🧲 Mole Concept in Combustion Reactions | JEE Chemistry | Doubtify JEE

🧮 Question:

100 grams of propane is completely reacted with 1000 grams of oxygen. The mole fraction of carbon dioxide in the resulting mixture is X × 10⁻². The value of x is:

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🖼️ Question Image:

🔬 Concept Overview:

This is a classic mole concept and stoichiometry problem involving the combustion of propane (C₃H₈). These kinds of questions appear frequently in JEE Mains and Advanced, especially in the Mole Concept, Gaseous State, and Chemical Reactions chapters.

We are asked to find the mole fraction of CO₂ in the final gaseous mixture formed after the complete combustion of propane in excess oxygen.

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⚗️ Step-by-Step Breakdown:

1. Balanced combustion reaction of propane:

$$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$$

2. Molar Mass of C₃H₈ = 44 g/mol

$$\Rightarrow \text{Moles of Propane} = \frac{100}{44} \approx 2.27 \, \text{mol}$$

3. Molar Mass of O₂ = 32 g/mol

$$\Rightarrow \text{Moles of Oxygen} = \frac{1000}{32} \approx 31.25 \, \text{mol}$$

4. From the equation:
   1 mol of C₃H₈ reacts with 5 mol of O₂
   So, 2.27 mol C₃H₈ requires 2.27 × 5 = 11.35 mol O₂

But we have 31.25 mol O₂, which is in excess. So propane is the limiting reagent.

5. From stoichiometry:
   1 mol C₃H₈ → 3 mol CO₂
   So, 2.27 mol C₃H₈ → 2.27 × 3 = 6.81 mol CO₂

6. Moles of H₂O formed = 2.27 × 4 = 9.08 mol

7. Oxygen left unreacted = 31.25 – 11.35 = 19.9 mol

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🧮 Total moles in product mixture:

• CO₂ = 6.81 mol

• H₂O = 9.08 mol

• O₂ (unreacted) = 19.9 mol

• Total = 6.81 + 9.08 + 19.9 = 35.79 mol

Mole fraction of CO₂:

$$\frac{\mathrm{<b>6.81</b>}}{\mathrm{<b>35.79</b>}}<b>\approx </b>\mathrm{<b>0.1902</b>}<b>=</b>\mathrm{<b>19.02</b>}<b>\times </b>{\mathrm{<b>10</b>}}^{<b>-</b>\mathrm{<b>2</b>}}<b>\Rightarrow </b>\boxed{{\displaystyle \mathrm{<b>X</b>}<b>=</b>\mathrm{<b>19</b><b><br></b>}}}$$

🔍 Why This Question is Important:

• It sharpens stoichiometric understanding using real-world combustion problems.

• Reinforces concepts like mole fraction, limiting reagent, and excess reactant.

• Very relevant for both JEE Main and Advanced level calculations.

🧠 Solution (Image):

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🎥 Video Solution:

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Tip: Don’t just memorize reactions – understand limiting reagents and gas volumes. Practice similar combustion questions to improve your speed in the exam!