Calculate the mass of aluminum and the volume of oxygen that would be required to obtain 27 g of alumina.

1. Let’s compose the reaction equation:

4 Al +3 O2 = 2 Al2O3

According to the equation: 4 mol of Al and 3 mol of O2 enter into the reaction, 2 mol of Al2O3 are formed.

Let’s find the mass of aluminum by the formula:

m (Al) = n * M = 4mol * 27g / mol = 108g

Let’s find the volume of O2 by the formula:

V (O2) = n * Vm = 3 mol * 22.4 l / mol = 67.2 l

Let’s find the mass of Al2O3 by the formula:

m (Al2O3) = n * M = 2 mol * (27 * 2 + 16 * 3) = 2 * 102 = 204 g

2. Let’s calculate the mass of Al:

x g Al – 27 g Al2O3

108 g Al – 204 g Al2O3

Hence, x = 108 * 27/204 = 14.3 g.

3. Let’s calculate the volume of О2:

xl O2 – 27 g Al2O3

67.2 l O2 – 204 g Al2O3

Hence, x = 67.2 * 27/204 = 8.9 liters.