When 39 g of zinc interact with hydrochloric acid, hydrogen will be released in volume (l)

1. Let’s compose the equation of interaction of zinc with hydrochloric acid:

Zn + 2HCl = ZnCl2 + H2 ↑;

2. Let’s calculate the chemical amount of zinc:

n (Zn) = m (Zn): M (Zn) = 39: 65 = 0.6 mol;

3. Determine the amount of released hydrogen:

n (H2) = n (Zn) = 0.6 mol;

4. Let’s calculate the volume of hydrogen:

V (H2) = n (H2) * Vm = 0.6 * 22.4 = 13.44 liters.

Answer: 13.44 liters.

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