QuestionApril 23, 2025

Consider the following equation. P_(4)(s)+F_(2)(g)arrow PF_(3)(g) Calculate the moles of F_(2) that will be required to produce 27.5 grams of PF_(3)

Consider the following equation. P_(4)(s)+F_(2)(g)arrow PF_(3)(g) Calculate the moles of F_(2) that will be required to produce 27.5 grams of PF_(3)
Consider the following equation. P_(4)(s)+F_(2)(g)arrow PF_(3)(g) Calculate the moles of F_(2) that will be required to produce 27.5 grams of PF_(3)

Solution
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Answer

0.313 moles of F_{2} required. Explanation 1. Calculate moles of PF_{3} Use the formula: \text{moles} = \frac{\text{mass}}{\text{molar mass}}. The molar mass of PF_{3} is 30.97 + 3 \times 19.00 = 87.97 g/mol. Moles of PF_{3} = \frac{27.5}{87.97}. 2. Determine moles of F_{2} From the balanced equation, 1 mole of PF_{3} requires 1 mole of F_{2}. Therefore, moles of F_{2} needed are equal to moles of PF_{3} calculated.

Explanation

1. Calculate moles of $PF_{3}$<br /> Use the formula: $\text{moles} = \frac{\text{mass}}{\text{molar mass}}$. The molar mass of $PF_{3}$ is $30.97 + 3 \times 19.00 = 87.97$ g/mol. Moles of $PF_{3} = \frac{27.5}{87.97}$.<br />2. Determine moles of $F_{2}$<br /> From the balanced equation, 1 mole of $PF_{3}$ requires 1 mole of $F_{2}$. Therefore, moles of $F_{2}$ needed are equal to moles of $PF_{3}$ calculated.
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