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.
FeSO4•NH4)2SO4•6H2O + H2C2O4•2H20 --> FeC2O4 + (NH4)2SO4 + H2SO4 + 8H2O
6FeC2O4 + 3H2O2 + 6K2C2O4•H2O -->
4K3[Fe(C2O4)3]•3H2O + 2Fe(OH)3 + 6H2O
2Fe(OH)3 + 3H2C2O4•2H2O + 3K2C2O4•H2O ---> 2K3[Fe(C2O4)3]•3H2O + 9H2O
To calculate the theoretical yield of the iron complex, we need to first determine the limiting reactant. In this case, the limiting reactant is the one that runs out first, and it will determine the maximum amount of product that can be obtained.
Given that 4.0 grams of ferrous ammonium sulfate (FeSO4*(NH4)2SO4*6H2O) is used, we need to convert this mass into moles. The molar mass of FeSO4*(NH4)2SO4*6H2O is 392.14 g/mol.
4.0 g * (1 mol / 392.14 g) = 0.0102 mol
Now, let's look at the balanced chemical equation to determine the stoichiometry of the reaction. From the equation:
FeSO4*(NH4)2SO4*6H2O + H2C2O4*2H2O → FeC2O4 + (NH4)2SO4 + H2SO4 + 8H2O
The stoichiometric ratio between ferrous ammonium sulfate and the iron complex (FeC2O4) is 1:1. So, the theoretical yield of the iron complex is 0.0102 mol.
To convert this into grams, we can use the molar mass of the iron complex, which is 367.94 g/mol.
0.0102 mol * (367.94 g / mol) = 3.76 g
Therefore, the theoretical yield of the iron complex is 3.76 grams.