Calculate the mass of aluminum that is required for the reduction by the method

Calculate the mass of aluminum that is required for the reduction by the method of aluminothermy ore weighing 100 g containing 87% manganese oxide.

Given:

m (MnO2) = 100 g

w% (MnO2) = 87%

To find:

m (Al) -?

Decision:

1) We compose the reaction equation corresponding to the condition of the problem:

4Al + 3MnO2 = 3Mn + 2Al2O3;

2) Find the mass of manganese oxide in the ore:

m (MnO2) = 100 g * 0.87 = 87 g;

3) Find the amount of manganese oxide:

n (MnO2) = m: M = 87 g: 87 g / mol = 1 mol;

4) We compose logical equality:

if 3 mol of MnO2 requires 4 mol of Al,

then 1 mol of MnO2 will require x mol of Al,

then x = 0.75 mol.

5) Find the mass of aluminum required for the reaction with the ore:

m (Al) = n * M = 0.75 mol * 27 g / mol = 20.25 g;

Answer: m (Al) = 20.25 g.