What volume of hydrogen will be released when phosphoric acid interacts with magnesium weighing 6 grams?

The interaction of magnesium with phosphoric acid occurs in accordance with the following chemical reaction equation:

3Mg + 2H3PO4 = Mg3 (PO4) 2 + 3H2;

When 1 metal dissolves, 1 mol of hydrogen gas is released.

Let’s calculate the chemical amount of a substance in 6 grams of metal.

N Mg = 6/24 = 0.25 mol;

This amount of hydrogen will be released.

Let’s calculate its volume.

One mole of gas under normal conditions fills a volume of 22.4 liters.

The volume of hydrogen will be:

V H2 = 0.25 x 22.4 = 5.6 liters;