JEE Main · 2023 · Shift-IIhardMOLE-115

When a hydrocarbon A undergoes combustion in the presence of air, it requires 9.5 equivalents of oxygen and produces 3…

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

Question

When a hydrocarbon A undergoes combustion in the presence of air, it requires 9.59.5 equivalents of oxygen and produces 33 equivalents of water. What is the molecular formula of A?

Options
  1. a

    \ceC8H6\ce{C8H6}

    ✓
  2. b

    \ceC9H9\ce{C9H9}

  3. c

    \ceC6H6\ce{C6H6}

  4. d

    \ceC9H6\ce{C9H6}

Correct Answera

\ceC8H6\ce{C8H6}

Detailed Solution

🧠 Water count fixes H, oxygen count fixes C For \ceCxHy\ce{C_xH_y} combustion: \ceCxHy+(x+y/4) \ceO2→x \ceCO2+(y/2) \ceH2O\ce{C_xH_y} + (x + y/4)\,\ce{O2} \rightarrow x\,\ce{CO2} + (y/2)\,\ce{H2O}.

🗺️ Read off the numbers 33 equivalents of \ceH2O\ce{H2O} means y/2=3  ⟹  y=6y/2 = 3 \implies y = 6.

9.59.5 equivalents of \ceO2\ce{O2} means: x+y/4=9.5  ⟹  x+1.5=9.5  ⟹  x=8.x + y/4 = 9.5 \implies x + 1.5 = 9.5 \implies x = 8.

So the formula is \ceC8H6\ce{C8H6} (phenylacetylene).

⚡ Quick option scan y=6y = 6 knocks out (b). Of the remaining options, only \ceC8H6\ce{C8H6} gives 8+1.5=9.58 + 1.5 = 9.5.

Answer: (a)\boxed{\text{Answer: (a)}}

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