Mole Fraction of CO2 After Combustion Calculation

❓ Question

100 g of propane is completely reacted with 1000 g of oxygen. The mole fraction of carbon dioxide in the resulting mixture is

$$x \times 10^{-2}$$

The value of $x$ is ______.

(Nearest integer)

Atomic weights:

$$H=1.008,\ C=12.00,\ O=16.00$$

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🖼 Question Image

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✍️ Short Explanation

This is a stoichiometry + mole fraction problem.

👉 First balance combustion reaction
👉 Find limiting reagent
👉 Calculate final moles of gases
👉 Apply mole fraction formula.

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🔷 Step 1 — Balanced Reaction 💯

Combustion of propane:

$$C_3H_8+5O_2\rightarrow3CO_2+4H_2O$$

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🔷 Step 2 — Calculate Initial Moles

Molar mass of propane:

$$=3(12)+8(1.008)$$
$$=36+8.064$$
$$=44.064$$

Moles of propane:

$$=\frac{100}{44.064}$$
$$\approx2.27$$

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Moles of oxygen:

$$=\frac{1000}{32}$$
$$=31.25$$

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🔷 Step 3 — Identify Limiting Reagent

From reaction:

$$1\text{ mol propane requires }5\text{ mol }O_2$$

For:

$$2.27\text{ mol propane}$$

Required oxygen:

$$=2.27\times5$$
$$\approx11.35$$

Available oxygen:

$$31.25$$

So oxygen is excess.

Propane is limiting reagent.

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🔷 Step 4 — Calculate Final Moles

CO$_2$ formed:

$$=3\times2.27$$
$$\approx6.81$$

Oxygen left:

$$31.25-11.35$$
$$\approx19.90$$

Water formed:

$$=4\times2.27$$
$$\approx9.08$$

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🔷 Step 5 — Mole Fraction of CO${}_{\mathrm{2}}$

Resulting gaseous mixture contains:

$$CO_2 + O_2$$

(Water assumed condensed)

Total moles:

$$6.81+19.90$$
$$=26.71$$

Mole fraction of CO$_2$:

$$=\frac{6.81}{26.71}$$
$$\approx0.255$$
$$=25.5\times10^{-2}$$

Nearest integer:

$$\boxed{26}$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ Water vapour ko gaseous mixture mein include kar lena

❌ Limiting reagent galat identify kar dena

❌ Combustion equation balance na karna

Remember:

$$\text{Mole fraction}= \frac{\text{moles of component}}{\text{total moles}}$$

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✅ Final Answer

$$\boxed{26}$$

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📚 Related Topics

• Coordination Compounds Hybridization and Magnetism
• First Order Growth and Second Order Decay Trick
• Mole Concept Combustion Analysis Shortcut