# Calculate the mass of iron that can be obtained from 325 kg. Magnetic iron ore mass of impurities is 15%

**Calculate the mass of iron that can be obtained from 325 kg. Magnetic iron ore mass of impurities is 15%. CO was used as a reducing agent.**

Let’s implement the solution:

1. We write down the equation according to the problem statement:

m = 325 kg; 15% impurity;

Fe3O4 + 4CO = 3Fe + 4CO2 – OBP, iron obtained;

X g -?

2. Let’s calculate the molar mass of substances:

M (Fe3O4) = 231.4 g / mol;

M (Fe) = 55.8 g / mol.

3. Determine the mass of ore without impurities, as well as the amount:

m (Fe3O4) = 326000 (1 – 0.15) = 276250 g;

Y (Fe3O4) = m / M = 276250 / 231.4 = 1193.8 mol.

4. Proportion:

1193.8 mol (Fe3O4) – X mol (Fe);

-1 mol -3 mol from here, X mol (Fe) = 1193 * 3/1 = 3581.5 mol.

5. Find the mass of the product:

m (Fe) = Y * M = 3581.5 * 55.8 = 199847.7 g. = 199 kg. 848 BC

Answer: received iron weighing 199 kg. 848 BC