To reduce manganese from manganese (IV) oxide by alumothermy, 10.8 g

To reduce manganese from manganese (IV) oxide by alumothermy, 10.8 g of aluminum were used. Determine the mass of the manganese formed.

Let’s find the amount of aluminum substance Al.

M (Al) = 27 g / mol.

n = 10.8 g: 27 g / mol = 0.4 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

3 MnO2 + 4Al = 2Al2O3 + 3Mn

For 4 mol of Al, there is 3 mol of Mn. Substances are in quantitative ratios

4: 3. The amount of Mn substance is 4/3 less than Al.

n (Mn) = 4 / 3n (Al).

4 mol Al – 3 mol Mn,

0.4 mol – x mol Mn,

X = (0.4 mol × 3 mol): 4 mol = 0.3 mol.

Let’s find the mass Mn.

M (Mn) = 55 g / mol.

m = n × M.

m = 55 g / mol × 0.3 mol = 16.5 g.

Answer: 16.5 g.