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QJEEM23Chbec2: JEE Main 2023 Chemistry - Class 11 JEEM23Chbec2 Solution: The volume of hydrogen liberated at STP by treating of…
The volume of hydrogen liberated at STP by treating of magnesium with excess of hydrochloric acid is _________
Given : Molar volume of gas is at STP.
Molar mass of magnesium is
Step-by-Step Verified Solution
The reaction between magnesium and hydrochloric acid is as follows :
`Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)`
From the equation, we see that 1 mole of magnesium (Mg) produces 1 mole of hydrogen gas (H2).
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 H2, 0.1 mol of Mg would produce 0.1 mol of H2.
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 H2 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 = 2.24 102 10-2 L, so the answer is 224 × 10-2 L.
`Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)`
From the equation, we see that 1 mole of magnesium (Mg) produces 1 mole of hydrogen gas (H2).
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 H2, 0.1 mol of Mg would produce 0.1 mol of H2.
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 H2 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 = 2.24 102 10-2 L, so the answer is 224 × 10-2 L.