4.0 gram of ferrous ammonium sulphate, FeSO4*(NH4)2SO4 * 6H2O, is used. Since the oxalate is in excess , Calculate the theoretical yield of the iron complex.

See your post above.

To calculate the theoretical yield of the iron complex, we first need to determine the limiting reactant between ferrous ammonium sulfate (FeSO4*(NH4)2SO4 * 6H2O) and oxalate.

The molar mass of ferrous ammonium sulfate can be calculated by adding up the atomic masses of its constituent elements:
FeSO4*(NH4)2SO4 * 6H2O:
Fe = 55.85 g/mol
S = 32.07 g/mol
O = 16.00 g/mol (4 oxygens)
N = 14.01 g/mol (2 nitrogens)
H = 1.01 g/mol (8 hydrogens)
2 x (N + 4 x H + S + 4 x O) = 2 x (14.01 g/mol + 4 x 1.01 g/mol + 32.07 g/mol + 4 x 16.00 g/mol) = 392.27 g/mol

Now, we can calculate the number of moles of ferrous ammonium sulfate:
moles = mass / molar mass
moles = 4.0 g / 392.27 g/mol = 0.0102 mol

Next, we need to determine the stoichiometry of the reaction between ferrous ammonium sulfate and oxalate. From the balanced chemical equation, we can see that the ratio of moles of ferrous ammonium sulfate to iron complex is 1:1.

Therefore, the moles of the iron complex formed will also be 0.0102 mol.

Finally, we can calculate the theoretical yield of the iron complex by multiplying the moles by its molar mass:
theoretical yield = moles of iron complex * molar mass of iron complex
Assuming the molar mass of the iron complex is given, you can calculate the theoretical yield.