JEE Main · 2019 · Shift-ImediumMOLE-194

At 300 K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of O2 for complete combustion, and 40 mL of…

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

Question

At 300 K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of \ceO2\ce{O2} for complete combustion, and 40 mL of \ceCO2\ce{CO2} is formed. The formula of the hydrocarbon is:

Options
  1. a

    \ceC4H8\ce{C4H8}

  2. b

    \ceC4H10\ce{C4H10}

  3. c

    \ceC4H6\ce{C4H6}

  4. d

    \ceC4H7Cl\ce{C4H7Cl}

Correct Answerc

\ceC4H6\ce{C4H6}

Detailed Solution

🧠 Volume ratios are coefficients (gases at same T, P) For gases under the same conditions, 11 volume = 11 mole. So 10mL10\,\text{mL} fuel : 55mL\ceO255\,\text{mL}\,\ce{O2} : 40mL\ceCO240\,\text{mL}\,\ce{CO2} becomes 1:5.5:41 : 5.5 : 4 in moles.

🗺️ Step by step For \ceCxHy\ce{CxHy}: \ceCxHy+(x+y/4)O2>xCO2+(y/2)H2O\ce{CxHy + (x + y/4)O2 -> xCO2 + (y/2)H2O}.

From \ceCO2\ce{CO2} ratio: x=4x = 4.

From \ceO2\ce{O2} ratio: x+y/4=5.5    4+y/4=5.5    y=6.x + y/4 = 5.5 \implies 4 + y/4 = 5.5 \implies y = 6.

So the hydrocarbon is \ceC4H6\ce{C4H6}.

⚠️ Common mistake Don't compare 5555 to 4040 directly — they are both with respect to the 10mL10\,\text{mL} of fuel. Always write coefficients per 1mL1\,\text{mL} of hydrocarbon first.

Answer: (c)\boxed{\text{Answer: (c)}}

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