Calculate the volume of `H_(2)` liberated at `27^(@)C` and 760 mm of Hg pressure by action by 0.6 g magnesium with excess of dil HCl.

Explanation

Magnesium reacts with dilute hydrochloric acid as
`Mg+2HCl toMgCl_(2)+H_(2)`
No. of moles of Mg = `0.6/24` = 0.025
No. of moles of `H_(2)` = 0.025 `" "(therefore1" mole Mg liberates")`
Ideal gas equation PV = nRT
P = 760 mm = 1 atm T = 27 + 273 = 300K
V = ? `" "` n = 0.025
R = 0.0821
Substituting these values in ideal gas equation.
Volume of `H_(2),V=(nRT)/P`
= `(0.025xx0.0821xx300)/1`
= 0.615 L