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.
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
1 answer
0 answers