JEE Main · 2022 · Shift-IImediumMOLE-129

Consider the reaction: 4HNO3(l) + 3KCl(s) - Cl2(g) + NOCl(g) + 2H2O(g) + 3KNO3(s) The amount of HNO3 required to…

Some Basic Concepts (Mole Concept) · Class 11 · JEE Main Previous Year Question

Question

Consider the reaction: \ce4HNO3(l)+3KCl(s)−>Cl2(g)+NOCl(g)+2H2O(g)+3KNO3(s)\ce{4HNO3(l) + 3KCl(s) -> Cl2(g) + NOCl(g) + 2H2O(g) + 3KNO3(s)}

The amount of \ceHNO3\ce{HNO3} required to produce 110.0 g110.0\,\text{g} of \ceKNO3\ce{KNO3} is: (Atomic masses: H = 1, O = 16, N = 14, K = 39)

Options
  1. a

    32.2 g32.2\,\text{g}

  2. b

    69.4 g69.4\,\text{g}

  3. c

    91.5 g91.5\,\text{g}

    ✓
  4. d

    162.5 g162.5\,\text{g}

Correct Answerc

91.5 g91.5\,\text{g}

Detailed Solution

🧠 The 4:34:3 ratio of acid to nitrate The given equation needs 44 moles of \ceHNO3\ce{HNO3} to make 33 moles of \ceKNO3\ce{KNO3}. So acid moles =(4/3)×nitrate moles= (4/3) \times \text{nitrate moles}.

🗺️ Three steps Moles of \ceKNO3=110/101≈1.089 mol\ce{KNO3} = 110 / 101 \approx 1.089\,\text{mol}. Moles of \ceHNO3\ce{HNO3} needed =(4/3)×1.089≈1.452 mol= (4/3) \times 1.089 \approx 1.452\,\text{mol}. Mass of \ceHNO3=1.452×63≈91.5 g\ce{HNO3} = 1.452 \times 63 \approx 91.5\,\text{g}.

⚠️ Common mistake Treating it as a 1:11:1 ratio and getting ≈68.6 g\approx 68.6\,\text{g} (close to option (b)). Always read the coefficients in the balanced equation — here it's 4:34:3, not 1:11:1.

Answer: (c)\boxed{\text{Answer: (c)}}

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Consider the reaction: 4HNO3(l) + 3KCl(s) - Cl2(g) + NOCl(g) + 2H2O(g) + 3KNO3(s) The amount of… (JEE Main 2022) | Canvas Classes