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

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???

To calculate the theoretical yield for K3[Fe(C2O4)3]*3H2O, you need to determine the limiting reactant and use the stoichiometric ratios from the balanced chemical equations.

In this case, you've provided a series of chemical reactions, so let's analyze them step by step.

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

From your solution, you correctly calculated that the mass of FeSO4∙(NH4)2SO4∙6H2O is 4.01 g, which corresponds to 0.01 mol based on its molar mass.

However, you made an error in calculating the moles of FeC2O4. The stoichiometric ratio between FeSO4∙(NH4)2SO4∙6H2O and FeC2O4 is 1:1. Therefore, the number of moles of FeC2O4 is also 0.01 mol, not 0.01 mol * 4/6 = 0.00667 mol.

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

Now that you have 0.01 mol of FeC2O4, you can determine the maximum possible yield of K3[Fe(C2O4)3]∙3H2O.

The stoichiometric ratio between FeC2O4 and K3[Fe(C2O4)3]∙3H2O is 6:4, so for 0.01 mol of FeC2O4, the theoretical yield of K3[Fe(C2O4)3]∙3H2O is 0.01 mol * 4/6 = 0.00667 mol.

You correctly converted the moles of K3[Fe(C2O4)3]∙3H2O to grams using its molar mass of 491.258 g/mol. The result is 0.00667 mol * 491.258 g/mol = 3.28 g K3[Fe(C2O4)3]∙3H2O.

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

Based on the previous step, you have 0.01 mol of FeC2O4, which is also the moles of Fe(OH)3 since the stoichiometric ratio between them is 1:1.

Using this value, you can calculate the theoretical yield of K3[Fe(C2O4)3]∙3H2O. The stoichiometric ratio between Fe(OH)3 and K3[Fe(C2O4)3]∙3H2O is 1:1, so the theoretical yield is 0.01 mol * 1/1 = 0.01 mol.

Again, you correctly converted the moles of K3[Fe(C2O4)3]∙3H2O to grams using its molar mass of 491.258 g/mol. The result is 0.01 mol * 491.258 g/mol = 4.91 g K3[Fe(C2O4)3]∙3H2O.

Finally, you added the individual yields from steps 2 and 3, which is correct. The sum of the theoretical yields is 3.28 g + 4.91 g = 8.19 g K3[Fe(C2O4)3]∙3H2O.

Therefore, the correct theoretical yield of K3[Fe(C2O4)3]∙3H2O is 8.19 g.