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: 3.84

Answer

20.96g $H_{2}SO_{4}$

Work Step by Step

a. $2NH_{3}(aq)+1H_{2}SO_{4}(aq)$ -> $ 1(NH_{4})_{2}SO_{4}(aq)$ b. Mass of $NH_{3}$ = $\frac{(20.3g(NH_{4})_{2}SO_{4})\times((NH_{4})_{2}SO_{4})\times(2molNH_{3})\times(17.04molNH_{3})}{(132.17g(NH_{4})_{2}SO_{4})\times(1mol(NH_{4})_{2}SO_{4})\times(1molNH_{3})}$ = 5.23 g $NH_{3}$ Mass of $H_{2}SO_{4}$= $\frac{(20.3g(NH_{4})_{2}SO_{4})\times(1mol(NH_{4})_{2}SO_{4})\times(1mol(NH_{4})_{2}SO_{4})\times(1molH_{2}SO_{4})\times(98.09gH_{2}SO_{4})}{(132.17g(NH_{4})_{2}SO_{4})\times(1mol(NH_{4})_{2}SO_{4})\times(1molH_{2}SO_{4})}$ = 15.07 g 15.07 + 5.89 = 20.96g $H_{2}SO_{4}$
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