Question

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

Answer 1

Calculate the theoretical yield for K3[Fe(C2O4)3]*3H2O; 491.258 g/mol

Mass of Ferrous Ammonium Sulfate Hexahyrdreate: 4.01 g, 392.17 g/mol

[1] FeSO4∙(NH4)2SO4∙ 6H2O + H2C2O4∙ 2H2O ---> FeC2O4 + (NH4)2SO4 + H2SO4 + 8H2O

[2] 6 FeC2O4 + 3H2O2 + 6K2C2O4∙ H2O --->
4K3[Fe(C2O4)3]∙ 3 H2O + 2 Fe(OH)3 + 6 H2O

[3] 2 Fe(OH)3 + 3 H2C2O4∙ 2H2O + 3 K2C2O4∙ H2O ----> 2 K3[Fe(C2O4)3]∙ 3H2O + 9 H2O


4.01g * 1mol/392.17g = 0.01 mol FeSO4∙(NH4)2SO4∙ 6H2O

0.01 mol * 1/1 = 0.01 mol FeC2O4

0.01 mol * 4/6 = 0.00667 mol K3[Fe(C2O4)3]*3H2O

0.00667 mol * 491.258g/1mol = 3.28 g K3[Fe(C2O4)3]*3H2O

0.01 mol * 2/6 = 0.0033 mol Fe(OH)3

0.0033 * 2/2 = 0.0033 mol K3[Fe(C2O4)3]*3H2O

0.0033 mol * 491.258g/1mol = 1.62 g K3[Fe(C2O4)3]*3H2O

3.28 + 1.62 = 4.9 g K3[Fe(C2O4)3]*3H2O


The steps are all correct over all answer is wrong however it should be 5.011 rounded off. Make sure you plugin the correct number following the procedure.