The volume of hydrogen released during the interaction of 0.25 mol of magnesium with
| September 30, 2021 | education
The volume of hydrogen released during the interaction of 0.25 mol of magnesium with an excess of hydrochloric acid is …
Reaction of magnesium with hydrochloric acid.
Mg + 2HCl = MgCl2 + H2.
The number of moles of magnesium and hydrogen will be equal.
n (Mg) = n (H2) = 0.25 mol.
The volume of the generated hydrogen.
V (H2) = n • Vm = 0.25 mol • 22.4 mol / L = 5.6 L.
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