Question

calculate the volume of hydrogen released when Zinc reacted with 294g sulphuric acid ​

Answer 1

Answer:

calculate the volume of hydrogen released when Zinc reacted with 294g sulphuric acid

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Answer 2

Given:

Mass of sulfuric acid, M = 294 gm

To Find:

The volume of hydrogen gas released upon the reaction of zinc with the given mass of sulfuric acid.

Calculation:

- The reaction of zinc with sulfuric acid can be given as:

Zn + H₂SO₄ → ZnSO₄ + H₂

- According to the above given reaction:

Moles of hydrogen gas produced when zinc reacts with 98 gm of Sulfuric acid = 1 mole

⇒ Moles of hydrogen gas produced when zinc reacts with 294 gm of Sulfuric acid = (294/98) = 3 moles

- Volume of 3 mole hydrogen gas at standard conditions = 3 × 22.4

⇒ V = 67.2 L

- Hence, the volume of hydrogen gas released when zinc reacts with 294 gm of sulfuric acid is 67.2 L.