Balance the combustion reaction of methanol (CH3OH) with oxygen (O2).

Answer

To balance the combustion reaction of methanol (CH3OH) with oxygen (O2), we follow these steps: 1. **Write the unbalanced equation:** CH3OH (l) + O2 (g) → CO2 (g) + H2O (l) 2. **Identify the number of atoms of each element on both sides before balancing:** - Reactants: - C: 1 in CH3OH - H: 4 in CH3OH (3 from CH3 and 1 from OH) - O: 1 from CH3OH and 2 from O2 (total = 3) - Products: - C: 1 in CO2 - H: 2 in H2O - O: 2 in CO2 and 1 in H2O (total = 3) 3. **Balance the hydrogen atoms:** Since there are 4 H in methanol and only 2 in the product side (H2O), we need to balance this by putting a coefficient of 2 before H2O: CH3OH (l) + O2 (g) → CO2 (g) + 2 H2O (l) 4. **Re-calculate the number of atoms for each side:** - Reactants: - C: 1 - H: 4 - O: 1 from CH3OH + 2 from O2 = 3 - Products: - C: 1 in CO2 - H: 4 (2 x 2 from 2 H2O) - O: 2 from CO2 + 2 from 2 H2O = 4 5. **Balance the oxygen atoms:** Now we have 3 O in the reactants and 4 in the products. This means we still need 1 more O in the reactants: - Since O2 molecules contribute 2 O each, we need a total of 2 O2 to give us 4 O (2 O2 = 4 O). Place a coefficient of 2 before O2: CH3OH (l) + 1.5 O2 (g) → CO2 (g) + 2 H2O (l) 6. **Convert fractional coefficients to whole numbers:** We can multiply the whole equation by 2 to eliminate the fraction: 2 CH3OH (l) + 3 O2 (g) → 2 CO2 (g) + 4 H2O (l) ### Final Balanced Equation: 2 CH3OH (l) + 3 O2 (g) → 2 CO2 (g) + 4 H2O (l)