P4 + F2 = PF3

What mass of F2 is needed to produce 120 gm of PF3, if the reaction has 81.1 % yield?

To find the mass of F2 needed to produce 120 grams of PF3, we need to use the balanced equation and the given yield.

From the balanced equation:
P4 + 5 * F2 → 4 * PF3

The molar mass of PF3 is 87.98 g/mol, which means that 120 grams of PF3 is equal to 120 / 87.98 = 1.3642 moles of PF3.

Since the reaction has a yield of 81.1%, we need to calculate the theoretical yield (the mass of PF3 that would be obtained if the reaction went to completion) by dividing the actual yield by the percentage yield:
Theoretical yield = actual yield / percentage yield = 1.3642 mol / 0.811 = 1.6832 mol

From the balanced equation, we can see that 4 moles of PF3 are produced from 5 moles of F2. Therefore, 1.6832 mol of PF3 would require:
(1.6832 mol / 4 mol) * 5 mol = 2.103 moles of F2.

Finally, we can calculate the mass of F2 needed using its molar mass, which is 38.00 g/mol:
Mass of F2 = number of moles * molar mass = 2.103 mol * 38.00 g/mol = 79.914 g.

Therefore, approximately 79.914 grams of F2 are needed to produce 120 grams of PF3, assuming an 81.1% yield.

To determine the mass of F2 needed to produce 120 grams of PF3, we first need to calculate the moles of PF3 produced. Then, we can use the stoichiometry of the reaction to determine the moles of F2 required. Finally, we can convert the moles of F2 to grams.

Step 1: Calculate the moles of PF3 produced.
To calculate the moles of PF3, we can divide the given mass of PF3 (120 grams) by its molar mass. The molar mass of PF3 can be calculated by adding the molar masses of phosphorus (P) and three fluorine atoms (F).

Molar mass of PF3:
(1 × molar mass of P) + (3 × molar mass of F)

The molar mass of phosphorus (P) is approximately 31.0 g/mol, and the molar mass of fluorine (F) is approximately 19.0 g/mol.

Molar mass of PF3:
(1 × 31.0 g/mol) + (3 × 19.0 g/mol)

Step 2: Calculate the moles of F2 required.
Since the stoichiometry of the reaction is given as P4 + F2 = PF3, we can see that the ratio between P4 and PF3 is 1:1. This means that the moles of PF3 produced are equal to the moles of P4 present in the reaction.

Now, we can determine the moles of F2 required by using the stoichiometry, which states that one mole of PF3 is produced from two moles of F2.

So, the moles of F2 required would be half the moles of PF3 produced.

Step 3: Convert moles of F2 to grams.
To convert the moles of F2 to grams, we need to multiply the moles of F2 by the molar mass of F2.

The molar mass of F2 is approximately 38.0 g/mol.

Finally, we can calculate the mass of F2 required by multiplying the moles of F2 by the molar mass of F2.

Note: Since the problem states that the reaction has an 81.1% yield, we multiply the calculated mass of F2 by 0.811 to account for the yield.

Let's calculate the values step by step:

Step 1: Calculate the molar mass of PF3:
Molar mass of PF3 = (1 × molar mass of P) + (3 × molar mass of F) = (1 × 31.0 g/mol) + (3 × 19.0 g/mol) = 31.0 g/mol + 57.0 g/mol = 88.0 g/mol

Step 2: Calculate the moles of PF3 produced:
Moles of PF3 = mass of PF3 / molar mass of PF3 = 120 g / 88.0 g/mol

Step 3: Calculate the moles of F2 required:
Moles of F2 = moles of PF3 / 2

Step 4: Calculate the mass of F2 required (accounting for yield):
Mass of F2 = moles of F2 × molar mass of F2 × yield (81.1%)
= moles of F2 × 38.0 g/mol × 0.811

Substitute the values from the previous steps into the equation to get the final answer.

Just follow the steps.

http://www.jiskha.com/science/chemistry/stoichiometry.html
Use (120/0.811 = ?) for mass PF3.