Let's analyze each step to compute the theoretical yield of TiFâ‚„ for the given reactants: 0.227 g of Ti and 0.296 g of Fâ‚‚.
1. Molecular Masses:
- Molar mass of Ti (Titanium): 47.87 g/mol
- Molar mass of Fâ‚‚ (Fluorine gas, where Fâ‚‚ consists of 2 Fluorine atoms): [tex]\(2 \times 18.998\)[/tex] g/mol = 37.996 g/mol
- Molar mass of TiFâ‚„ (Titanium tetrafluoride): 47.87 g/mol (for Ti) + [tex]\(4 \times 18.998\)[/tex] g/mol (for 4 Fluorine atoms) = 123.862 g/mol
2. Given Masses:
- Mass of Ti = 0.227 g
- Mass of Fâ‚‚ = 0.296 g
3. Calculate Moles of Each Reactant:
- Moles of Ti: [tex]\(\frac{0.227 \text{ g}}{47.87 \text{ g/mol}} = 0.004742\)[/tex] moles
- Moles of Fâ‚‚: [tex]\(\frac{0.296 \text{ g}}{37.996 \text{ g/mol}} = 0.007790\)[/tex] moles
4. Determine Limiting Reactant:
- According to the balanced equation [tex]\( \text{Ti} + 2 \text{F}_2 \rightarrow \text{TiF}_4 \)[/tex]:
- 1 mole of Ti reacts with 2 moles of Fâ‚‚ to produce 1 mole of TiFâ‚„.
- Moles of Ti needed to react with available Fâ‚‚: [tex]\( \frac{0.007790 \text{ moles of F}_2}{2} = 0.003895 \text{ moles of Ti needed} \)[/tex]
- Since 0.004742 moles of Ti are available, Ti is in excess. So, Fâ‚‚ is the limiting reactant.
5. Calculate Moles of TiFâ‚„ Produced:
- Based on the limiting reactant (Fâ‚‚), the moles of TiFâ‚„ produced: [tex]\( 0.007790 \text{ moles of F}_2 \div 2 = 0.003895 \)[/tex] moles of TiFâ‚„.
6. Calculate the Mass of TiFâ‚„ Produced (Theoretical Yield):
- Mass of TiFâ‚„: [tex]\( 0.003895 \text{ moles} \times 123.862 \text{ g/mol} \approx 0.482 \text{ g}\)[/tex].
Thus, for part C:
The theoretical yield of TiFâ‚„ when using 0.227 g of Ti and 0.296 g of Fâ‚‚ is 0.48 grams (rounded to two significant figures).
