To balance the chemical equation for the combustion of propane (C₃H₈), we need to ensure that the number of atoms for each element is equal on both sides of the equation.
The unbalanced equation is:
\[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]
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Start by balancing the carbon (C) atoms:
- Propane has 3 carbon atoms, so we need 3 CO₂ molecules: \[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO}_2 + \text{H}_2\text{O} \]
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Next, balance the hydrogen (H) atoms:
- Propane has 8 hydrogen atoms, which means we need 4 H₂O molecules to balance the H: \[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \]
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Now, count the oxygen (O) atoms needed on the product side:
- From 3 CO₂, we have \(3 \times 2 = 6\) O.
- From 4 H₂O, we have \(4 \times 1 = 4\) O.
- Total O atoms needed = 6 + 4 = 10 O.
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Now, find the number of O₂ molecules needed:
- Since O₂ has 2 oxygen atoms per molecule, we need \(10 \div 2 = 5\) O₂ molecules: \[ \text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \]
The balanced equation is:
\[ \text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \]
So, the coefficients placed in the underlined spaces are:
- C₃H₈: 1
- O₂: 5
- CO₂: 3
- H₂O: 4
Thus, the final answer with coefficients would be:
\[\underline{1}\text{ C}_3\text{H}_8 + \underline{5}\text{ O}_2 \rightarrow \underline{3}\text{ CO}_2 + \underline{4}\text{ H}_2\text{O}\]