## Chemistry: A Molecular Approach (3rd Edition)

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