What is the theoretical yield of Fe2(SO4)3 if 20.00 g of FePO4 reacts with an excess of Na2SO4?
Those don't react but if they did it would be this.
2FePO4 + 3Na2SO4 ==> 2Na3PO4 + Fe2(SO4)3
mols FePO4 = grams/molar mass = ?
Then using the coefficients in the balanced equation, convert mols FePO4 to mols Fe2(SO4)3. That's
mols FePO4 x (1 mol Fe2(SO4)3/2 mols FePO4) = ?mols FePO4 x (1/2) = ?
Then grams Fe2(SO4)3 = mols x molar mass. That is the theoretical yield.
To find the theoretical yield of Fe2(SO4)3, we need to use the stoichiometry of the balanced chemical equation relating FePO4 and Fe2(SO4)3.
The balanced equation for the reaction is:
3 FePO4 + 2 Na2SO4 -> Fe2(SO4)3 + 2 Na3PO4
From the equation, we can see that 3 moles of FePO4 react with 2 moles of Na2SO4 to produce 1 mole of Fe2(SO4)3.
First, we need to find the number of moles of FePO4 in 20.00 g. To do this, we divide the mass of FePO4 by its molar mass:
Molar mass of FePO4 = (1x Atomic mass of Fe) + (1x Atomic mass of P) + (4x Atomic mass of O)
= (1x 55.845 g/mol) + (1x 30.974 g/mol) + (4x 16.00 g/mol)
= 55.845 + 30.974 + 64.00
= 150.819 g/mol
Number of moles of FePO4 = Mass of FePO4 / Molar mass of FePO4
= 20.00 g / 150.819 g/mol
≈ 0.132 moles
According to the stoichiometry of the balanced equation, the molar ratio between FePO4 and Fe2(SO4)3 is 3:1. So, if 0.132 moles of FePO4 react, the theoretical yield of Fe2(SO4)3 can be calculated using the stoichiometric ratio:
Theoretical yield of Fe2(SO4)3 = (0.132 moles FePO4) * (1 mole Fe2(SO4)3 / 3 moles FePO4)
= 0.044 moles
Finally, to find the mass of Fe2(SO4)3, we multiply the molar mass of Fe2(SO4)3 by the number of moles:
Molar mass of Fe2(SO4)3 = (2x Atomic mass of Fe) + (3x Atomic mass of S) + (12x Atomic mass of O)
= (2x 55.845 g/mol) + (3x 32.06 g/mol) + (12x 16.00 g/mol)
= 111.690 + 96.18 + 192.00
≈ 399.870 g/mol
Mass of Fe2(SO4)3 = Theoretical yield of Fe2(SO4)3 * Molar mass of Fe2(SO4)3
= 0.044 moles * 399.870 g/mol
≈ 17.595 g
Therefore, the theoretical yield of Fe2(SO4)3 is approximately 17.595 grams.
To find the theoretical yield of Fe2(SO4)3, we need to first write the balanced equation for the reaction between FePO4 and Na2SO4. The balanced equation is as follows:
FePO4 + 3Na2SO4 -> Fe2(SO4)3 + 2Na3PO4
From the balanced equation, we can see that 1 mole of FePO4 reacts with 3 moles of Na2SO4 to produce 1 mole of Fe2(SO4)3. This means that the mole ratio of FePO4 to Fe2(SO4)3 is 1:1.
Next, we need to calculate the number of moles of FePO4. We can use the molar mass of FePO4 to convert the given mass to moles:
molar mass FePO4 = atomic mass of Fe + 4 * atomic mass of O + atomic mass of P
= (55.845 g/mol) + 4 * (15.999 g/mol) + (30.974 g/mol)
= 150.824 g/mol
moles of FePO4 = mass of FePO4 / molar mass FePO4
= 20.00 g / 150.824 g/mol
= 0.1325 mol
Since the mole ratio of FePO4 to Fe2(SO4)3 is 1:1, the number of moles of Fe2(SO4)3 produced will also be 0.1325 mol.
Finally, we can calculate the theoretical yield of Fe2(SO4)3 using the molar mass of Fe2(SO4)3:
molar mass Fe2(SO4)3 = 2 * (atomic mass of Fe) + 3 * (atomic mass of S) + 12 * (atomic mass of O)
= 2 * (55.845 g/mol) + 3 * (32.06 g/mol) + 12 * (15.999 g/mol)
= 399.877 g/mol
theoretical yield of Fe2(SO4)3 = moles of Fe2(SO4)3 * molar mass Fe2(SO4)3
= 0.1325 mol * 399.877 g/mol
= 52.94 g
Therefore, the theoretical yield of Fe2(SO4)3 is 52.94 grams.