JEE Main · 2023 · Shift-IIhardMOLE-169

When hydrocarbon A undergoes complete combustion it requires 11 equivalents of O2 and produces 4 equivalents of water.…

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

Question

When hydrocarbon A undergoes complete combustion it requires 1111 equivalents of \ceO2\ce{O2} and produces 44 equivalents of water. Molecular formula of A is:

Options
  1. a

    \ceC11H8\ce{C11H8}

  2. b

    \ceC11H4\ce{C11H4}

  3. c

    \ceC5H8\ce{C5H8}

  4. d

    \ceC9H8\ce{C9H8}

    ✓
Correct Answerd

\ceC9H8\ce{C9H8}

Detailed Solution

🧠 The combustion template For \ceCxHy\ce{C_xH_y}: \ceCxHy+(x+y/4)O2−>xCO2+(y/2)H2O.\ce{C_xH_y + (x + y/4) O2 -> x CO2 + (y/2) H2O}. Match the coefficients to the given equivalents.

🗺️ Two equations Water: y/2=4⇒y=8y/2 = 4 \Rightarrow y = 8. Oxygen: x+y/4=11⇒x+2=11⇒x=9x + y/4 = 11 \Rightarrow x + 2 = 11 \Rightarrow x = 9.

So the formula is \ceC9H8\ce{C9H8}.

⚡ Speed scan Check (d): \ceC9H8\ce{C9H8}. \ceO2\ce{O2} needed =9+8/4=11= 9 + 8/4 = 11. Water =8/2=4= 8/2 = 4. Both match.

Answer: (d)\boxed{\text{Answer: (d)}}

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