Question

the synthesis of FeC2O4*2H2O, Which reagent would be the limiting reagent? (NH4)2[Fe(H2O)6](SO4)2 IS 3.42g H2C2O4*2H2O is 1.98g 2. using your answers above, calculate the theoretical yield (in grams) of potassium trioxalatofer-rate(iii) trihydrate, K3[Fe(C2O4)3]*3H2O

Answer 1

1. First write down the balanced equation for the preparation of FeC₂O₄.2H₂O

The balanced equation is:

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

Ferrous Ammonium oxalic acid Ferrous oxalate ammonium

sulphate hexahydrate dihydrate sulfate

mass of Fe(NH₄)₂(SO₄)₂.6H₂O = 3.42 g

mass of H₂C₂O₄.2H₂O = 1.98 g

Calculate the moles of Fe(NH₄)₂(SO₄)₂.6H₂O = 3.42 g * 1 mol Fe(NH₄)₂(SO₄)₂.6H₂O / 392.14 g = 0.00872 mol

Now calculate the moles of H₂C₂O₄.2H₂O = 1.98 g * 1mol H₂C₂O₄.2H₂O / 126.07 g = 0.0157 mol

There are less moles of Fe(NH₄)₂(SO₄)₂.6H₂O , so it is a limiting reagent.

According to the balanced equation, there is 1:1 molar ratio between Fe(NH₄)₂(SO₄)₂.6H₂O and FeC₂O₄.2H₂O.

Hence, Theoretical yield of FeC₂O₄.2H₂O = 0.00872 mol

2. The balanced equation for the preparation of K₃Fe(C₂O₄)₃.3H₂O from FeC₂O₄.2H₂O

2FeC₂O₄.2H₂O(s) + H₂C₂O₄(aq) + H₂O₂(aq)+3K₂C₂O₄(aq) 2 K₃[Fe(C₂O₄)₃].3H₂O(s)

There is 1:1 molar ratio between FeC₂O₄.2H₂O and K₃[Fe(C₂O₄)₃].3H₂O

Hence, Theoretical yield of K₃Fe(C₂O₄)₃.3H₂O (considering FeC₂O₄.2H₂O as limiting reagent) = 0.00872 mol

Theoretical yield of K₃Fe(C₂O₄)₃.3H₂O in grams = 0.00872 mol * 491.23 g K₃Fe(C₂O₄)₃.3H₂O / 1 mol = 4.28 g