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 - Page 109: 3.67


9.0 moles of $H_{2}$

Work Step by Step

$3H_{2}+N_{2}$ -> $2NH_{3}$ mole of $H_{2}$ = $\frac{(6molNH_{3})\times(3molH_{2})}{2molNH_{3}}$ =9.0 moles of $H_{2}$
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