In an experiment 4.0 g of ferrous ammonium sulphate, FeSO4.(NH4)2SO4.6H2O, is used. Since the oxalate in in excess, calculate the theoretical yield of the iron complex?

What's the experiment? What's the reaction? What complex?

3.2

To calculate the theoretical yield of the iron complex, we first need to determine the molar mass of the ferrous ammonium sulfate (FeSO4.(NH4)2SO4.6H2O).

The molar mass of FeSO4 is:
Iron (Fe) = 55.845 g/mol
Sulfur (S) = 32.06 g/mol
Oxygen (O) = 16.00 g/mol

So, the molar mass of FeSO4 is:
(1 * Fe) + (1 * S) + (4 * O) = (1 * 55.845) + (1 * 32.06) + (4 * 16.00) = 278.05 g/mol

The molar mass of (NH4)2SO4 is:
Nitrogen (N) = 14.01 g/mol
Hydrogen (H) = 1.0079 g/mol
Sulfur (S) = 32.06 g/mol
Oxygen (O) = 16.00 g/mol

So, the molar mass of (NH4)2SO4 is:
(2 * (1 * N + 4 * H)) + (1 * S) + (4 * O) = (2 * (1 * 14.01 + 4 * 1.0079)) + (1 * 32.06) + (4 * 16.00) = 132.14 g/mol

The molar mass of 6H2O is:
(6 * (2 * H + 1 * O)) = (6 * (2 * 1.0079 + 1 * 16.00)) = 108.17 g/mol

So, the molar mass of the entire compound FeSO4.(NH4)2SO4.6H2O is:
(1 * FeSO4) + (2 * (NH4)2SO4) + (6 * H2O) = (1 * 278.05) + (2 * 132.14) + (6 * 108.17) = 392.14 g/mol

Now, we can calculate the moles of the compound used in the experiment by dividing the given mass (4.0 g) by the molar mass:
moles = mass / molar mass = 4.0 g / 392.14 g/mol = 0.01019 mol

Since the oxalate is in excess, it means that the compound FeSO4.(NH4)2SO4.6H2O is not the limiting reagent. Therefore, all of the moles of ferrous ammonium sulfate will be fully converted into the iron complex.

So, the theoretical yield of the iron complex is also 0.01019 mol.

To calculate the theoretical yield of the iron complex, we need to find the molar mass of ferrous ammonium sulfate and determine the ratio of FeSO4.(NH4)2SO4.6H2O to Fe complex in the balanced chemical equation.

First, let's find the molar mass of FeSO4.(NH4)2SO4.6H2O:

- The molar mass of FeSO4 can be calculated as:
- Atomic mass of Fe = 55.845 g/mol
- Atomic mass of S = 32.06 g/mol
- Atomic mass of O = 16.00 g/mol (4 oxygen atoms)
- Total molar mass of FeSO4 = 55.845 + 32.06 + (16.00 * 4) = 151.908 g/mol

- The molar mass of NH4 can be calculated as:
- Atomic mass of N = 14.01 g/mol
- Atomic mass of H = 1.008 g/mol (4 hydrogen atoms)
- Total molar mass of NH4 = 14.01 + (1.008 * 4) = 18.036 g/mol

- The molar mass of (NH4)2SO4 can be calculated as:
- Total molar mass of (NH4)2SO4 = (18.036 * 2) + (32.06 + (16.00 * 4)) = 132.139 g/mol

- Finally, the molar mass of FeSO4.(NH4)2SO4.6H2O can be calculated by adding the molar masses of FeSO4 and (NH4)2SO4 along with the molar mass of water (6 water molecules):
- Molar mass of H2O = (1.008 * 2) + (16.00 * 1) = 18.016 g/mol (water)
- Total molar mass of FeSO4.(NH4)2SO4.6H2O = 151.908 + 132.139 + (18.016 * 6) = 392.135 g/mol

Now that we have the molar mass of the compound, we can calculate the number of moles of the compound used in the experiment:

- Moles of FeSO4.(NH4)2SO4.6H2O = Mass of compound / Molar mass of compound
- Moles of FeSO4.(NH4)2SO4.6H2O = 4.0 g / 392.135 g/mol = 0.0102 mol (rounded to four decimal places)

Next, we need to determine the molar ratio between FeSO4.(NH4)2SO4.6H2O and the iron complex. From the balanced chemical equation, we can deduce that:

1 mol of FeSO4.(NH4)2SO4.6H2O produces 1 mol of iron complex.

Therefore, the theoretical yield of the iron complex is equal to the number of moles of FeSO4.(NH4)2SO4.6H2O used in the experiment, which is 0.0102 mol.

Please note that this calculation represents the theoretical maximum yield. In reality, experimental conditions and limitations may cause the actual yield to differ.