Calculate the volume of oxygen spent on the reaction with 2.7 g of aluminum and the mass of the resulting oxide?

1. Let’s compose the equation of aluminum oxidation:

4Al + 3O2 = 2Al2O3;

2. find the chemical amount of aluminum:

n (Al) = m (Al): M (Al) = 2.7: 27 = 0.1 mol;

3. Determine the amount of oxygen consumed and the resulting oxide:

n (O2) = n (Al) * 3: 4 = 0.1 * 3: 4 = 0.075 mol;

n (Al2O3) = n (Al): 2 = 0.1: 2 = 0.05 mol;

4.Calculate the volume of oxygen:

V (O2) = n (O2) * Vm = 0.075 * 22.4 = 1.68 l;

5.Calculate the mass of the oxide:

m (Al2O3) = n (Al2O3) * M (Al2O3);

M (Al2O3) = 2 * 27 + 3 * 16 = 102 g / mol;

m (Al2O3) = 0.05 * 102 = 5.1 g.

Answer: 1.68 l; 5.1 g