## Chemistry: Molecular Approach (4th Edition)

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 $TiCl_4$
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 $TiCl_4$