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 $H_{2}O$
Chemistry 12th Edition
by Chang, Raymond; Goldsby, Kenneth
Published by
McGraw-Hill Education
ISBN 10:
0078021510
ISBN 13:
978-0-07802-151-0
Chapter 3 - Mass Relationships in Chemical Reactions - Questions & Problems - Page 109: 3.68
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