Answer
See the answer below.
Work Step by Step
From stoichiometry, number of moles of oxygen: 2.0
$PV=nRT$
$V=2.0\ mol×0.082057\ atm.L/mol.K×273\ K/1\ atm$
$V=44.8\ L$
Chemistry and Chemical Reactivity (9th Edition)
by Kotz, John C.; Treichel, Paul M.; Townsend, John R.; Treichel, David A.
Published by
Cengage Learning
ISBN 10:
1133949649
ISBN 13:
978-1-13394-964-0
Chapter 10 Gases and Their Properties - Study Questions - Page 403e: 65
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
