The reaction between ethane (C₂H₆) and oxygen gas (O₂) to produce carbon dioxide (CO₂) and water (H₂O) is a combustion reaction. This type of reaction typically involves a hydrocarbon reacting with oxygen to produce carbon dioxide and water, usually releasing energy in the form of heat and light.
Evidence for the Reaction
The main piece of evidence for the occurrence of a combustion reaction includes:
- The production of gases (CO₂ and H₂O vapor).
- Heat and light are produced, which can be observed as flames.
- The reaction consumes oxygen, which can be measured by a decrease in oxygen gas in the reaction environment.
Predicting and Balancing the Reaction
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Write the unbalanced chemical equation for the combustion of ethane: \[ \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]
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Count the number of atoms of each element in the reactants and products:
- Reactants: 2 Carbon (C) in C₂H₆, 6 Hydrogen (H) in C₂H₆, O in O₂
- Products: Each CO₂ has 1 Carbon and each H₂O has 2 Hydrogens.
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Balance the equation:
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Balance the carbon atoms: \[ \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O} \]
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Balance the hydrogen atoms (6 H in C₂H₆ will yield 3 H₂O): \[ \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \]
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Now balance the oxygen atoms. We have:
- On the product side: \(2 \times 2 = 4\) O from CO₂ and \(3 \times 1 = 3\) O from H₂O, totaling 7 O atoms.
- Therefore, we need \(\frac{7}{2}\) or \(3.5\) O₂ (since O₂ has 2 O atoms). \[ \text{C}_2\text{H}_6 + \frac{7}{2} \text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \]
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To eliminate the fraction, multiply the entire equation by 2: \[ 2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} \]
Final Balanced Equation:
\[ 2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} \]
This balanced equation indicates that two moles of ethane react with seven moles of oxygen to produce four moles of carbon dioxide and six moles of water.
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