Calculate the mass of oxygen required to burn 3g of aluminum.

The combustion reaction of aluminum is described by the following chemical reaction equation.

4Al + 3O2 = 2Al2O3;

Four moles of aluminum react with three moles of aluminum oxide.

Let’s find the amount of substance in 3 grams of aluminum.

N Al = 3/27 = 0.111 = 11.1%;

The amount of oxygen will be

N O2 = N Al x 3/4 = 0.111 x 3/4 = 0.08325 mol;

The mass of the substance will be.

M O2 = 16 x 2 = 32 grams / mol;

m O2 = 0.08325 x 32 = 2.664 grams;