Question

- Determine the theoretical yield in grams of iron(II) oxalate dihydrate, FeC204.2H20.

Answer 1

Ans-4b: The balanced chemical reaction for this compexation is given below:

Fe(NH4)2(SO4)2.6H2O + H2C2O4 = FeC2O4.2H2O + (NH4)2SO4 + H2SO4 + 4H2O

The Molar masses of th reactants and products:

Fe(NH4)2(SO4)2.6H2O   = 392.13 g/mol

Oxalic acid = 90.03 g/mol

FeC2O4.2H2O                  = 179.89 g/mol

It is given:

Ferrous ammonium sulfate hexahydrate weight = 2.5 g

Calculated moles of it = 2.5 g / 392.13 g/mol = 0.0064 moles

10 mL of 1.5 M Oxalic acid

Calculate the wt. of oxalic acid as below from its molar mass:

= 90.03 g/mol * 1.5 mol = 135.045 g

The 1000 mL has 135.045 g of oxalic acid

10 mL will have = (135.045/1000) * 10 = 1.35 g of Oxalic acid

Moles of it = 1.35/90.03 = 0.0150 moles

The above complex formation reaction is taking place in 1:1 mole ratio, and giving one mole of the ferrous oxalate dihydrate complex, hence from the above calculation it is clear that the ferrous ammonium sulfate hexahydrate is a limiting reagent and the theoretical yield should be calculated w.r.t. this reagent. Oxalic acid is in excess.

Therefore 0.0064 moles of Ferrous ammonium sulfate hexahydrate will give 0.0064 moles of ferrous oxalate dehydrate (FeC2O4.2H2O).

Molar mass of FeC2O4.2H2O = 179.89 g/mol

1 mole = 179.89 g

So, 0.0064 moles = 179.89 * 0.0064 = 1.15 g

Theoretical yield of FeC2O4.2H2O = 1.15 g, obtained from 2.5 g of the Fe(NH4)2(SO4)2.6H2O