Stoichiometry Trick: How Much Water Forms When Butane Burns? 🔥

 

❓ 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 ________.

---

🖼️ Question Image

---

✍️ Short Solution

We solve by:

1. Converting given masses to moles.

2. Identifying the limiting reagent using reaction stoichiometry.

3. Using stoichiometry to find moles (and mass) of H₂O produced.

4. Converting mass of liquid water to volume (density ≈ 1.00 g·mL⁻¹).

---

Step 1 — Molar masses & moles

• Molar mass (approx):
  $M_{\mathrm{C_4H_{10}}}=4(12.01)+10(1.008)=58.12\ \text{g·mol}^{-1}$
  

• $M_{\mathrm{O_2}}=32.00\ \text{g·mol}^{-1}$
  $M_{\mathrm{H_2O}}=18.015\ \text{g·mol}^{-1}$
  

• Moles of butane:

  $$n_{\mathrm{C_4H_{10}}}=\frac{174000\ \text{g}}{58.12\ \text{g·mol}^{-1}}\approx 2993.81\ \text{mol}$$
  

• Moles of oxygen:

  $$n_{{\mathrm{O}}_2}=\frac{320000\text{ g}}{32.00{\text{g}\text{<span style="color: black;">mol</span>}}^{-1}}=10000.00\text{ mol}$$

---

Step 2 — Limiting reagent

Reaction requires $\tfrac{13}{2}=6.5$ mol O₂ per 1 mol C₄H₁₀.

• O₂ required to fully consume available butane:

  $$6.5\times 2993.8059 \approx 19459.74\ \text{mol O}_2$$

We have only 10000 mol O₂ ⇒ O₂ is limiting.

---

Step 3 — Moles of H₂O formed

From reaction: 6.5 mol O₂ → 5 mol H₂O
So per mole O₂, H₂O formed = $5\mathrm{/}6.5=0.76923077$

Moles H₂O produced:

$$n_{{\mathrm{H}}_2\mathrm{O}}=10000\times \frac{5}{6.5}\approx 7692.308\text{ mol}$$

Mass of H₂O:

$$m_{{\mathrm{H}}_2\mathrm{O}}=7692.308\times 18.015\approx 138576.9\text{ g}=138.5769\text{ kg}$$

---

Step 4 — Volume of liquid water

Using density of water ≈ 1.00 g·mL⁻¹ (i.e. 1 kg = 1 L):

$$\text{Volume}=\frac{138576.9\text{ g}}{1000{\text{g}\text{<span style="color: black;"></span>}\text{L}}^{-1}}\approx 138.5769\text{ L}$$

Nearest integer: 139 L.

---

✅ Final Answer

$$\boxed{\text{Volume of water produced} \approx 139\ \text{L}}$$