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
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.
To calculate the theoretical yield of the iron complex, we need to use stoichiometry and the given amounts of the reactants. Here's a step-by-step guide on how to calculate it:
Step 1: Convert the given mass of ferrous ammonium sulphate to moles.
The molar mass of ferrous ammonium sulphate, FeSO4*(NH4)2SO4 * 6H2O, can be calculated by adding the molar masses of its components.
FeSO4: 55.85 g/mol (Fe) + 32.07 g/mol (S) + 4 x 16.00 g/mol (O) = 151.91 g/mol
(NH4)2SO4: 2 x (14.01 g/mol (N) + 4 x 1.01 g/mol (H)) + 32.07 g/mol (S) + 4 x 16.00 g/mol (O) = 132.14 g/mol
6H2O: 6 x (2 x 1.01 g/mol (H) + 16.00 g/mol (O)) = 108.00 g/mol
Total molar mass = 151.91 g/mol + 132.14 g/mol + 108.00 g/mol = 392.05 g/mol
Now, divide the given mass of 4.0 gram by the molar mass:
4.0 g / 392.05 g/mol = 0.0102 mol of ferrous ammonium sulphate
Step 2: Use stoichiometry to determine the moles of the iron complex produced.
From the balanced chemical equation, we can see that the stoichiometric ratio between ferrous ammonium sulphate and the iron complex (FeC2O4) is 1:1. This means that for every 1 mol of ferrous ammonium sulphate, we obtain 1 mol of the iron complex.
Therefore, the moles of the iron complex produced will be the same as the moles of ferrous ammonium sulphate: 0.0102 mol.
Step 3: Convert the moles of the iron complex to grams.
To convert from moles to grams, we need to use the molar mass of the iron complex.
FeC2O4: 55.85 g/mol (Fe) + 2 x 12.01 g/mol (C) + 4 x 16.00 g/mol (O) = 143.94 g/mol
Multiply the moles of the iron complex by the molar mass:
0.0102 mol x 143.94 g/mol = 1.47 grams
Therefore, the theoretical yield of the iron complex is 1.47 grams.
To calculate the theoretical yield of the iron complex, we need to determine the limiting reactant.
First, let's convert the mass of ferrous ammonium sulphate to moles:
Given:
Mass of ferrous ammonium sulphate = 4.0 grams
Molar Mass of ferrous ammonium sulphate (FeSO4*(NH4)2SO4 * 6H2O):
Fe = 55.845 g/mol
S = 32.06 g/mol
O = 16.00 g/mol
N = 14.01 g/mol
H = 1.01 g/mol
Molar Mass of (NH4)2SO4 * 6H2O:
2(N) + 8(H) + S + 4(O) + 6(H) + 12(O) = 132.14 g/mol
Total Molar Mass of ferrous ammonium sulphate:
55.845 + 32.06 + 4(16.00) + 2(14.01) + 132.14 = 392.13 g/mol
Moles of ferrous ammonium sulphate = Mass / Molar Mass
Moles of ferrous ammonium sulphate = 4.0 g / 392.13 g/mol ≈ 0.0102 mol
Now, let's calculate the moles of oxalate used in the reactions:
Molar Mass of H2C2O4•2H2O:
2(H) + 2(C) + 4(O) + 2(H) + 16(O) = 90.03 g/mol
Moles of H2C2O4•2H2O = Moles of ferrous ammonium sulphate (1:1 stoichiometric ratio)
Moles of H2C2O4•2H2O = 0.0102 mol
Now, let's calculate the moles and theoretical yield of the iron complex:
Using the given balanced equations, the stoichiometric ratios are as follows:
1 mol FeSO4*(NH4)2SO4 * 6H2O → 1 mol FeC2O4
Moles of FeC2O4 = Moles of ferrous ammonium sulphate = 0.0102 mol
Finally, let's convert moles of FeC2O4 to grams:
Molar Mass of FeC2O4:
Fe = 55.845 g/mol
C = 12.01 g/mol
O = 16.00 g/mol
Molar Mass of FeC2O4 = 55.845 + 2(12.01) + 4(16.00) = 143.87 g/mol
Theoretical yield of iron complex = Moles of FeC2O4 × Molar Mass of FeC2O4
Theoretical yield of iron complex = 0.0102 mol × 143.87 g/mol ≈ 1.47 grams
Therefore, the theoretical yield of the iron complex is approximately 1.47 grams.