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: 5.46b


$\Delta H = -6.237 kJ $

Work Step by Step

2 KClO3 --> 2 KCl + 3O2 $\Delta H = -89.4 kJ $ From the coefficients we see that for the production of 2 moles of KCl, that 89.4 kJ of heat are released. To find how much is released for 10.4 g of KCl, we need to convert this to moles so we must find the molecular mass of KCl: K has a mass of 39.083 g/mol Cl has a mass of 35.45 g/mol KCl has a mass of 74.533 g/mol Then we have 10.4 g * (1mol KCl/ 74.533 g) * (-89.4 kJ/ 2 mole KCl) = -6.237 kJ
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