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
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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
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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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