Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 2 Atoms, Molecules, and Ions - Study Questions: 93

Answer

$XeF_2$

Work Step by Step

Atomic weights (g/mol): $Xe: 131.29,\ F: 18.998$ Molar mass $F_2$: 38.00 g/mol Mass of Xe: 0.526 g Mass of the product: 0.678 g Mass of fluorine $F_2$: 0.678 - 0.526 = 0.152 g Number of moles: Xe: $0.526/131.29=3.99\times10^{-3}\ mol$ $F_2$: $0.152/38.00=4.00\times10^{-3}\ mol\ F_2 = 8.0\times10^{-3}\ mol\ F$ Normalizing by the smallest amount of moles: Xe: 1.0 F: 2.0 Empirical formula of the product: $XeF_2$
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.