Question:
The number of moles of CO$_2$ produced when 2 moles of C$_2$H$_6$ are completely burnt is:
The number of moles of CO$_2$ produced when 2 moles of C$_2$H$_6$ are completely burnt is:
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Always balance the combustion reaction to find correct mole ratios of reactants and products.
Updated On: Mar 14, 2026
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The Correct Option is A
Approach Solution - 1
To determine the number of moles of CO2 produced when 2 moles of C2H6 are completely burnt, we must first consider the balanced chemical equation for the combustion of ethane (C2H6).
The balanced equation for the complete combustion of ethane is:
2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
From the equation, we can see that 2 moles of C2H6 produce 4 moles of CO2 when completely burnt in oxygen.
Given that we start with 2 moles of C2H6, using stoichiometry as per the balanced equation, we calculate the moles of CO2 produced:
- According to the balanced equation, the molar ratio of C2H6 to CO2 is 2:4, or 1:2.
- Thus, 2 moles of C2H6 produce \(2 \times 2 = 4\) moles of CO2.
Therefore, the number of moles of CO2 produced is 4.
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Approach Solution -2
Step 1: Write the unbalanced combustion reaction of ethane (C\(_2\)H\(_6\)) \[ \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \] Step 2: Balance the equation Balance the carbon (C) atoms: \[ \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow 2\text{CO}_2 + \text{H}_2\text{O} \] Balance the hydrogen (H) atoms: \[ \text{C}_2\text{H}_6 + \text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \] Balance the oxygen (O) atoms: On the right, there are \(2 \times 2 = 4\) O atoms in CO\(_2\) and \(3 \times 1 = 3\) O atoms in H\(_2\)O, totaling 7 O atoms. So, we need \(\frac{7}{2}\) O\(_2\) on the left: \[ \text{C}_2\text{H}_6 + \frac{7}{2} \text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \] Step 3: Multiply the entire equation by 2 to remove the fraction \[ 2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O} \] Step 4: Analyze the molar ratio From the equation: \[ 2\text{C}_2\text{H}_6 \rightarrow 4\text{CO}_2 \] This means: \[ 1 \text{ mol of C}_2\text{H}_6 \rightarrow 2 \text{ mol of CO}_2 \] Step 5: Apply the ratio to the given amount We are given 2 moles of C\(_2\)H\(_6\), so: \[ 2 \times 2 = 4 \text{ moles of CO}_2 \]
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