Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 28 - Some Nonmetals and Metalloids - Exercises - The Noble Gases - Page 1080: 8

Answer

Balanced equation: XeF$_4$(s) + F$_2$(g) --> XeF$_6$(s) ? g XeF$_6$ = 1.85g XeF$_4$ $\times$ $\frac{1 mol XeF_4} {207 g XeF_4}$ $\times$ $\frac{1 mol XeF_6} {1 mol XeF_4}$ $\times$ $\frac{245g XeF_6} {1 mol XeF_6}$ = 2.19 g XeF$_6$

Work Step by Step

Balanced equation: XeF$_4$(s) + F$_2$(g) --> XeF$_6$(s) ? g XeF$_6$ = 1.85g XeF$_4$ $\times$ $\frac{1 mol XeF_4} {207 g XeF_4}$ $\times$ $\frac{1 mol XeF_6} {1 mol XeF_4}$ $\times$ $\frac{245g XeF_6} {1 mol XeF_6}$ = 2.19 g XeF$_6$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.