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
Given : Molar volume of gas is 22.4 L at STP.
Molar mass of magnesium is 24 g mol–1
Correct Answer: 224
Solution and Explanation
Given:
- Mass of magnesium (\( w \)) = 2.4 g
- Molar mass of magnesium (\( M \)) = 24 g/mol
- Molar volume of gas at STP = 22.4 L
Step 1: Write the Balanced Chemical Equation
The reaction of magnesium with hydrochloric acid is:
\[ \text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \]
Step 2: Calculate the Moles of Magnesium
The number of moles of magnesium (\( n \)) is given by:
\[ n = \frac{w}{M} = \frac{2.4 \, \text{g}}{24 \, \text{g/mol}} = 0.1 \, \text{mol}. \]
Step 3: Calculate the Moles of Hydrogen Gas Produced
From the balanced equation, 1 mole of \( \text{Mg} \) produces 1 mole of \( \text{H}_2 \). Therefore:
\[ \text{Moles of } \text{H}_2 = 0.1 \, \text{mol}. \]
Step 4: Calculate the Volume of Hydrogen Gas at STP
The volume of 1 mole of gas at STP is 22.4 L. Thus, the volume of \( 0.1 \, \text{mol} \) of \( \text{H}_2 \) is:
\[ V = n \times 22.4 \, \text{L/mol} = 0.1 \, \text{mol} \times 22.4 \, \text{L/mol} = 2.24 \, \text{L} = 224 \times 10^{-2} \, \text{L}. \]
Final Answer:
The volume of hydrogen liberated at STP is \( 224 \times 10^{-2} \, \text{L} \).
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