Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 5 - Thermochemistry - Exercises - Page 206: 5.46a


$\Delta H = -40.528kJ$

Work Step by Step

We can use a simple proportion calc to find the answer: If the formation of 3 moles of $O_2$ has a $\Delta H = -89.4kJ$, 1.36 moles will have: $\frac{3}{-89.4} = \frac{1.36}{x}$ $3x = 1.36 * - 89.4$ $3x = -121.584$ $x = -40.528kJ$
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