The volume of 5.0 m hcl is required to react completely with 3.00 g of magnesium is 20 L.
Mg + 2HCl -----MgCl2 + H2
Given,
Molar mass of Mg = 24g/mol
Mole = Given mass/ Molar mass
Mole= 3/24
= 0.125 mol
From the given equation we get to know that
Mol ratio of Mg and HCl is 1:2.
Therefore,
mol of HCl = (2/1) × mol of Mg
=2× 0.125
= 0.25
Molarity = m× V
= M/m
= 5/0.25
= 20L
Thus, we find that the volume of 5.0 m hcl is required to react completely with 3.00 g of magnesium is 20 L.
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