To balance the equation for the combustion of propane (C₃H₈), we start with the unbalanced equation:
\[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]
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Count the number of atoms of each element in the reactants and products.
- Reactants: C = 3, H = 8, O = 2 (in O₂)
- Products: C = 1 (in CO₂), H = 2 (in H₂O)
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Balance the carbon atoms first. There are 3 carbon atoms in propane, so we need 3 CO₂:
\[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO}_2 + \text{H}_2\text{O} \]
- Balance the hydrogen atoms next. There are 8 hydrogen atoms in propane, which means we need 4 H₂O to balance the hydrogen:
\[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \]
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Count the oxygen atoms now. We have:
- From 3 CO₂: \(3 \times 2 = 6\) O
- From 4 H₂O: \(4 \times 1 = 4\) O
- Total oxygen atoms needed: \(6 + 4 = 10\) O
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On the left side, we have O₂. Since each O₂ molecule contains 2 oxygen atoms, we need 5 O₂ to provide 10 oxygen atoms:
\[ 5 \text{O}_2 \]
- Now we can write the fully balanced equation:
\[ \text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \]
This equation is now balanced with:
- Carbons: 3 on both sides
- Hydrogens: 8 on both sides
- Oxygens: 10 on both sides
Therefore, the balanced equation for the combustion of propane is:
\[ \text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \]