Certainly! Let's go step-by-step to solve the problem of finding the theoretical yield of TiF₄ given 2.3 grams of Titanium (Ti) and 1.8 grams of Fluorine gas (F₂).
### Step 1: Write the Balanced Chemical Equation
The balanced reaction equation is:
[tex]\[ \text{Ti (s)} + 2 \text{F}_2 \text{(g)} \rightarrow \text{TiF}_4 \text{(s)} \][/tex]
### Step 2: Determine the Molar Mass of Reactants and Product
- Molar mass of Titanium (Ti): 47.87 g/mol
- Molar mass of Fluorine gas (F₂): 2 18.998 g/mol = 37.996 g/mol
- Molar mass of Titanium Tetrafluoride (TiF₄): 47.87 g/mol + 4 18.998 g/mol = 123.862 g/mol
### Step 3: Calculate the Moles of Each Reactant
- Moles of Ti:
[tex]\[ \text{moles of Ti} = \frac{\text{mass of Ti}}{\text{molar mass of Ti}} = \frac{2.3 \text{ g}}{47.87 \text{ g/mol}} \approx 0.048 \][/tex]
- Moles of F₂:
[tex]\[ \text{moles of F}_2 = \frac{\text{mass of F}_2}{\text{molar mass of F}_2} = \frac{1.8 \text{ g}}{37.996 \text{ g/mol}} \approx 0.047 \][/tex]
### Step 4: Determine the Limiting Reactant
According to the balanced equation, 1 mole of Ti reacts with 2 moles of F₂. Therefore:
- The moles of Ti required to react with the given F₂:
[tex]\[ \text{moles of Ti required} = \frac{\text{moles of F}_2}{2} = \frac{0.047}{2} \approx 0.0235 \][/tex]
Compare the moles of Ti required (0.0235) with the moles of Ti available (0.048):
- Since 0.0235 < 0.048, F₂ is the limiting reactant.
### Step 5: Calculate the Moles of TiF₄ Produced
Since F₂ is the limiting reactant, we use its moles to find the moles of TiF₄ produced:
- From the balanced equation, 2 moles of F₂ produce 1 mole of TiF₄. Hence, 0.047 moles of F₂ produce:
[tex]\[ \text{moles of TiF}_4 = \frac{0.047}{2} \approx 0.0235 \][/tex]
### Step 6: Calculate the Theoretical Yield in Grams
- The theoretical yield of TiF₄:
[tex]\[ \text{mass of TiF}_4 = \text{moles of TiF}_4 \times \text{molar mass of TiF}_4 = 0.0235 \times 123.862 \approx 2.91 \][/tex]
### Step 7: Express the Answer Using Two Significant Figures
Rounded to two significant figures, the theoretical yield of TiF₄ is 2.9 grams.
Thus, the theoretical yield of Titanium Tetrafluoride (TiF₄) from 2.3 grams of Titanium and 1.8 grams of Fluorine gas is:
[tex]\[ \boxed{2.9 \text{ grams}} \][/tex]
