The reaction between magnesium and hydrochloric acid is as follows :
`Mg(s) + 2HCl(aq) → MgCl
2(aq) + H
2(g)`
From the equation, we see that 1 mole of magnesium (Mg) produces 1 mole of hydrogen gas (H
2).
The molar mass of magnesium (Mg) is given as 24 g/mol. Therefore, 2.4 g of magnesium would correspond to (2.4 g)/(24 g/mol) = 0.1 mol of Mg.
Since 1 mole of Mg produces 1 mole of H
2, 0.1 mol of Mg would produce 0.1 mol of H
2.
At STP (Standard Temperature and Pressure), the molar volume of a gas is 22.4 L/mol.
Therefore, the volume of 0.1 mol of H
2 would be (0.1 mol)*(22.4 L/mol) = 2.24 L.
So, the volume of hydrogen liberated at STP by treating 2.4 g of magnesium with excess of hydrochloric acid is 2.24 L.
In the terms of the question where the volume is expressed as ______ × 10
-2 L, we convert 2.24 L into the desired form, i.e., 2.24 L = 2.24
10
2 10
-2 L, so the answer is 224 × 10
-2 L.
