Answer
$9.4mol\ TiCl_{4}$
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:
$18.8mol\ Cl_{2}\times\frac{1mol\ TiCl_{4}}{2mol\ Cl_{2}}= 9.4mol\ TiCl_{4}$
Chemistry: A Molecular Approach (3rd Edition)
by Tro, Nivaldo J.
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
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