JEE Main 2025 · 3 April · evening · Single correct · +4 / −1

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.

  1. (a) 0.24 mg
  2. (b) 235.7 g
  3. (c) 2.444 g
  4. (d) 236 mg
Answer and solution

Answer (d) 236 mg

Solution

Chemistry ma’am · 3 steps

One Mg gives one H2, so find the moles of hydrogen and you have the moles of magnesium.

  1. Balanced equation. Mg + 2HCl → MgCl₂ + H₂. One mole of Mg gives one mole of H₂, so the two mole counts are equal.
  2. Moles of hydrogen. At STP one mole of gas occupies 22.4 L = 22400 mL. So n(H₂) = 220/22400 = 9.82 × 10⁻³ mol.

    m(Mg) = 220/22400 × 24 = 0.2357 g

  3. So. 0.2357 g = 235.7 mg ≈ 236 mg, which is option (d).

The idea to keep. Gas volume at STP divided by 22.4 L gives moles. Use the balanced equation's mole ratio, then multiply by molar mass. Watch the units on the volume and on the final mass.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Mole Concept & Stoichiometry

Asked about 210 times in JEE Main since 2020. This question is from 2025.