Butane Combustion Water Yield Calculation

 

❓ Question

Butane reacts with oxygen to produce carbon dioxide and water following the equation:

$${\mathrm{C}}_4{\mathrm{H}}_{10}(g)+\frac{13}{2}{\mathrm{O}}_2(g)⟶4\mathrm{C}{\mathrm{O}}_2(g)+5{\mathrm{H}}_2\mathrm{O}(l)$$

If 174.0 kg of butane is mixed with 320.0 kg of O₂, the volume of liquid water formed (in liters, nearest integer) is ________.

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 Mole concept
👉 Limiting reagent
👉 Stoichiometry.

Main idea:

First identify limiting reagent, then calculate moles of water formed.

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🔷 Step 1 — Calculate Moles of Butane 💯

Molar mass of butane:

$$C_4H_{10}$$
$$=4(12)+10(1)$$
$$=58\text{ g/mol}$$

Given mass:

$$174.0\text{ kg}=174000\text{ g}$$

Moles of butane:

$$n=\frac{174000}{58}$$
$$=3000\text{ mol}$$

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🔷 Step 2 — Calculate Moles of Oxygen

Molar mass of oxygen:

$$O_2=32\text{ g/mol}$$

Given:

$$320.0\text{ kg}=320000\text{ g}$$

Moles of oxygen:

$$n=\frac{320000}{32}$$
$$=10000\text{ mol}$$

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🔷 Step 3 — Find Limiting Reagent

Reaction:

$$1\text{ mol }C_4H_{10}$$

requires:

$$\frac{13}{2}=6.5\text{ mol }O_2$$

For $3000$ mol butane, oxygen needed:

$$3000\times6.5$$
$$=19500\text{ mol}$$

But available oxygen:

$$10000\text{ mol}$$

Thus:

$$\boxed{ O_2\text{ is limiting reagent} }$$

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🔷 Step 4 — Calculate Water Formed

From equation:

$$\frac{13}{2}O_2 \rightarrow 5H_2O$$

Thus:

$$6.5\text{ mol }O_2 \rightarrow5\text{ mol }H_2O$$

For $10000$ mol $O_2$:

$$H_2O= 10000\times\frac5{6.5}$$
$$= \frac{50000}{6.5}$$
$$\approx7692.3\text{ mol}$$

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🔷 Step 5 — Convert into Volume

Mass of water:

$$7692.3\times18$$
$$\approx138461.5\text{ g}$$

Density of water:

$$1\text{ g/mL}$$

Hence:

$$138461.5\text{ mL}$$
$$=138.46\text{ L}$$

Nearest integer:

$$\boxed{ 138\text{ L} }$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ Limiting reagent check na karna

❌ kg to g conversion bhool jaana

❌ Water density use na karna

Remember:

Always:

$$\boxed{ \text{First find limiting reagent} }$$

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✅ Final Answer

$$\boxed{ 138\text{ L} }$$

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