How do you balance the combustion reaction of C3H7OH with O2?

To balance the combustion reaction of propanol (C₃H₇OH) with oxygen (O₂), we start with the unbalanced equation:

C₃H₇OH + O₂ → CO₂ + H₂O

1. **Identify the number of atoms**: In the reactants, we have 3 carbon atoms, 8 hydrogen atoms, and 1 oxygen atom from propanol. In the products, we produce carbon dioxide (CO₂) and water (H₂O).

2. **Balance the carbon**: To balance the carbon, we need 3 CO₂ in the products:

C₃H₇OH + O₂ → 3CO₂ + H₂O

3. **Balance the hydrogen**: Since there are 8 hydrogen atoms in propanol, we need 4 water molecules to balance the hydrogen:

C₃H₇OH + O₂ → 3CO₂ + 4H₂O

4. **Count the oxygen**: Now, count the total number of oxygen atoms in the products. We have 3 from CO₂ (3 x 2 = 6) and 4 from water (4 x 1 = 4), which gives us a total of 10 oxygen atoms.

5. **Balancing the oxygen**: In the reactants, we start with 1 oxygen from C₃H₇OH. Therefore, we need 9 more from O₂, which can be calculated as follows: 9/2 O₂ = 9 O atoms. This gives us a total of 10 oxygen atoms needed. To avoid using a fraction, we can double all coefficients.

6. **Final balanced equation**: The balanced equation becomes:

2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O

This results in an equal number of each type of atom on both sides of the equation, completing the balancing process.