help me out! i dont know how to do this Problem?

Question

P4(s) + 6 Cl2(g) --> 4 PCl3(l)
If the reaction proceeds in 65% yield, how many grams of P4 are required if 137.5 grams of PCl3 are desired?

Answer 1

There is more than one way to do this, here's one way:

The first term is the weight fraction of P in PCl3 (like a % but not multiplied by 100%); then it is multiplied by the desired weight of PCl3 you want to make; the final term is the "fudge factor" to account for only 65% yield for the RXN.

30.97/[30.97 + (3 x 35.45)] x 137.5g x 1/0.65 = 47.71g

Answer 2

If the reaction is only going to yield 65% then on a theoretical 100% yield basis you will need enough P4 to produce 137.5/65%g or 211.5g.

Now just work back:
Number moles PCl3 = mass/MW
=211.5/(30.97+35.45*3)
= 1.540moles

As 1 mole of P4 produces 4 moles of PCl3 you will need to start with 1.54/4 or 0.3851 moles of P4

Convert this to grams = 0.3851 * (30.97*4) = 47.71g

Answer 3

I have no idea. Good luck!