Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 3 - Sections 3.1-3.12 - Exercises - Problems by Topic - Page 132: 62a

Answer

$989g\ HNO_{3}$

Work Step by Step

$15.7mol\ HNO_{3}\times\frac{63.01g\ HNO_{3}}{1mol\ HNO_{3}}= 989g\ HNO_{3}$
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