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.

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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 determine the limiting reagent. Since the oxalate is in excess, it means that the ferrous ammonium sulfate is the limiting reagent.

To calculate the theoretical yield, we need to convert the given mass of ferrous ammonium sulfate to moles.

1. Calculate the molar mass of ferrous ammonium sulfate:
FeSO4(NH4)2SO4 6H2O = (1 * atomic mass of Fe) + (1 * atomic mass of S) + (4 * atomic mass of O) + (2 * atomic mass of N) + (8 * atomic mass of H) + (2 * atomic mass of S) + (6 * atomic mass of O)
FeSO4(NH4)2SO4 6H2O = (1 * 55.845) + (1 * 32.06) + (4 * 16) + (2 * 14.01) + (8 * 1.008) + (2 * 32.06) + (6 * 16)
FeSO4(NH4)2SO4 6H2O = 278.01 g/mol

2. Calculate the moles of ferrous ammonium sulfate used:
moles = mass / molar mass
moles = 4.0 g / 278.01 g/mol
moles ≈ 0.01438 mol

Since the molar ratio between ferrous ammonium sulfate and the iron complex is 1:1, the moles of iron complex formed will also be 0.01438 mol.

3. Calculate the mass of the iron complex:
mass = moles * molar mass
mass = 0.01438 mol * molar mass of iron complex

Unfortunately, the molar mass of the iron complex is not provided, so we are unable to calculate the exact theoretical yield without that information.

To calculate the theoretical yield of the iron complex, you first need to determine the limiting reactant. In this case, the limiting reactant is the one that will be completely consumed, which means it will determine the amount of product that can be formed.

The reaction between ferrous ammonium sulphate and oxalic acid can be represented as follows:

FeSO4(NH4)2SO4 6H2O + H2C2O4 → [Fe(C2O4)(H2O)2] + 2(NH4)2SO4 + H2SO4

From the balanced equation, we can see that one mole of ferrous ammonium sulphate reacts with one mole of oxalic acid to form one mole of the iron complex.

First, we need to convert the given mass of ferrous ammonium sulphate to moles:

4.0 g FeSO4(NH4)2SO4 6H2O × (1 mol FeSO4(NH4)2SO4 6H2O / molar mass of FeSO4(NH4)2SO4 6H2O)

The molar mass of FeSO4(NH4)2SO4 6H2O can be calculated as follows:

1(Fe) + 1(S) + 4(O) + 2(N) + 8(H) + 8(H) + 4(O) + (2(N) + 8(H)) + 4(O) + (6(H) + 16(O)) = 392.14 g/mol

Now, we can calculate the number of moles of ferrous ammonium sulphate:

4.0 g / 392.14 g/mol ≈ 0.0102 mol

Since the balanced equation tells us that one mole of ferrous ammonium sulphate reacts with one mole of oxalic acid to form one mole of the iron complex, we can conclude that the theoretical yield of the iron complex is also 0.0102 moles.

If you need to find the mass of the iron complex, you can multiply the number of moles by the molar mass of the iron complex.