Question:medium

The volume of hydrogen liberated at STP by treating 2.4 g of magnesium with excess of hydrochloric acid is _________ × 10–2 L
Given : Molar volume of gas is 22.4 L at STP.
Molar mass of magnesium is 24 g mol–1

Updated On: Feb 26, 2026
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Correct Answer: 224

Solution and Explanation

To find the volume of hydrogen gas liberated, we start by using the chemical reaction between magnesium (Mg) and hydrochloric acid (HCl):
Mg + 2HCl → MgCl2 + H2
From the reaction, 1 mole of Mg produces 1 mole of H2.
Given:
  • Molar mass of Mg = 24 g/mol
  • Molar volume of gas at STP = 22.4 L/mol
Mass of Mg given = 2.4 g
Moles of Mg = 2.4 g/24 g/mol = 0.1 mol
Since 1 mole of Mg produces 1 mole of H2, 0.1 mol of Mg will liberate 0.1 mol of H2.
Volume of H2 at STP = Moles of H2 × Molar volume = 0.1 mol × 22.4 L/mol = 2.24 L
Expressing 2.24 L in the required format:
2.24 L = 224 × 10−2 L
The computed value, 224, fits precisely within the specified range of 224 to 224. Therefore, the volume of hydrogen liberated is 224 × 10−2 L at STP.
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