A chemist reacts 112 g of iron with 72 g of water. State which reactant is the limiting reagent. Calculate what is the maximum yield in grams of Fe3O4 that could be produced in this reaction. For the substance that is not the limiting reagent, calculate the amount in excess.

To determine the limiting reagent, we need to compare the moles of each reactant.

Step 1: Calculate the moles of iron (Fe):
Molar mass of Fe = 55.8 g/mol
Moles of Fe = mass of Fe / molar mass of Fe
Moles of Fe = 112 g / 55.8 g/mol
Moles of Fe ≈ 2.01 mol

Step 2: Calculate the moles of water (H2O):
Molar mass of H2O = 18 g/mol
Moles of H2O = mass of H2O / molar mass of H2O
Moles of H2O = 72 g / 18 g/mol
Moles of H2O = 4 mol

Step 3: Determine the mole ratio of the reactants (Fe:H2O)
From the balanced chemical equation, we know that the ratio is 3:4. That is, for every 3 moles of Fe, we need 4 moles of H2O.

Step 4: Compare the moles of each reactant to the ratio.
For Fe: 2.01 mol
For H2O: 4 mol

Since the ratio is 3:4, and we have more than enough moles of water, the limiting reagent is iron (Fe).

Now, let's calculate the maximum yield of Fe3O4 that could be produced.

Step 5: Determine the mole ratio of the product (Fe3O4) to the limiting reagent (Fe)
From the balanced chemical equation, we know that the ratio is 1:1. That is, for every 1 mole of Fe, we get 1 mole of Fe3O4.

Step 6: Calculate the moles of Fe3O4 that can be produced.
Moles of Fe3O4 = moles of limiting reagent (Fe) = 2.01 mol

Step 7: Calculate the maximum yield of Fe3O4 in grams.
Molar mass of Fe3O4 = 231.53 g/mol (55.8 g/mol for Fe x 3 + 16 g/mol for O x 4)
Maximum yield = moles of Fe3O4 x molar mass of Fe3O4
Maximum yield = 2.01 mol x 231.53 g/mol
Maximum yield ≈ 465.2 g

Therefore, the maximum yield of Fe3O4 that could be produced in this reaction is approximately 465.2 grams.

Finally, let's calculate the amount of the substance in excess, which in this case is water (H2O).

Step 8: Calculate the amount of the excess reactant.
Excess = moles of H2O - moles required (based on limiting reagent)
Excess = 4 mol - (2.01 mol x (4 mol/3 mol))
Excess = 4 mol - 2.68 mol
Excess ≈ 1.32 mol

Step 9: Calculate the amount of H2O in excess in grams.
Mass of excess H2O = moles of excess H2O x molar mass of H2O
Mass of excess H2O ≈ 1.32 mol x 18 g/mol
Mass of excess H2O ≈ 23.76 g

Therefore, the mass of water (H2O) in excess is approximately 23.76 grams.

To determine the limiting reagent, we need to compare the moles of each reactant and see which one is consumed completely. The reactant that produces fewer moles of the desired product will be the limiting reagent.

Let's find the moles of each reactant:

1. Iron (Fe):

The molar mass of iron (Fe) is 55.85 g/mol.
Moles of Fe = Mass of Fe / Molar mass of Fe
Moles of Fe = 112 g / 55.85 g/mol
Moles of Fe = 2.00 mol

2. Water (H2O):

The molar mass of water (H2O) is 18.02 g/mol.
Moles of H2O = Mass of H2O / Molar mass of H2O
Moles of H2O = 72 g / 18.02 g/mol
Moles of H2O = 3.99 mol

From the balanced chemical equation, we know that the reaction is:

4 Fe + 3 H2O -> Fe3O4 + 4 H2

From the balanced equation, we see that 4 moles of Fe react with 3 moles of H2O to produce 1 mole of Fe3O4.

Therefore, based on the stoichiometry, the moles of Fe3O4 produced would theoretically be:

Moles of Fe3O4 = (2.00 mol Fe) / (4 mol Fe3O4) = 0.50 mol Fe3O4

The molar mass of Fe3O4 is 231.53 g/mol, so the maximum yield of Fe3O4 in grams would be:

Maximum yield of Fe3O4 = Moles of Fe3O4 × Molar mass of Fe3O4
Maximum yield of Fe3O4 = 0.50 mol × 231.53 g/mol
Maximum yield of Fe3O4 = 115.76 g

Since water (H2O) is in excess, all of the 72 g of water will react, leaving some iron (Fe) unreacted. To calculate the amount of excess iron, we subtract the moles of Fe that reacted from the total moles of Fe:

Excess iron = Total moles of Fe - Moles of Fe used
Excess iron = 2.00 mol - 2.00 mol (because 2.00 mol of Fe is completely consumed)
Excess iron = 0.00 mol

Since the moles of excess iron is 0.00 mol, there is no excess iron left.