Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 3 - Mass Relationships in Chemical Reactions - Questions & Problems: 3.68

Answer

19.6 mol $H_{2}O$

Work Step by Step

$2CH_{3}OH + 3O_{2}$ -> $2CO_{2}+ 4H_{2}O$ (excess $O_{2}$) mol of $H_{2}O$ = $\frac{(9.8molCH_{3}OH)\times(4molH_{2}O)}{2 mol CH_{3}OH}$ =19.6 mol $H_{2}O$
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