Limiting Reactant help

I have this homework problem and I was hoping that someone could do it so i would be able to check my answer, thanks

10 grams of P2O5 react with 10 grams of 3H2O
P2O5 + 3H2O -> 2H3PO4

1. Identify the limiting reactant
2. Calculate theoretical yield of the product in grams
3. Calculate the excess in grams

P2O5 + 3 H2O → 2 H3PO4

(10 g P2O5) / (141.9447 g P2O5/mol) = 0.07045 mol P2O5
(10 g H2O) / (18.0153 g H2O/mol) = 0.5551 mol H2O

0.07045 mole of P2O5 would react completely with 0.07045 x (3/1) = 0.21135 mole of H2O, but there is more H2O present than that, so H2O is in excess and P2O5 is the limiting reactant.

(0.07045 mol P2O5) x (2/1) x (97.9953 g H3PO4/mol) = 14 g H3PO4

((0.5551 mol H2O initially) - (0.21135 mol H2O reacted)) x (18.0153 g H2O/mol) =
6.2 g H2O excess