Determine the mass of the liberated metal in the interaction of 0.56 liters of hydrogen and with aluminum oxide.

1. Let’s compose the reaction equation:

Al2O3 + 3 H2 = 2 Al + 3 H2O

According to the equation:

3 mol of hydrogen enters into the reaction;
2 mol of aluminum is formed.
Let’s find the volume of hydrogen by the formula:

V (H2) = Vm * n = 22.4 l / mol * 3 mol = 67.2 l

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

m (Al) = n * M = 2 mol * 27 g / mol = 54 g

2. Let’s calculate the mass of the released aluminum, making up the proportion:

0.56 l H2 – x g Al

67.2 l H2 – 54 g Al

Hence, x = 0.56 * 54 / 67.2 = 0.45 g.