To balance the chemical equation for the combustion of propane (C₃H₈), we start with the unbalanced equation:
\[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]
Steps to Balance:
-
Count the number of atoms of each element in the reactants and products:
-
Reactants:
- C: 3 (from C₃H₈)
- H: 8 (from C₃H₈)
- O: 2 (from O₂)
-
Products:
- C: 1 (from CO₂)
- H: 2 (from H₂O)
- O: 3 total (2 from CO₂ + 1 from H₂O)
-
-
Balance the carbon (C) atoms: We have 3 carbon atoms in C₃H₈, so we need 3 CO₂ molecules. \[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO}_2 + \text{H}_2\text{O} \]
-
Balance the hydrogen (H) atoms: We have 8 hydrogen atoms in C₃H₈, so we need 4 H₂O molecules (since each H₂O contributes 2 H atoms). \[ \text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \]
-
Now count the oxygen (O) atoms:
- On the right side, we have 3 CO₂ (6 O) + 4 H₂O (4 O) = \(6 + 4 = 10\) O atoms.
- On the left side, O₂ contributes oxygen. If we use x O₂, we have \(2x\) oxygen atoms.
Set up the equation: \[ 2x = 10 \] Solving for x gives: \[ x = 5 \]
So, we need 5 O₂ molecules.
-
Write the balanced equation: Putting it all together, we have: \[ \text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \]
Final Balanced Equation:
\[ \text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \]
1 answer