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 110: 3.71

Answer

20.8 g $NCl_{3}$

Work Step by Step

$3NaClO+NH_{3}$ -> $NCl_{3}+3NaOH$ Mass of $NCl_{3}$ = $\frac{(2.94gNH{3})\times(1molNH_{3})\times(1molNCl_{3}}{(17.04gNH_{3})\times(1molNH_{3})\times(1molNCl_{3})}$ $\approx$20.7663380… =20.8 g $NCl_{3}$
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