To solve the problem, we need to follow a few steps:
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Determine the molar mass of PF3.
- The atomic masses are approximately as follows:
- Phosphorus (P): 30.97 g/mol
- Fluorine (F): 19.00 g/mol
- The molar mass of PF3 = 30.97 g/mol + (3 × 19.00 g/mol) = 30.97 g/mol + 57.00 g/mol = 87.97 g/mol.
- The atomic masses are approximately as follows:
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Calculate the number of moles of PF3 produced.
- Using the mass given, the number of moles of PF3 can be calculated as: \[ \text{moles of PF3} = \frac{85 \text{ g}}{87.97 \text{ g/mol}} \approx 0.966 \text{ moles} \]
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Determine the theoretical yield of PF3 based on the given yield percentage.
- The reaction has a 64.9% yield, thus the theoretical moles of PF3 (\( n_{\text{theoretical}} \)) required to obtain the 0.966 moles of PF3 produced can be calculated as: \[ n_{\text{theoretical}} = \frac{0.966 \text{ moles}}{0.649} \approx 1.489 \text{ moles} \]
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Determine the number of moles of P4 required.
- The reaction shows that 1 mole of P4 produces 4 moles of PF3. Thus, the number of moles of P4 required is: \[ n_{P4} = \frac{n_{\text{theoretical PF3}}}{4} = \frac{1.489}{4} \approx 0.37225 \text{ moles} \]
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Calculate the mass of P4 required.
- The molar mass of P4 (where P has a mass of 30.97 g/mol) is: \[ \text{molar mass of P4} = 4 \times 30.97 \text{ g/mol} \approx 123.88 \text{ g/mol} \]
- Now, we can calculate the mass of P4: \[ \text{mass of P4} = n_{P4} \times \text{molar mass of P4} = 0.37225 \text{ moles} \times 123.88 \text{ g/mol} \approx 46.11 \text{ g} \]
So, the mass of P4 needed to produce 85 g of PF3 with a 64.9% yield is approximately 46.11 grams.
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