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JEE Chemistry 2025 Question with Solution

Mass of magnesium required to produce 220 mL220 \, \text{mL} of hydrogen gas at STP on reaction with excess of dil. HCl is Given : Molar mass of Mg is 24 g mol−124 \, \text{g mol}^{-1}.

  • A

    235.7 g

  • B

    0.24 mg

  • C

    236 mg

  • D

    2.444 g

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Volume of hydrogen gas = 220 mL220 \, \text{mL} at STP. Molar mass of Mg = 24 g mol−124 \, \text{g mol}^{-1}.

Find: Mass of magnesium required.

The balanced chemical equation is

Mg+2HCl→MgCl2+H2\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_{2} + \text{H}_2

From the equation, 11 mole of Mg produces 11 mole of H2\text{H}_2.

At STP, 11 mole of any gas occupies 22.4 L=22400 mL22.4 \, \text{L} = 22400 \, \text{mL}. Therefore,

Moles of H2=22022400=0.00982 mol\text{Moles of H}_2 = \frac{220}{22400} = 0.00982 \, \text{mol}

Hence, moles of Mg required = 0.00982 mol0.00982 \, \text{mol}.

Mass of Mg=0.00982×24=0.23568 g\text{Mass of Mg} = 0.00982 \times 24 = 0.23568 \, \text{g}

Converting into milligrams,

0.23568 g=235.68 mg≈236 mg0.23568 \, \text{g} = 235.68 \, \text{mg} \approx 236 \, \text{mg}

Therefore, the correct option is C.

Detailed Calculation

Given:

  • Hydrogen gas produced = 220 mL220 \, \text{mL}
  • At STP, molar volume = 22400 mL mol−122400 \, \text{mL mol}^{-1}
  • Molar mass of Mg = 24 g mol−124 \, \text{g mol}^{-1}

Find: Mass of Mg required.

Using stoichiometry of the reaction,

Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)\text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g)

So, moles of Mg used = moles of H2\text{H}_2 formed.

Now,

Moles of H2=220 mL22400 mL/mol=222240=111120 mol\text{Moles of H}_2 = \frac{220 \, \text{mL}}{22400 \, \text{mL/mol}} = \frac{22}{2240} = \frac{11}{1120} \, \text{mol}

Therefore,

Mass of Mg=111120×24=2641120 g≈0.2357 g\text{Mass of Mg} = \frac{11}{1120} \times 24 = \frac{264}{1120} \, \text{g} \approx 0.2357 \, \text{g}

Converting grams to milligrams,

0.2357 g=235.7 mg0.2357 \, \text{g} = 235.7 \, \text{mg}

This is closest to 236 mg236 \, \text{mg} among the given options. Therefore, the correct option is C.

Common mistakes

  • Using 22.4 mL22.4 \, \text{mL} instead of 22.4 L22.4 \, \text{L} as the molar volume at STP is incorrect. At STP, 11 mole of gas occupies 22.4 L=22400 mL22.4 \, \text{L} = 22400 \, \text{mL}. Always convert the given gas volume into consistent units before calculating moles.

  • Ignoring the 1:11:1 mole ratio between Mg and H2\text{H}_2 leads to a wrong mass. The balanced equation shows that 11 mole of Mg produces 11 mole of hydrogen gas. Use the stoichiometric coefficients directly from the balanced reaction.

  • Stopping at 0.23568 g0.23568 \, \text{g} and matching it to an option in grams is incorrect because the suitable option is given in milligrams. Convert grams to milligrams by multiplying by 10001000 before selecting the answer.

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