JEE Main · 2022 · Shift-ImediumMOLE-138

SO2Cl2 + 2H2O - H2SO4 + 2HCl. 16 moles of NaOH is required for complete neutralisation of the resultant acidic mixture.…

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

Question

\ceSO2Cl2+2H2O>H2SO4+2HCl\ce{SO2Cl2 + 2H2O -> H2SO4 + 2HCl}. 16 moles of NaOH is required for complete neutralisation of the resultant acidic mixture. The number of moles of \ceSO2Cl2\ce{SO2Cl2} used is:

Options
  1. a

    1616

  2. b

    88

  3. c

    44

  4. d

    22

Correct Answerc

44

Detailed Solution

🧠 Each \ceSO2Cl2\ce{SO2Cl2} gives 4\ceH+4\,\ce{H+} The hydrolysis \ceSO2Cl2+2H2O>H2SO4+2HCl\ce{SO2Cl2 + 2H2O -> H2SO4 + 2HCl} produces:

  • 11 mole of \ceH2SO4\ce{H2SO4}2\ceH+2\,\ce{H+}
  • 22 moles of \ceHCl\ce{HCl}2\ceH+2\,\ce{H+}
  • Total: 4\ceH+4\,\ce{H+} per \ceSO2Cl2\ce{SO2Cl2}.

🗺️ Two steps \ceNaOH\ce{NaOH} used =16= 16 moles → 1616 moles of \ceH+\ce{H+} neutralised. Moles of \ceSO2Cl2=16/4=4\ce{SO2Cl2} = 16 / 4 = 4.

⚠️ Common mistake Counting only the \ceH2SO4\ce{H2SO4} and ignoring the \ceHCl\ce{HCl}. Both acids take part — count every \ceH+\ce{H+} produced.

Answer: (c)\boxed{\text{Answer: (c)}}

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SO2Cl2 + 2H2O - H2SO4 + 2HCl. 16 moles of NaOH is required for complete neutralisation of the… (JEE Main 2022) | Canvas Classes