## Chemistry: Molecular Approach (4th Edition)

a. - Calculate or find the molar mass for $Ti$: $Ti$ : 47.87 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$5.0 \space g \times \frac{1 \space mole}{ 47.87 \space g} = 0.10 \space mole$$ - Calculate or find the molar mass for $F_2$: $F_2$ : ( 19.00 $\times$ 2 )= 38.00 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$5.0 \space g \times \frac{1 \space mole}{ 38.00 \space g} = 0.13 \space mole$$ Find the amount of product if each reactant is completely consumed. $$0.10 \space mole \space Ti \times \frac{ 1 \space mole \ TiF_4 }{ 1 \space mole \space Ti } = 0.10 \space mole \space TiF_4$$ $$0.13 \space mole \space F_2 \times \frac{ 1 \space mole \ TiF_4 }{ 2 \space moles \space F_2 } = 0.065 \space mole \space TiF_4$$ Since the reaction of $F_2$ produces less $TiF_4$ for these quantities, it is the limiting reactant. - Calculate or find the molar mass for $TiF_4$: $TiF_4$ : ( 47.87 $\times$ 1 )+ ( 19.00 $\times$ 4 )= 123.87 g/mol - Using the molar mass as a conversion factor, find the mass in g: $$0.065 \space mole \times \frac{ 123.87 \space g}{1 \space mole} = 8.1 \space g$$ b. - Using the molar mass as a conversion factor, find the amount in moles: $$2.4 \space g \times \frac{1 \space mole}{ 47.87 \space g} = 0.050 \space mole$$ - Using the molar mass as a conversion factor, find the amount in moles: $$1.6 \space g \times \frac{1 \space mole}{ 38.00 \space g} = 0.042 \space mole$$ Find the amount of product if each reactant is completely consumed. $$0.050 \space mole \space Ti \times \frac{ 1 \space mole \ TiF_4 }{ 1 \space mole \space Ti } = 0.050 \space mole \space TiF_4$$ $$0.042 \space mole \space F_2 \times \frac{ 1 \space mole \ TiF_4 }{ 2 \space moles \space F_2 } = 0.021 \space mole \space TiF_4$$ Since the reaction of $F_2$ produces less $TiF_4$ for these quantities, it is the limiting reactant. - Using the molar mass as a conversion factor, find the mass in g: $$0.021 \space mole \times \frac{ 123.87 \space g}{1 \space mole} = 2.6 \space g$$ c. - Using the molar mass as a conversion factor, find the amount in moles: $$0.233 \space g \times \frac{1 \space mole}{ 47.87 \space g} = 0.00487 \space mole$$ - Using the molar mass as a conversion factor, find the amount in moles: $$0.288 \space g \times \frac{1 \space mole}{ 38.00 \space g} = 0.00758 \space mole$$ Find the amount of product if each reactant is completely consumed. $$0.00487 \space mole \space Ti \times \frac{ 1 \space mole \ TiF_4 }{ 1 \space mole \space Ti } = 0.00487 \space mole \space TiF_4$$ $$0.00758 \space mole \space F_2 \times \frac{ 1 \space mole \ TiF_4 }{ 2 \space moles \space F_2 } = 0.00379 \space mole \space TiF_4$$ Since the reaction of $F_2$ produces less $TiF_4$ for these quantities, it is the limiting reactant. - Using the molar mass as a conversion factor, find the mass in g: $$0.00379 \space mole \times \frac{ 123.87 \space g}{1 \space mole} = 0.469 \space g$$