The volume of hydrogen liberated at STP by treating $2.4 \mathrm{~g}$ of magnesium with excess of hydrochloric acid is _________ $\times ~10^{-2} \mathrm{~L}$
Given : Molar volume of gas is $22.4 \mathrm{~L}$ at STP.
Molar mass of magnesium is $24 \mathrm{~g} \mathrm{~mol}^{-1}$
Answer (integer)
224
Solution
The reaction between magnesium and hydrochloric acid is as follows :
<br/><br/>`Mg(s) + 2HCl(aq) → MgCl<sub>2</sub>(aq) + H<sub>2</sub>(g)`
<br/><br/>From the equation, we see that 1 mole of magnesium (Mg) produces 1 mole of hydrogen gas (H<sub>2</sub>).
<br/><br/>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.
<br/><br/>Since 1 mole of Mg produces 1 mole of H<sub>2</sub>, 0.1 mol of Mg would produce 0.1 mol of H<sub>2</sub>.
<br/><br/>At STP (Standard Temperature and Pressure), the molar volume of a gas is 22.4 L/mol. <br/><br/>Therefore, the volume of 0.1 mol of H<sub>2</sub> would be (0.1 mol)*(22.4 L/mol) = 2.24 L.
<br/><br/>So, the volume of hydrogen liberated at STP by treating 2.4 g of magnesium with excess of hydrochloric acid is 2.24 L.
<br/><br/>In the terms of the question where the volume is expressed as ______ × 10<sup>-2</sup> L, we convert 2.24 L into the desired form, i.e., 2.24 L = 2.24 $\times$ 10<sup>2</sup> $\times$ 10<sup>-2</sup> L, so the answer is 224 × 10<sup>-2</sup> L.
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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