How many liters of hydrogen will be released during the interaction of 50 g of magnesium with 30% impurities

How many liters of hydrogen will be released during the interaction of 50 g of magnesium with 30% impurities with a solution of hydrochloric acid weighing 150 g with a mass fraction of (Na Cl) 30%?

Magnesium dissolves in hydrochloric acid solution. The reaction is described by the following chemical equation:

Mg + 2HCl = MgCl2 + H2 ↑;

Let’s find the molar amount of magnesium substance.

To do this, we divide its mass by the weight of 1 mole of metallic magnesium.

M Mg = 24 grams / mol;

N Mg = 50 x 0.7 / 24 = 1.458 mol;

Let’s calculate the amount of hydrochloric acid.

M HCl = 1 + 35.5 = 36.5 grams / mol;

N HCl = 150 x 0.3 / 36.5 = 1.233 mol;

Therefore, no more than 1.233 / 2 = 0.6164 mol of hydrogen can be synthesized.

Let’s calculate the volume of hydrogen.

For this purpose, we multiply the amount of the substance by the volume of 1 mole of gas (which is 22.4 liters).

V H2 = 0.6164 x 22.4 = 13.81 liters;