What volume of hydrogen will be released during the interaction of hydrochloric acid with 337.5

What volume of hydrogen will be released during the interaction of hydrochloric acid with 337.5 kg of technical aluminum containing 80 pure aluminum?

The reaction of aluminum with hydrochloric acid is described by the following chemical reaction equation.

Al + 3HCl = AlCl3 + 1.5 H2;

When 1 mol of aluminum is dissolved, 1.5 mol of hydrogen is released. This requires 3 mol of hydrochloric acid.

The mass of pure aluminum will be 337,500 x 0.8 = 270,000 grams;

Let’s determine the amount of substance contained in 270,000 grams of aluminum.

M Al = 27 grams / mol;

N Al = 270,000/27 = 10,000 mol;

Reaction with 10,000 moles of aluminum will produce 15,000 moles of hydrogen.

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

Let’s determine the volume of hydrogen.

V H2 = 15,000 x 22.4 = 336,000 liters = 336 m3;



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