write a balance equation for the following.
when octane is burned in an excess of oxygen, carbon dioxide and water vapor produced
2C8H18 + 25O2 ==> 16CO2 + 18H2O
To write a balanced equation for the combustion of octane (C8H18) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O), you need to consider the number of each type of atom on both sides of the equation.
1. Start by writing the chemical formula for octane: C8H18.
2. Identify the reactants and products in the reaction:
Reactants: Octane (C8H18) and Oxygen (O2)
Products: Carbon Dioxide (CO2) and Water Vapor (H2O)
3. Write the unbalanced equation by combining the components on both sides:
C8H18 + O2 → CO2 + H2O
4. Now, you need to balance the equation by adjusting the coefficients in front of each compound to ensure that the number of atoms is equal on both sides.
To balance the carbon atoms, you need to place a coefficient of 8 in front of CO2 on the product side:
C8H18 + O2 → 8CO2 + H2O
For hydrogen atoms, you need 18 hydrogen atoms on the product side to match the 18 hydrogen atoms in octane. To achieve this, place a coefficient of 9 in front of H2O:
C8H18 + O2 → 8CO2 + 9H2O
Finally, balance the oxygen atoms by adding the appropriate number of O2 molecules on the reactant side. In this case, you need 25 O2 molecules:
C8H18 + 25O2 → 8CO2 + 9H2O
Now, the equation is balanced, meaning that the number of each type of atom is the same on both sides.
To summarize, the balanced equation for the combustion of octane in excess oxygen is:
C8H18 + 25O2 → 8CO2 + 9H2O