Find the volume of hydrogen released during the interaction of zinc with 500 ml of sulfuric

Find the volume of hydrogen released during the interaction of zinc with 500 ml of sulfuric acid solution with a density of 1.14 g / ml.

Zinc metal reacts with sulfuric acid to release hydrogen gas and form zinc sulfate (zinc sulfate). The synthesis is described by the following equation:

Zn + H2SO4 = ZnSO4 + H2;

The metal reacts with acid in equal molar amounts. During the reaction, the same amounts of salt and hydrogen are synthesized.

Let’s calculate the available chemical amount of sulfuric acid. To do this, divide its weight by the weight of 1 mole of acid. A solution with a density of 1.14 g / ml contains 227.9 grams of sulfuric acid in 1 liter.

M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol; N H2SO4 = 227.9 / 2/98 = 1.163 mol;

The same amount of hydrogen and zinc sulfate will be synthesized.

Let’s calculate the volume of this amount of hydrogen gas. To do this, multiply the amount of substance by the volume of 1 mole of ideal gas (22.4 liters).

N H2 = 1.163 mol; V H2 = 1.163 x 22.4 = 26.05 liters;