An ore contains Fe3O4 and no other iron. The
iron in a 38.8-gram sample of the ore is all
converted by a series of chemical reactions to
Fe2O3. The mass of Fe2O3 is measured to be
25 g. What was the mass of Fe3O4 in the
sample of ore?
See your previous post.
31
To find the mass of Fe3O4 in the sample of ore, we can use the principle of conservation of mass. The total mass of iron before and after the reaction should be the same.
First, let's calculate the molar mass of Fe3O4 and Fe2O3:
Molar mass of Fe3O4 = (molar mass of Fe) x 3 + (molar mass of O) x 4
= 55.845 g/mol x 3 + 16.00 g/mol x 4
= 231.54 g/mol
Molar mass of Fe2O3 = (molar mass of Fe) x 2 + (molar mass of O) x 3
= 55.845 g/mol x 2 + 16.00 g/mol x 3
= 159.69 g/mol
Next, we need to determine the number of moles of Fe2O3 formed in the reaction. We can use the given mass of Fe2O3 and its molar mass:
Number of moles of Fe2O3 = mass of Fe2O3 / molar mass of Fe2O3
= 25 g / 159.69 g/mol
≈ 0.1567 mol
Since the reaction follows the stoichiometry of 1 mol Fe3O4 → 1 mol Fe2O3, the number of moles of Fe3O4 is also equal to 0.1567 mol.
Finally, we can find the mass of Fe3O4 using its molar mass and the number of moles:
Mass of Fe3O4 = number of moles of Fe3O4 x molar mass of Fe3O4
= 0.1567 mol x 231.54 g/mol
≈ 36.28 g
Therefore, the mass of Fe3O4 in the sample of ore is approximately 36.28 grams.