What mass of F2 is needed to produce 116 g of PF3 if the reaction has a 77.4% yield?

Well, to calculate the mass of F2 needed, we need to first determine the stoichiometry of the reaction between F2 and PF3. From the balanced equation:

2 F2 + 3 PF3 -> 6 F + P2F6

We can see that 2 moles of F2 react with 3 moles of PF3 to produce 6 moles of F.

Now, let's calculate the number of moles of PF3 using the given mass and its molar mass (which is around 87 g/mol):

116 g PF3 / 87 g/mol ≈ 1.333 moles of PF3

Since the reaction has a 77.4% yield, only 77.4% of the PF3 will be produced, which is:

0.774 * 1.333 moles of PF3 ≈ 1.033 moles of PF3

Using the stoichiometry of the balanced equation above, we can determine the amount of F2 required:

1.033 moles PF3 × (2 moles F2 / 3 moles PF3) ≈ 0.6887 moles F2

Finally, to convert moles of F2 to grams, we'll use its molar mass (around 38 g/mol):

0.6887 moles F2 × 38 g/mol ≈ 26.17 g F2

Therefore, approximately 26.17 grams of F2 would be needed to produce 116 grams of PF3 with a 77.4% yield. But hey, just remember, in chemistry, sometimes you just gotta go with the flow and not be too square about things!

To determine the mass of F2 needed to produce 116 g of PF3 with a 77.4% yield, we need to use the stoichiometry of the reaction between F2 and PF3.

The balanced equation for the reaction is:
3 F2 + 2 PF3 -> 2 P + 6 F

From the balanced equation, we can see that 2 moles of PF3 are produced for every 3 moles of F2.

First, we need to calculate the number of moles of PF3 produced:

Moles of PF3 = Mass of PF3 / Molar mass of PF3

The molar mass of PF3 is calculated as follows:
Molar mass of P = 30.97 g/mol
Molar mass of F = 18.998 g/mol
Total molar mass of PF3 = (30.97 g/mol) + 3 * (18.998 g/mol) = 87.964 g/mol

Moles of PF3 = 116 g / 87.964 g/mol = 1.32 mol

Since the yield of the reaction is given as 77.4%, we need to calculate the moles of F2 required based on the moles of PF3 produced:

Moles of F2 = Moles of PF3 * (3 mol F2 / 2 mol PF3)

Moles of F2 = 1.32 mol * (3 mol F2 / 2 mol PF3) = 1.98 mol

Now, we can calculate the mass of F2:

Mass of F2 = Moles of F2 * Molar mass of F2

The molar mass of F2 is calculated as follows:
Molar mass of F = 18.998 g/mol
Total molar mass of F2 = 2 * (18.998 g/mol) = 37.996 g/mol

Mass of F2 = 1.98 mol * 37.996 g/mol = 75.17 g

Therefore, approximately 75.17 g of F2 is needed to produce 116 g of PF3 with a yield of 77.4%.

P + 3F2 ==> 2PF3

mols PF3 = 116/molar mass PF3 = about 1.3.
Convert mols PF3 to mols F2.
1.3 mols PF3 x (3 mols F2/2 mols PF3) = 1.3 x (3/2) = 1.95 mols F2.
How many g is that? 1.95 x molar mass F2 = about 74 g F2 (You shoud redo all of this more accurately.)
Since this is only 77.4% yield, then
74/0.774 = g F2 needed to produce 116 g F2.

You could go another route. You could have said 77.4% of what number is 116 which gives 116/0.774 = about 149.9 and use that number for grams PF3, then calculate grams F2 needed to produce that. You should get the same answer either way.

Sorry bro I don't know