What mass of manganese can be obtained from 52.2 g of manganese oxide by reducing 40 g of aluminum?

Based on the data, we will compose the equation of aluminothermy:
3MnO2 + 4Al = 3Mn + 2Al2O3 – manganese is released;

Calculations by formulas:
M (MnO2) = 86.9 g / mol;

M (Al) = 26.9 g / mol;

M (Mn) = 54.9 g / mol.

Let’s determine the amount of starting materials:
Y (MnO2) = m / M = 52.2 / 86.9 = 0.6 mol (deficient substance);

Y (Al) = m / M = 40 / 26.9 = 1.5 mol (substance in excess);

Y (Mn) = 0.6 mol since the amount of substances is 3 mol.

Calculations are made for the substance in deficiency.

Find the mass of the metal:
m (Mn) = Y * M = 0.6 * 54.9 = 32.9 g

Answer: manganese was obtained with a mass of 32.9 g.