Fe3O4 can be reduced by treatment with hydrogen to yield iron metal and water vapor write the balanced equation. this process required 36kcal for every 1.00 mol of Fe3O4 reduced. how much energy in kcal is required to produce 64g of iron. how many grams of hydrogen are needed to produce 71g of iron
Fe3O4 + 4H2 + 36 kcal ==> 3Fe + 4H2O
36 kcal x 64/3*atomic mass Fe = ? kcal required.
mols Fe needed = grams/atomic mass = ?
Use the coefficients in the balanced equation to convert mols Fe to mols H2.
Now convert g H2 to mols H2. g = mols x molar mass.
To balance the equation for the reduction of Fe3O4 with hydrogen, we need to know the correct stoichiometric coefficients. The equation can be written as follows:
Fe3O4 + 4H2 -> 3Fe + 4H2O
Now, let's calculate the energy required to produce 64g of iron.
First, we need to convert the mass of iron (in grams) to moles. The molar mass of iron (Fe) is approximately 55.85 g/mol. So, the number of moles of iron produced is:
64g / 55.85 g/mol ≈ 1.15 mol
Since the process requires 36 kcal for every 1.00 mol of Fe3O4 reduced, we can calculate the energy required to produce 1.15 moles of iron:
1.15 mol × 36 kcal/mol = 41.4 kcal
Therefore, 41.4 kcal of energy is required to produce 64g of iron.
Next, let's determine the amount of hydrogen needed to produce 71g of iron.
We follow a similar approach as before, starting by converting the mass of iron (71g) to moles:
71g / 55.85 g/mol ≈ 1.27 mol
From the balanced equation, we can see that 1.27 moles of iron requires 4 times as many moles of hydrogen. So, the moles of hydrogen required is:
1.27 mol × 4 = 5.08 mol
Finally, to calculate the mass of hydrogen needed, we multiply the moles by the molar mass of hydrogen (approximately 2.02 g/mol):
5.08 mol × 2.02 g/mol ≈ 10.26 g
Therefore, approximately 10.26 grams of hydrogen are required to produce 71g of iron.