Ti(s) + 2 F_{2} → TiF_{4}

Calculate the amount of TiF_{4} produced from each reactant and determine the limiting reactant:

mass Ti (molar mass) → moles Ti (mole-to-mole comparison) → moles TiF_{4} (molar mass) → mass TiF_{4}

molar mass Ti = 47.867 g/mol

molar mass TiF_{4} = 123.867 g/mol

$\mathbf{mass}\mathbf{}{\mathbf{TiF}}_{\mathbf{4}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Ti}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{Ti}}}{\mathbf{47}\mathbf{.}\mathbf{867}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Ti}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{TiF}}_{\mathbf{4}}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{Ti}}}\mathbf{\times}\frac{\mathbf{123}\mathbf{.}\mathbf{867}\mathbf{}\mathbf{g}\mathbf{}{\mathbf{TiF}}_{\mathbf{4}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{TiF}}_{\mathbf{4}}}}$

**mass TiF _{4} = 6.47 g**

mass F_{2} (molar mass) → moles F_{2} (mole-to-mole comparison) → moles TiF_{4} (molar mass) → mass TiF_{4}

For the reaction

Ti(s) + 2F_{2} → TiF_{4}

compute the theoretical yield of the product (in grams) for each of the following initial amounts of reactants.

2.5 g Ti, 1.7 g F_{2}

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