What volume of H2 is required for Fe reduction from 100 g of ore containing 70% FE2O3?

To solve, we write down the equation of the process:

Fe2O3 + 3H2 = 2Fe + 3H2O – OBP, iron is released;
Let’s make the calculations:
M (Fe2O3) = 159.6 g / mol;

M (H2) = 2 g / mol.

3. Determine the mass, the amount of the original substance:

W = m (practical) / m (theoretical) * 100;

m (Fe2O3) = 100 / 0.70 = 142.85 g (theoretical weight);

Y (Fe2O3) = m / M = 142.85 / 159.6 = 0.89 mol.

Proportion:
0.89 mol (Fe2O3) – X mol (H2);

-1 mol – 3 mol from here, X mol (H2) = 0.89 * 3/1 = 2.67 mol.

We find the volume of H2:
V (H2) = 2.67 * 22.4 = 59.8 L

Answer: you need 59.8 liters of hydrogen