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))?
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))?
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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:
\[
(0.172 \, \text{mol}) \times \left(\frac{-5316.6 \, \text{kJ}}{2 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5a0011f3-a94e-4ba4-97f0-d2315dab4566%2Fc1d640f9-76ae-4677-8c75-df97f650f602%2Fvh573b_processed.jpeg&w=3840&q=75)
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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