Ethanol (C,H,O) is combusted in air according to the following reaction: C,H,O(1) + 0,(g) → CO,.(g) + H,O(1) How many grams of water would be produced by the complete combustion of 6.45 moles of ethanol in the presence of excess oxygen?
Ethanol (C,H,O) is combusted in air according to the following reaction: C,H,O(1) + 0,(g) → CO,.(g) + H,O(1) How many grams of water would be produced by the complete combustion of 6.45 moles of ethanol in the presence of excess oxygen?
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter5: Principles Of Chemical Reactivity: Energy And Chemical Reactions
Section5.8: Product- Or Reactant-favored Reactions And Thermodynamics
Problem 2.1ACP
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![### Combustion of Ethanol in Air
Ethanol \((\text{C}_2\text{H}_6\text{O})\) is burned in the presence of air according to the following chemical reaction:
\[
\text{C}_2\text{H}_6\text{O}(l) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l)
\]
**Problem:**
How many grams of water would be produced by the complete combustion of 6.45 moles of ethanol in the presence of excess oxygen?
#### Explanation:
1. **Balanced Equation:**
To solve this problem, first ensure the reaction equation is balanced. The balanced equation for the combustion of ethanol is:
\[
\text{C}_2\text{H}_6\text{O}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)
\]
2. **Moles of Water Produced:**
From the balanced equation, we see that 1 mole of ethanol produces 3 moles of water when completely combusted.
Therefore, if we have 6.45 moles of ethanol, the moles of water produced would be:
\[
6.45 \text{ moles } \text{C}_2\text{H}_6\text{O} \times 3 \text{ moles } \text{H}_2\text{O} \text{ per 1 mole } \text{C}_2\text{H}_6\text{O} = 19.35 \text{ moles } \text{H}_2\text{O}
\]
3. **Converting Moles to Grams:**
The molar mass of water \((\text{H}_2\text{O})\) is:
\[
2 \times 1.01 \text{ g/mol (H)} + 16.00 \text{ g/mol (O)} = 18.02 \text{ g/mol}
\]
Thus, the grams of water produced by 19.35](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F57d9716f-2245-4f13-9c98-221b78bc13ff%2Ffc95a5f7-c06f-4324-a77a-4cecf10826ec%2Fgejfro_processed.png&w=3840&q=75)
Transcribed Image Text:### Combustion of Ethanol in Air
Ethanol \((\text{C}_2\text{H}_6\text{O})\) is burned in the presence of air according to the following chemical reaction:
\[
\text{C}_2\text{H}_6\text{O}(l) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l)
\]
**Problem:**
How many grams of water would be produced by the complete combustion of 6.45 moles of ethanol in the presence of excess oxygen?
#### Explanation:
1. **Balanced Equation:**
To solve this problem, first ensure the reaction equation is balanced. The balanced equation for the combustion of ethanol is:
\[
\text{C}_2\text{H}_6\text{O}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)
\]
2. **Moles of Water Produced:**
From the balanced equation, we see that 1 mole of ethanol produces 3 moles of water when completely combusted.
Therefore, if we have 6.45 moles of ethanol, the moles of water produced would be:
\[
6.45 \text{ moles } \text{C}_2\text{H}_6\text{O} \times 3 \text{ moles } \text{H}_2\text{O} \text{ per 1 mole } \text{C}_2\text{H}_6\text{O} = 19.35 \text{ moles } \text{H}_2\text{O}
\]
3. **Converting Moles to Grams:**
The molar mass of water \((\text{H}_2\text{O})\) is:
\[
2 \times 1.01 \text{ g/mol (H)} + 16.00 \text{ g/mol (O)} = 18.02 \text{ g/mol}
\]
Thus, the grams of water produced by 19.35
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