How much iron can be obtained by reducing 1 ton of magnetite containing 72% of iron scale Fe3O4

univerkov | September 2, 2021 | education

How much iron can be obtained by reducing 1 ton of magnetite containing 72% of iron scale Fe3O4 by aluminothermal method. if the practical product yield is 86% of the theoretically possible?

Solution:
m (Fe3O4) = 1000 * 0.72 = 720kg
v (Fe3O4) = 720/232 = 3.1 mol
3Fe3O4 + 8AI = 4AI2O3 + 9Fe
v (Fe) = 3 * v (Fe3O4) = 3 * 3.1 = 9.3 mol
m (Fe) = 9.3 * 56 = 520.8g
m (Fe) = 0.86 * 520.8 = 447.9 g
Answer: 447.9 g

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