Question

Please find the theoretical percent of Fe in K₃[Fe(C₂O₄)₃] x 3H₂O

Here is some information I already found:

Concentration of Solution A	.02096 M
Total Volume of Solution A	.2 L
Moles of Fe	4.192 x 10-3 mol
Molecular Weight of Fe	55.847 g/mol
Mass of Fe in Solution A	.234110624 g
Mass of K3[Fe(C2O4)3] x 3H2O in Solution A	.93 g
Weight Percent in K3[Fe(C2O4)3] x 3H2O	25.17318537634409%
Theoretical % Fe in K3[Fe(C2O4)3] x 3H2O	???????
Percent Error	???????

Answer 1

Fe (NH4)2(SO4)2 (S) + H2C2O2 (aq) --->FeC2O4 (H2O)2(S) + (NH4)2SO4(aq) + H2SO4(aq) + 4H20(l)

FeC2O4 (H2O) 2(S) + 3K2C2O4(aq) + H2O2 (aq) ---> 2K3Fe(C2O4)3(H20)3

The first reaction involved the formation of iron (ii) oxalate dehydrates by reaction of iron (ii) ammonium sulfate hexahydrate with oxalic acid.

The second reaction involved a reaction between iron (ii) ammonium sulfate hexahydrate with potassium oxalate and hydrogen peroxide to give potassium trioxalatoferrate

How many moles of FeC2O4 (H2O)2(S)?

If 1.0g in 1000 ml

then 13 ml

= (13 * 1.0)/ 1000

= 0.013 moles

For reaction 2, how many moles of K3Fe (C2O4)3(H20)3?

2 FeC2O4(H2O)2 (aq) + 3 K2C2O4 (aq) + H2C2O4 (aq)---> 2 K3Fe(C2O4)3(H2O)3(aq)

Mole ratio is 2: 2 which the same as 1: 1 is.

Moles for 2FeC2O4 (H2O)2 (aq) = 0.013 moles, hence,

Moles for 2K3Fe (C2O4)3(H2O)3(aq) = 0.013 moles

How many grams of K3Fe (C2O4)3(H2O)3?

The number of moles of K3Fe (C2O4)3(H2O)3(aq) = 0.013 moles.

Molar mass of K3Fe (C2O4)3(H2O)3(aq) = 491.25 g/mol

Mass = no. of moles * molar mass

the percentage yield of the green crystals, K3Fe (C2O4)3(H2O)3?

Percentage yield = Actual yield/ Theoretical yield * 100

Actual yield = 1.456 g

Theoretical yield = 6.38625 g

% Yield = 1.456/6.38625 * 100

= 22.8%

= 0.013moles * 491.25 grams/moles

= 6.38625 grams

percent error

percent error = (experimental - theoretical)/ theoretical * 100

Experimental yield = 1.456 g

Theoretical yield = 6.38625 g

= (1.456 – 6.38625)/6.38625 * 100

= 77.2%