4.0g of ferrous ammonium sulphate, FeSO4 ∙ (NH4)2SO4 ∙ 6H2O is used. Since the oxalate is in excess, calculate the theoretical yield of the iron complex.
NOTE: We assume 100% conversion efficiency, going from 1 reaction to the next.
3 separate equations in a reaction series:
 FeSO4∙(NH4)2SO4∙ 6H2O + H2C2O4∙ 2H2O ---> FeC2O4 + (NH4)2SO4 + H2SO4 + 8H2O
 6 FeC2O4 + 3H2O2 + 6K2C2O4∙ H2O ---> 4 K3[Fe(C2O4)3]∙ 3 H2O + 2 Fe(OH)3 + 6 H2O
 2 Fe(OH)3 + 3 H2C2O4∙ 2H2O + 3 K2C2O4∙ H2O ----> 2 K3[Fe(C2O4)3]∙ 3H2O + 9 H2O
I found the molar mass of ammonium sulphate hydrate compound to be:
m[FeSO4∙(NH4)2 SO4∙6H2O] = 392.21 g/mol
Because we assume that the oxalate is in excess, that means the ammonium sulphate hydrate compound is obviously the limiting reagent. So I use molar mass and the grams provided to get the moles, which is:
mol[FeSO4∙(NH4)2 SO4∙6H2O] = 0.010198618 mol
Now, I get the moles of each product of the r'xn using the mol[FeSO4∙(NH4)2 SO4∙6H2O], but I am experiencing dilemma:
mol[FeC2O4] = mol[(NH4)2SO4] = mol[H2SO4]
^ All of these products have the same amount of moles produced.
My question is: which one do I use to proceed to the next reaction?