What is the maximum amount of iron that can be obtained from 454g of iron (III) oxide?
To determine the maximum amount of iron that can be obtained from 454g of iron (III) oxide, we need to understand the concept of the mole and the molar mass.
1. Start by determining the molar mass of iron (III) oxide (Fe2O3).
- Iron (Fe) has a molar mass of 55.85 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
- Since there are two iron atoms and three oxygen atoms in iron (III) oxide, the molar mass of Fe2O3 is:
Molar mass of Fe2O3 = (2 * 55.85 g/mol) + (3 * 16.00 g/mol) = 159.70 g/mol.
2. Calculate the number of moles of Fe2O3 in 454g.
- The number of moles (n) can be calculated using the formula:
n = mass (in grams) / molar mass.
- For Fe2O3: n = 454g / 159.70 g/mol ≈ 2.85 mol.
3. Determine the molar ratio between Fe2O3 and Fe.
- From the balanced chemical equation for the reaction between Fe2O3 and Fe:
Fe2O3 + 3Fe → 2Fe2O3
- The ratio is 1:3, which means 1 mol of Fe2O3 produces 3 mol of Fe.
4. Calculate the number of moles of Fe that can be obtained.
- Multiply the number of moles of Fe2O3 (2.85 mol) by the molar ratio:
Number of moles of Fe = 2.85 mol * 3 ≈ 8.55 mol.
5. Finally, calculate the mass of Fe that can be obtained from the calculated moles.
- The mass (m) can be calculated using the formula:
m = n * molar mass.
- For Fe: m = 8.55 mol * 55.85 g/mol ≈ 477.54 g.
Therefore, the maximum amount of iron that can be obtained from 454g of iron (III) oxide is approximately 477.54 grams.
oops. Typo
454 g x (2*mm Fe/mm Fe2O3) = ?
mm = molar mass
454 g x (3*mm Fe/mm Fe2O3) = ?