Chemistry: The Molecular Nature of Matter and Change 7th Edition

Published by McGraw-Hill Education
ISBN 10: 007351117X
ISBN 13: 978-0-07351-117-7

Chapter 5 - Problems - Page 249: 5.156

Answer

Pfinal = 0.323 atm and PI[sub]2[/sub] = 0.0332 atm

Work Step by Step

I2(s) + 7F2(g) --> 2IF7(g) Initially, nF[sub]2[/sub] = PV/RT = (350 x 1/760)(2.50)/(0.0821)(250) = 0.0561 mol F2(g) nI[sub]2[/sub] = 2.50/126.9*2 = 0.0098 mol I2(s) The F2 gas is the limiting reactant. Therefore, 0.0561 mol F2 x (1 mol I2 / 7 mol F2) = 8.01x10-3 mol I2 consumed in reaction. 0.0561 mol F2 x (2 mol IF7 / 7 mol F2) = 0.0160 mol IF7 produced. nI[sub]2[/sub] remaining = 0.0098 - 8.01x10-3 = 0.0018 mol I2(g) remaining. Pfinal = ntotalRT/V = (nI[sub]2 remaining[/sub] + nIF[sub]7[/sub])RT/V = (0.0117 + 0.0160)(0.0821)(550)/2.50 = 0.323 atm. PI[sub]2[/sub] = nI[sub]2[/sub]RT/V = (0.0018)(0.0821)(550)/2.50 = 0.0332 atm.
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.