help! P4(s) + F2(g) PF3(g)?

Question

How many grams of F2 are needed to produce 120. g of PF3 if the reaction has a 74.5 % yield?

Answer 1

You need to balance the reaction first.

P4 + 6 F2 --> 4 PF3

If you need 120 g from a reaction whose efficiency is 74.5%, then your theoretical yield needs to be 161.07 g, which is 1.83 moles (MW PF3 = 88 g/mol).

You need 6 moles of F2 for every 4 moles of PF3 produced, so you need 2.75 moles, or 104.5 grams (MW F2 = 38 g/mol).