How much hydrogen is required to react with aluminum oxide to obtain aluminum weighing 2.7 g?

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

Al2O3 + 3H2 = 2Al + 3H2O;

1 mole of metal oxide reacts with 3 moles of hydrogen. In this case, 2 mol of metal and 3 mol of water are synthesized.

Let’s calculate the chemical amount of a substance in 2.7 grams of aluminum.

M Al = 27 grams / mol;

N Al = 2.7 / 27 = 0.1 mol;

To obtain such an amount of metal, 0.1 / 2 x 3 = 0.15 mol of hydrogen is required.

Let’s calculate its volume.

Under normal conditions, one liter of ideal gas assumes a volume of 22.4 liters.

V H2 = 0.15 x 22.4 = 3.36 liters;