Answer
The theorectical yield of the product for that initial amount of reactants is equal to 0.469 g $TiF_4$.
Work Step by Step
1. Calculate the number of moles of $Ti$:
47.87* 1 = 47.87g/mol
$0.233g \times \frac{1 mol}{ 47.87g} = 4.87 \times 10^{-3}mol (Ti)$
- The balanced reaction is:
$Ti(s) + 2F_2(g) -- \gt TiF_4$
- Therefore:
The ratio of $Ti$ to $TiF_4$ is 1 to 1:
$4.87 \times 10^{-3} mol (Ti) \times \frac{ 1 mol(TiF_4)}{ 1 mol (Ti)} = 4.87 \times 10^{-3}mol (TiF_4)$
2. Calculate the number of moles of $F_2$:
19.00* 2 = 38.00g/mol
$0.288g \times \frac{1 mol}{ 38.00g} = 7.58 \times 10^{-3}mol (F_2)$
- 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:
$7.58 \times 10^{-3} mol (F_2) \times \frac{ 1 mol(TiF_4)}{ 2 mol (F_2)} = 3.79 \times 10^{-3}mol (TiF_4)$
Since the number of moles of $TiF_4$ produced by $F_2$ is lower, it is the limiting reactant, so that reaction will produce: $ 0.00379$ moles of $TiF_4$.
3. Calculate the mass of $TiF_4$:
47.87* 1 + 19.00* 4 = 123.87g/mol
$3.79 \times 10^{-3} mol \times \frac{ 123.87 g}{ 1 mol} = 0.469g (TiF_4)$
