`PF_(3)` reacts with `XeF_(4)` to give `PF_(5)`
`underset((g))(2PF_(3))tounderset((s))(XeF_(4))tounderset((g))(2PF_(5))+underset((g))(Xe)`
If 100.0gm of `PF_(3)` and 50.0 gm `XeF_(4)` react, then which of the following statement is true?

Options

A. `XeF_(4)` is the limiting reagent
B. `PF_(3)` is the limiting reagent
C. 1.127 mol of `PF_(5)` are produced
D. 0.382 mol of `PF_(5)` are produced

Answer

To solve the problem, we need to analyze the reaction between PF₃ and XeF₄, determine the number of moles of each reactant, and identify the limiting reagent.

### Step 1: Write the balanced chemical equation.
The balanced equation for the reaction is:
\[ 2 \text{PF}_3 (g) + \text{XeF}_4 (s) \rightarrow 2 \text{PF}_5 (g) + \text{Xe} (g) \]

### Step 2: Calculate the molar masses of the reactants.
- Molar mass of PF₃:
- P: 30.97 g/mol
- F: 19.00 g/mol
- Molar mass of PF₃ = 30.97 + (3 × 19.00) = 30.97 + 57.00 = 87.97 g/mol

- Molar mass of XeF₄:
- Xe: 131.29 g/mol
- F: 19.00 g/mol
- Molar mass of XeF₄ = 131.29 + (4 × 19.00) = 131.29 + 76.00 = 207.29 g/mol

### Step 3: Calculate the number of moles of each reactant.
- Moles of PF₃:
\[ \text{Moles of PF}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{100.0 \, \text{g}}{87.97 \, \text{g/mol}} \approx 1.13 \, \text{mol} \]

- Moles of XeF₄:
\[ \text{Moles of XeF}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{50.0 \, \text{g}}{207.29 \, \text{g/mol}} \approx 0.24 \, \text{mol} \]

### Step 4: Determine the limiting reagent.
According to the balanced equation, 2 moles of PF₃ react with 1 mole of XeF₄. Therefore, we can find out how many moles of PF₃ are required for the available moles of XeF₄:
- Required moles of PF₃ for 0.24 moles of XeF₄:
\[ \text{Required moles of PF}_3 = 2 \times \text{moles of XeF}_4 = 2 \times 0.24 = 0.48 \, \text{mol} \]

Since we have 1.13 moles of PF₃ available, which is more than the 0.48 moles required, XeF₄ is the limiting reagent.

### Step 5: Conclusion
Based on the calculations, the true statement is that **XeF₄ is the limiting reagent** in this reaction.