How do you balance the following equation: C6H14 + O2 —> CO2 +H2O
The Correct Answer and Explanation is:
To balance the combustion reaction of hexane (C6H14) with oxygen (O2) producing carbon dioxide (CO2) and water (H2O), we need to ensure that the number of atoms for each element is the same on both sides of the equation.
The unbalanced equation is:
C6H14 + O2 → CO2 + H2O
Step-by-step process:
- Balance Carbon (C):
- Hexane (C6H14) has 6 carbon atoms. On the product side, carbon appears in CO2.
- So, to balance carbon, place a coefficient of 6 in front of CO2:
- Balance Hydrogen (H):
- Hexane (C6H14) has 14 hydrogen atoms. On the product side, hydrogen appears in H2O.
- To balance hydrogen, place a coefficient of 7 in front of H2O, because each molecule of H2O has 2 hydrogen atoms:
- Balance Oxygen (O):
- Now, count the oxygen atoms on the product side. On the right side, there are 6 CO2 molecules, each containing 2 oxygen atoms, so that’s a total of 6×2=126 \times 2 = 126×2=12 oxygen atoms from CO2.
- Additionally, there are 7 H2O molecules, each containing 1 oxygen atom, so that’s a total of 7 oxygen atoms from H2O.
- Therefore, the total oxygen on the product side is 12+7=1912 + 7 = 1912+7=19 oxygen atoms.
- To balance oxygen on the left side, we need 19 oxygen atoms. Since each O2 molecule provides 2 oxygen atoms, place a coefficient of 9.5 in front of O2:
- Final adjustment (optional):
- To avoid fractional coefficients, multiply the entire equation by 2 to get whole numbers:
Final balanced equation:
2C6H14+19O2→12CO2+14H2O2C6H14 + 19O2 \rightarrow 12CO2 + 14H2O2C6H14+19O2→12CO2+14H2O
Explanation:
- Carbon is balanced by ensuring 6 CO2 molecules for each hexane molecule.
- Hydrogen is balanced by adjusting the number of H2O molecules to match the 14 hydrogens in C6H14.
- Oxygen is balanced by adjusting the O2 molecules to provide the required 19 oxygen atoms, which come from both CO2 and H2O.
This equation now adheres to the law of conservation of mass, with an equal number of atoms for each element on both sides.
