Unbalanced reaction question for Chem

Consider the following unbalanced reaction.

P4(s) + F2(g) --> PF3(g)

How many grams of F2 are needed to produce 113 g of PF3, if the reaction has a 75.6% yield?

So when balanced, it's:

P4(s) + 6F2(g) --> 4PF3(g)

with a 75.6% yield, you would actually need to shoot for:
113g * 100 / 75.6 = 149.47g of PF3

149.47g PF3 * (1 mol PF3 / 87.964g PF3) * (6 mol F2 / 4 mol PF3) = 2.54884 mol F2
2.54884 mol F2 * (37.996g F2 / 1 mol F2) = 96.8g of F2 is needed

The Balanced equation would be (P4) + 6(F2)--> 4(PF3)

Now convert 113g PF3 to moles. : 113g x (1 mol/ 88 (g/mol)) = 1.28 mol PF3
Now divide 1.28 mol PF3 by 75.6% to get the amount of product in a perfect world.
=1.69 mol PF 3
Now we use stoichiometry:
1.69 mol PF3 x (6 mol F2/ 4 mol PF3) = 2.54 mol F2 needed
Now change back to grams:
2.54 mol F2 x (38g / 1 mol F2) = 96.52g F2
So it should be about 96.52g F2

P4 + (6)F2 yields (4)PF3

(100)(113.6) / (x) = 75.6 therefore x= 150.26
150.26 x 1 mol / (97.97) x 6 mol/4mol x (19) / 1 mol = 43.41 g