Question

Determine the volume of hydrogen liberated by 3.2 gm of the metal with an acid, e.q. wt. of the metal is 32.

Answer 1

32 gm metal liberate 1 gm H₂ or 11.2 liters of H₂ at S.T.P.

3.2 g metal liberate = \(\frac{11.2 \times 3.2}{32}\) = 1.12 liters