate it, the solution has to be evaporated to dryness to drive off the volatile HCl and heated to high tempertures to drive off all the water.

Here, 25.0g AlCl₃ was mixed with 30.0g H₂SO₄. Eventually, 28.46g pure aluminum sulfate Al₂(SO₄)₃ was isolated. What is the percentage yield?

Any help for how to solve this question would be greatly appreciated :)

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