Chemistry: A Molecular Approach (3rd Edition)

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

Chapter 4 - Sections 4.1-4.9 - Exercises - Problems by Topic - Page 187: 41c


$9.4mol\ TiCl_{4}$

Work Step by Step

$\frac{12.4mol\ TiCl_{4}}{1}= 12.4mol$ $\frac{18.8mol\ Cl_{2}}{2}= 9.4$ As can be seen $Cl_{2}$ is the limiting reagent so: $18.8mol\ Cl_{2}\times\frac{1mol\ TiCl_{4}}{2mol\ Cl_{2}}= 9.4mol\ TiCl_{4}$
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