The chemical equation you provided is the combustion of methane (CHâ‚„):
\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]
This equation indicates that one molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water.
If you change \(2\text{O}_2\) to \(3\text{O}_2\), the equation would be:
\[ \text{CH}_4 + 3\text{O}_2 \rightarrow \text{CO}_2 + \text{?} \]
To balance the equation with 3 moles of \(O_2\), we need to recalculate the products. Each molecule of \(O_2\) provides oxygen atoms for the products:
- You start with 3 moles of \(O_2\), which gives you 6 oxygen atoms.
- The \(CO_2\) product requires 2 oxygen atoms (1 carbon dioxide molecule).
- That leaves you with \(6 - 2 = 4\) oxygen atoms for water.
- Each water molecule requires 1 oxygen atom, so you can form \(4 / 2 = 2\) water molecules.
Thus, the balanced reaction with \(3O_2\) becomes:
\[ \text{CH}_4 + 3\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]
However, since you've added more oxygen but didn't add more methane, the overall result in terms of products (the number of \(CO_2\) and \(H_2O\) molecules) doesn't actually change for this specific reaction under typical conditions. In the end, you still produce 1 molecule of \(CO_2\) and 2 molecules of \(H_2O\) when using \(CH_4\). The excess \(O_2\) would remain unreacted:
So the result is: 1 \(CO_2\) and 2 \(H_2O\), with 1 extra \(O_2\) unreacted.
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