What mass of manganese can be obtained by reducing 90 g of manganese oxide 4 containing

What mass of manganese can be obtained by reducing 90 g of manganese oxide 4 containing 5% impurities by the aluminothermal method?

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

3MnO2 + 4Al = 3Mn + 2Al2O3 – OBP, manganese is released;
Let’s calculate the molar masses of substances:
M (MnO2) = 86.9 g / mol;

M (Mn) = 54.9 g / mol.

Determine the mass of the original substance without impurities, its amount:
m (MnO2) = 90 * (1 – 0.05) = 85.5 g;

Y (MnO2) = m / M = 85.5 / 86.9 = 0.98 mol;

Y (Mn) = 0.98 mol since the amount of substances is 3 mol.

Find the mass of the product:
m (Mn) = Y * M = 0.98 * 54.9 = 54.35 g

Answer: manganese was obtained with a mass of 54.35 g.