Answer
The theorectical yield of the product for that initial amount of reactants is equal to 8.1 g $TiF_4$.
Work Step by Step
1. Calculate the number of moles of $Ti$:
47.87* 1 = 47.87g/mol
$5.0g \times \frac{1 mol}{ 47.87g} = 0.10mol (Ti)$
- The balanced reaction is:
$Ti(s) + 2F_2(g) -- \gt TiF_4$
- Therefore:
The ratio of $Ti$ to $TiF_4$ is 1 to 1:
$0.10 mol (Ti) \times \frac{ 1 mol(TiF_4)}{ 1 mol (Ti)} = 0.10mol (TiF_4)$
2. Calculate the number of moles of $F_2$:
19.00* 2 = 38.00g/mol
$5.0g \times \frac{1 mol}{ 38.00g} = 0.13mol (F_2)$
- The balanced reaction is:
$Ti(s) + 2F_2(g) -- \gt TiF_4$
- Therefore:
The ratio of $F_2$ to $TiF_4$ is 2 to 1:
$0.13 mol (F_2) \times \frac{ 1 mol(TiF_4)}{ 2 mol (F_2)} = 0.065mol (TiF_4)$
Since the number of moles of $TiF_4$ produced by $F_2$ is lower, it is the limiting reactant, so that reaction will produce: $ 0.065$ moles of $TiF_4$.
3. Calculate the mass of $TiF_4$:
47.87* 1 + 19.00* 4 = 123.87g/mol
$0.065 mol \times \frac{ 123.87 g}{ 1 mol} = 8.1g (TiF_4)$