Calculate the mass of aluminum oxide that is formed when burning aluminum in 4.48 liters of oxygen?

Given:

V (O2) = 4.48 L

Find:

m (Al2O3) -?

Solution:

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

4Al + 3O2 = 2Al2O3;

2) Find the amount of oxygen contained in 4.48 liters of gas:

n (O2) = V: Vm = 4.48 L: 22.4 L / mol = 0.2 mol;

3) We compose logical equality:

if 2 mol of Al2O3 requires 3 mol of O2,

then x mol of Al2O3 will require 0.2 mol of O2,

then x = 0.2 * 2: 3 = 0.13 mol.

4) Find the mass of aluminum oxide formed during the reaction:

m (Al2O3) = n * M = 0.13 mol * 102 g / mol = 13.26 g.

Answer: m (Al2O3) = 13.26 g.