Consider the following unbalanced reaction.?

Question

Consider the following unbalanced reaction.

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

How many grams of F2 are needed to produce 108 g of PF3 if the reaction has a 78.1% yield?

Answer 1

Balance the equation:

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

108 g PF3 = 78.1% of the "theoretical" yield, so you can figure out how much it would actually make by dividing:

108 / .781 = 138.28 g would "theoretically" be produced.

So, then work backwards to find how much F2 you need to start with:

138.28 g PF3 x ( 1 mol PF3 / 87.969 g/mol PF3) x ( 6 mol F2 / 4 mol PF3) x (38g F2 / 1 mol F2) = 89.6 F2 needed.

Answer 2

P4 + 6F2 →  4PF3

6F2 = 6*2*18.998 = 227.988

4PF3 = 4P +12F = 4*30.974 + 12*18.998 = 351.872

227.988g F2 yield 351.872g PF3

(227.988/351.872)108 = 69.976g F2 required to produce 108g of PF3

But yield is 78.1%

69.976*100/78.1 = 89.6g F2 required