Consider the following unbalanced reaction.

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

How many grams of F2 are needed to produce 114. g of PF3 if the reaction has a 78.1% yield?

i tried it and got this

114g PF3 x (mol PF3/87.97g PF3) x (6 mol F2/ mol PF3) x (38g F2/mol F2)=73.9gF2

then plugged it into the % yield equation and got 57.7

what did i do wrong??

You're ok as far as you went but you didn't do the % yield right.

If the yield is only 78.1%, then you know that you must take MORE, not less than 73.9.
73.9/0.781 = ??

Your approach to solving the problem seems correct, but there might be a mistake in your calculations.

Let's go through the steps again:

1. Calculate the number of moles of PF3:
114 g PF3 x (1 mol PF3 / 87.97 g PF3) = 1.297 mol PF3

2. Use the balanced equation to determine the number of moles of F2 required:
1.297 mol PF3 x (6 mol F2 / 4 mol PF3) = 1.9465 mol F2

3. Convert the moles of F2 to grams using the molar mass of F2:
1.9465 mol F2 x (38 g F2 / 1 mol F2) = 73.867 g F2

So, the correct answer for the grams of F2 needed to produce 114.0g of PF3 is approximately 73.9 g F2. It seems like you made a rounding error in your calculation, which could explain the slight difference in the value. The calculated value is correct, and there is no need to consider the yield at this stage since it is already factored into the stoichiometry calculations.

Your initial calculations to find the amount of F2 needed to produce 114 g of PF3 were correct. However, there seems to be an error in your final calculation when applying the percentage yield.

To calculate the actual yield of PF3, you need to multiply the theoretical yield (73.9 g F2) by the percentage yield (78.1%):

Actual yield = Theoretical yield × Percentage yield
= 73.9 g F2 × (78.1/100)
≈ 57.7 g F2

So, it appears that your initial calculation for the moles of F2 and the subsequent conversion to grams was correct. The discrepancy lies in applying the percentage yield.