The volume of HCl = 2.10⁴ mL
Given
5.0 M(mol/L) HCl solution
3 g of Mg
Required
Volume of HCl
Solution
Reaction
Mg + 2HCl ⇒ MgCl₂ + H₂
mol of Mg (Ar=24 g/mol) :
mol = mass : Ar
mol = 3 : 24
mol = 0.125
From the equation, mol ratio of Mg : HCl = 1 : 2, so mol HCl :
= 2/1 x mol Mg
= 2 x 0.125
= 0.25
Molarity : mol solute per liters solution
M = m x V
V = M : m
V = 5 mol/L : 0.25
V = 20 L = 2.10⁴ mL