Answer
a. The reaction produces 2 moles of $TiCl_4$
b. The reaction produces 7 moles of $TiCl_4$
c. The reaction produces 9.40 moles of $TiCl_4$
Work Step by Step
a.
Find the amount of product if each reactant is completely consumed.
$$ 4 \space moles \space Ti \times \frac{ 1 \space mole \ TiCl_4 }{ 1 \space mole \space Ti } = 4.0 \space moles \space TiCl_4 $$
$$ 4 \space moles \space Cl_2 \times \frac{ 1 \space mole \ TiCl_4 }{ 2 \space moles \space Cl_2 } = 2.0 \space moles \space TiCl_4 $$
Since the reaction of $ Cl_2 $ produces less $ TiCl_4 $ for these quantities, it is the limiting reactant.
The reaction produces 2 moles of $TiCl_4$
b.
Find the amount of product if each reactant is completely consumed.
$$ 7 \space moles \space Ti \times \frac{ 1 \space mole \ TiCl_4 }{ 1 \space mole \space Ti } = 7 \space moles \space TiCl_4 $$
$$ 17 \space moles \space Cl_2 \times \frac{ 1 \space mole \ TiCl_4 }{ 2 \space moles \space Cl_2 } = 8.5 \space moles \space TiCl_4 $$
Since the reaction of $ Ti $ produces less $ TiCl_4 $ for these quantities, it is the limiting reactant.
The reaction produces 7 moles of $TiCl_4$
c.
Find the amount of product if each reactant is completely consumed.
$$ 12.4 \space moles \space Ti \times \frac{ 1 \space mole \ TiCl_4 }{ 1 \space mole \space Ti } = 12.4 \space moles \space TiCl_4 $$
$$ 18.8 \space moles \space Cl_2 \times \frac{ 1 \space mole \ TiCl_4 }{ 2 \space moles \space Cl_2 } = 9.40 \space moles \space TiCl_4 $$
Since the reaction of $ Cl_2 $ produces less $ TiCl_4 $ for these quantities, it is the limiting reactant.
The reaction produces 9.40 moles of $TiCl_4$
