Hi, could someone please take a look at my solution to this problem and let me know if it's correct? I would really appreciate some help.

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

3 answers

The method looks ok to me but I disagree slightly with the numbers. I think most of that is that you threw away some significant figures here and there. For example, the first calculation of 0.01 should be 0.0102. The 0.0033 should be recalculated for three s.f. etc.

Thank you Dr. Bob

Why did you add 3.28???