Question

if you add 10.0 grams of P4 to 20.0 grams of O2 what is the theoretical yield of P2O5

Answer 1

First, you need to set up a balanced chemical equation for the reaction. Can you do that?

Answer 2

P4 + 5 O2 = 2 P2O5
Convert to moles:
(10.0 g P4) / (123.895 g P4/mol) = 0.080714 mol P4
(20.0 g O2) / (31.99886 g O2/mol) = 0.62502 mol O2
Determine the limiting reactant:
0.080714 mole of P4 would react completely with 0.080714 x (5/1) = 0.40357 mole of O2, but there is more O2 present than that, so O2 is in excess and P4 is the limiting reactant.
Do the stoichiometry:
(0.080714 mol P4) x (2 mol P2O5 / 1 mol P4) x (141.9445 g P2O5/mol) =
22.9 g P2O5 in theory