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: 3.77

Answer

$2.58\times10^4kg$ $NH_{3}$

Work Step by Step

$2NH_{3}+H_{2}SO_{4}$ -> $(NH_{4})_{2}SO_{4}$ mass of $NH_{3}$ (in Kg) = $\frac{(1.00\times10^8g(NH_{4})_{2}SO_{4})\times(1mol(NH_{4})_{2}SO_{4})\times(2molNH_{3})\times(17.04gNH_{3})}{(132.17g(NH_{4})_{2}SO_{4})\times(1mol(NH_{4})_{2}SO_{4})\times(1molNH_{3})\times(1\times10^3Kg)}$ =$2.58\times10^4kg$ $NH_{3}$
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