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Chemistry Question 51 – JEE-MAIN 2025

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Mass of magnesium required to produce 220 mL of hydrogen gas at STP on reaction with excess of dil. HCl is Given: Molar mass of Mg is 24 g mol−1.

The problem requires understanding the balanced chemical reaction and applying the ideal gas law concept at STP to determine the moles of hydrogen gas produced.

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Ninja Strategy•Magnitude Estimation

Quickly estimate the order of magnitude for the moles of gas and corresponding mass of Mg. 220 mL is about 0.01 moles of gas, so the mass of Mg should be around 0.01×24 g=0.24 g or 240 mg. Only option 2 is in this range.

Step 1: Write the balanced chemical equation and determine mole ratio.✦ Active

The reaction between magnesium and dilute hydrochloric acid is:

Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)

From the balanced equation, 1 mole of Mg produces 1 mole of H2 gas.

Step 2: Calculate moles of hydrogen gas produced at STP.○ Expand

Given volume of H2=220 mL=0.220 L. At STP, 1 mole of any gas occupies 22.4 L. Therefore, the moles of H2 produced are:

Moles of H2=Volume of H2Molar volume at STP=0.220 L22.4 L/mol≈0.009821 mol
Step 3: Calculate the mass of magnesium required.○ Expand

Since 1 mole of Mg produces 1 mole of H2, the moles of Mg required are 0.009821 mol. The molar mass of Mg is 24 g mol−1. The mass of Mg required is:

Mass of Mg=Moles of Mg×Molar mass of Mg

Mass of Mg=0.009821 mol×24 g/mol≈0.2357 g. Converting to milligrams: 0.2357 g×1000 mg/g=235.7 mg. This value is approximately 236 mg.

💡 Teacher's Secret Hint

Pay attention to unit conversions (mL to L, g to mg).

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