What is the theoretical yield (and limiting reagent) for this reaction?

Preparation of K3[Fe(ox)3].3H2O

- Reagents include:
2.5 g of Fe2(SO4)3 (Ferric (III) Sulfate)
5.0 g of BaC2O4 (Barium Oxalate)
2.7 g of K2C2O4.H2O (Potassium Oxalate Monohydrate)

The balanced equation is as follows:

Fe2(SO4)3 + 3 BaC2O4 + 3 K2C2O4 → 2 K3[Fe(C2O4)3] + 3 BaSO4

Given the following molar masses (M):
Fe2(SO4)3 = 399.88 g/mol
BaC2O4 = 225.34 g/mol
K2C2O4.H2O = 184.23 g/mol
PRODUCT = K3[Fe(ox)3].3H2O = 491.25 g/mol

Now the thing that is confusing me is the hydrate portion of the complex, do we write the hydrate parts in the equation or leave them out altogether as shown in the equation above?

If we leave them out, what is the limiting reagent of this reaction and therefore the theoretical yield of the product complex K3[Fe(ox)3].3H2O ?

I am confused, if we leave the hydrate portion out of the equation then how do we calculate the correct yield for the product which is a trihydrate?? 

Any help would be appreciated, thank you