Answer
a. 8.1 g
b. 2.6 g
c. 0.469 g
Work Step by Step
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:
$$ 0.00379 \space mole \times \frac{ 123.87 \space g}{1 \space mole} = 0.469 \space g$$
