How much manganese can be obtained from 1 ton of Mn3O4 by the aluminothermal method, if the starting material

How much manganese can be obtained from 1 ton of Mn3O4 by the aluminothermal method, if the starting material contains 10% impurities, the manganese yield is 92% of the theoretically possible?

Let’s write down given:

m (Mn3O4) = 1 t or 1000 kg

w (impurities) = 10% or 0.1

η (Mn) = 92% or 0.92

m (Mn) -?

Solution.

1) Find the mass of impurities:

m (impurities) = 1000kg * 0.1 = 100kg

2) Calculate the mass of pure Mn3O4:

m (Mn3O4) = 1000 kg – 100 kg = 900 kg

3) Let us write down the equation for the reaction of the interaction of Mn3O4 with Al and calculate the mass of the manganese formed;

8 Al + 3 Mn3O4 = 9 Mn + 4 Al2O3

3 mol 9 mol

M (Mn3O4) = 55 * 3 + 16 * 4 = 229 g / mol

M (Mn) = 55 g / mol

According to the reaction equation: from 3 * 229 g (Mn3O4), 9 * 55 g (Mn) is formed,

Then, according to the problem statement: from 900 kg (Mn3O4) x kg (Mn) is formed;

Hence:

x = 900kg * 9 * 55g / 3 * 229g

x = 648.47 kg (Mn) is the theoretical yield.

4) Calculate the practical yield of manganese:

η = m (practical) * 100% / m (theoretical)

m (practical) = η (Mn) * m (theoretical)

m (practical Mn) = 0.92 * 648.47 kg = 596.59 kg

Answer: 596.59 kg (Mn) was obtained.

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