8. Butane, which is the fuel used in cigarette lighters, combusts according to the following thermochemical equation: 2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H₂O(g) AH = -5316.6 kJ LR If 10.0 g of butane burns completely in oxygen and the heat is used to warm up 25.0 L of water at 22.0 °C, what will be the final temperature of the water (assuming that no heat is lost to the surroundings and that d = 1.00 g/mL for water))?

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### Thermochemistry of Butane Combustion

**Problem Statement:**
Butane, the fuel used in cigarette lighters, combusts according to the following thermochemical equation:

\[ 
2 \, \text{C}_4\text{H}_{10(g)} + 13 \, \text{O}_{2(g)} \rightarrow 8 \, \text{CO}_{2(g)} + 10 \, \text{H}_2\text{O}_{(g)} \quad \Delta H^\circ = -5316.6 \, \text{kJ} 
\]

If 10.0 g of butane burns completely in oxygen and the heat is used to warm up 25.0 L of water at 22.0 °C, what will be the final temperature of the water (assuming that no heat is lost to the surroundings and that \(d = 1.00 \, \text{g/mL} \) for water)?

**Analysis and Calculations:**

1. **Understanding the Reaction:**
   - The combustion of butane is an exothermic reaction, releasing a significant amount of energy (\(-5316.6 \, \text{kJ}\)).

2. **Heat Transfer to Water:**
   - 10.0 g of butane is used, and this energy will heat 25.0 L of water.
   - Convert 25.0 L of water to grams: \( 25,000 \, \text{g} \) (assuming \(d = 1.00 \, \text{g/mL} \)).

3. **Calculate Moles of Butane:**
   - Molar mass of butane (C\(_4\)H\(_{10}\)): approximately \(58.12 \, \text{g/mol}\).
   - Moles of butane in 10.0 g: 
     \[
     \frac{10.0 \, \text{g}}{58.12 \, \text{g/mol}} \approx 0.172 \, \text{mol}
     \]

4. **Energy Released:**
   - Total energy released by 0.172 mol of butane:
     \[
     (0.172 \, \text{mol}) \times \left(\frac{-5316.6 \, \text{kJ}}{2 \
Transcribed Image Text:### Thermochemistry of Butane Combustion **Problem Statement:** Butane, the fuel used in cigarette lighters, combusts according to the following thermochemical equation: \[ 2 \, \text{C}_4\text{H}_{10(g)} + 13 \, \text{O}_{2(g)} \rightarrow 8 \, \text{CO}_{2(g)} + 10 \, \text{H}_2\text{O}_{(g)} \quad \Delta H^\circ = -5316.6 \, \text{kJ} \] If 10.0 g of butane burns completely in oxygen and the heat is used to warm up 25.0 L of water at 22.0 °C, what will be the final temperature of the water (assuming that no heat is lost to the surroundings and that \(d = 1.00 \, \text{g/mL} \) for water)? **Analysis and Calculations:** 1. **Understanding the Reaction:** - The combustion of butane is an exothermic reaction, releasing a significant amount of energy (\(-5316.6 \, \text{kJ}\)). 2. **Heat Transfer to Water:** - 10.0 g of butane is used, and this energy will heat 25.0 L of water. - Convert 25.0 L of water to grams: \( 25,000 \, \text{g} \) (assuming \(d = 1.00 \, \text{g/mL} \)). 3. **Calculate Moles of Butane:** - Molar mass of butane (C\(_4\)H\(_{10}\)): approximately \(58.12 \, \text{g/mol}\). - Moles of butane in 10.0 g: \[ \frac{10.0 \, \text{g}}{58.12 \, \text{g/mol}} \approx 0.172 \, \text{mol} \] 4. **Energy Released:** - Total energy released by 0.172 mol of butane: \[ (0.172 \, \text{mol}) \times \left(\frac{-5316.6 \, \text{kJ}}{2 \
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