Asked by Serena
4.0 g of ferrous ammmonium sulphate, FeSO4(NH4)2SO4 6H2O, is used. Since the oxalate is in excess, calculate the theoretical yield of the iron complex.
Please help with this question.
Thank you.
Please help with this question.
Thank you.
Answers
Answered by
sana haddad
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.
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.