Consider the following unbalanced reaction.

P4(s) + F2(g) PF3(g)
How many grams of F2 are needed to produce 212 g of PF3 if the reaction has a 79.8% yield?

Follow the directions I gave you before for determining the amount of product (stoichiometry problem) except that calculation was for a 100% yield. For this one start with 212/0.798 = ? grams in order to end up with 212 with a 79.8% yield. I shall be happy to check your calculations if you care to post them or if you get stuck explain where you are in the process and what you're having trouble with.

To solve this problem, we'll follow these steps:

Step 1: Write and balance the chemical equation:
P4(s) + F2(g) → PF3(g)

Step 2: Calculate the molar mass of the compounds involved:
- Molar mass of P4 = 4 x atomic mass of P = 4 x 31.0 g/mol = 124.0 g/mol
- Molar mass of F2 = 2 x atomic mass of F = 2 x 19.0 g/mol = 38.0 g/mol
- Molar mass of PF3 = atomic mass of P + 3 x atomic mass of F = 31.0 g/mol + 3 x 19.0 g/mol = 31.0 g/mol + 57.0 g/mol = 88.0 g/mol

Step 3: Determine the stoichiometry of the balanced equation:
According to the balanced equation, 1 mol of P4 reacts with 6 mol of F2 to produce 4 mol of PF3.

Step 4: Convert the given mass of PF3 to moles:
Moles of PF3 = Mass of PF3 / Molar mass of PF3
Moles of PF3 = 212 g / 88.0 g/mol ≈ 2.4091 mol

Step 5: Determine the moles of F2 required using the stoichiometry of the balanced equation:
From the stoichiometry, 1 mol of P4 reacts with 6 mol of F2 to produce 4 mol of PF3.
So, the moles of F2 required = (2.4091 mol of PF3) x (6 mol of F2 / 4 mol of PF3) = 3.6137 mol of F2

Step 6: Convert the moles of F2 to grams:
Mass of F2 = Moles of F2 x Molar mass of F2
Mass of F2 = 3.6137 mol x 38.0 g/mol ≈ 137.20 g

However, we have to account for the reaction yield of 79.8% in this question. To do that, we multiply the calculated mass of F2 by the yield percentage:

Mass of F2 needed = Mass of F2 x Yield percentage
Mass of F2 needed = 137.20 g x 0.798 ≈ 109.44 g

Therefore, approximately 109.44 grams of F2 are needed to produce 212 grams of PF3 with a 79.8% yield.