Questions: Calculate the needed amount of reactant for the reaction described below by constructing a BCA table and determining the grams of F2 required. Complete Parts 1-2 before submitting your answer. P4(s) + 6 F2(g) → 4 PF3(g) A mixture of F2 and solid P4 produce 120.g of P3. Assume there are 1000. g of solid P4 available. Based on your knowledge of stoichiometry, set up the table below that represents 100% yield with the given reaction conditions.

Calculate the needed amount of reactant for the reaction described below by constructing a BCA table and determining the grams of F2 required. Complete Parts 1-2 before submitting your answer.

P4(s) + 6 F2(g) → 4 PF3(g)

A mixture of F2 and solid P4 produce 120.g of P3. Assume there are 1000. g of solid P4 available. Based on your knowledge of stoichiometry, set up the table below that represents 100% yield with the given reaction conditions.

Transcript text: Calculate the needed amount of reactant for the reaction described below by constructing a BCA table and determining the grams of $F_{2}$ required. Complete Parts 1-2 before submitting your answer. \[ \mathrm{P}_{4}(\mathrm{~s})+6 \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{PF}_{3}(\mathrm{~g}) \] A mixture of $F_{2}$ and solid $P_{4}$ produce 120.g of $P_{3}$. Assume there are 1000. g of solid $P_{4}$ available. Based on your knowledge of stoichiometry, set up the table below that represents $100 \%$ yield with the given reaction conditions.

Solution

Solution Steps

Step 1: Write the Balanced Chemical Equation

The balanced chemical equation for the reaction is: \[ \mathrm{P}_{4}(\mathrm{~s}) + 6 \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{PF}_{3}(\mathrm{~g}) \]

Step 2: Determine the Molar Masses

Calculate the molar masses of the reactants and products:

Molar mass of $ \mathrm{P}_{4} $: $ 4 \times 30.9738 \, \text{g/mol} = 123.8952 \, \text{g/mol} $
Molar mass of $ \mathrm{F}_{2} $: $ 2 \times 18.9984 \, \text{g/mol} = 37.9968 \, \text{g/mol} $
Molar mass of $ \mathrm{PF}_{3} $: $ 30.9738 + 3 \times 18.9984 \, \text{g/mol} = 87.9682 \, \text{g/mol} $

Step 3: Convert Grams of $ \mathrm{PF}_{3} $ to Moles

Given 120.0 g of $ \mathrm{PF}_{3} $: \[ \text{Moles of } \mathrm{PF}_{3} = \frac{120.0 \, \text{g}}{87.9682 \, \text{g/mol}} = 1.3646 \, \text{mol} \]

Step 4: Use Stoichiometry to Find Moles of $ \mathrm{F}_{2} $

From the balanced equation, 4 moles of $ \mathrm{PF}_{3} $ are produced from 6 moles of $ \mathrm{F}_{2} $: \[ \text{Moles of } \mathrm{F}_{2} = 1.3646 \, \text{mol} \times \frac{6 \, \text{mol} \, \mathrm{F}_{2}}{4 \, \text{mol} \, \mathrm{PF}_{3}} = 2.0469 \, \text{mol} \]

Step 5: Convert Moles of $ \mathrm{F}_{2} $ to Grams

\[ \text{Grams of } \mathrm{F}_{2} = 2.0469 \, \text{mol} \times 37.9968 \, \text{g/mol} = 77.78 \, \text{g} \]