Write a balanced chemical equation for the reaction that occurs when the hydrocarbon styrene, {\rm C}_{8}{\rm H}_{8}(l), is combusted in air.
C8H8 + 10 O2 ==> 8CO2 + 4H2O
To write a balanced chemical equation for the combustion of styrene, we need to consider that combustion reactions involve the reactants reacting with oxygen gas (O2) to produce carbon dioxide (CO2) and water (H2O).
The balanced equation for the combustion of styrene (C8H8) in air can be written as:
C8H8 + 12O2 → 8CO2 + 4H2O
This equation shows that 1 molecule of styrene (C8H8) reacts with 12 molecules of oxygen (O2) to produce 8 molecules of carbon dioxide (CO2) and 4 molecules of water (H2O).
To write a balanced chemical equation for the combustion of styrene, we need to understand the reaction that occurs when a hydrocarbon is combusted in air.
When a hydrocarbon undergoes combustion, it reacts with oxygen (O2) from the air to produce carbon dioxide (CO2) and water (H2O) as the main products. The general balanced chemical equation for the combustion of a hydrocarbon is:
Hydrocarbon + Oxygen -> Carbon Dioxide + Water
Now, we will use this general equation to write the balanced chemical equation for the combustion of styrene (C8H8):
C8H8 + O2 -> CO2 + H2O
To balance this equation, we need to ensure that there is the same number of atoms of each element on both sides of the equation.
Here is the balanced equation for the combustion of styrene:
2C8H8 + 17O2 -> 16CO2 + 8H2O
In this balanced equation, we have 2 molecules of styrene reacting with 17 molecules of oxygen to produce 16 molecules of carbon dioxide and 8 molecules of water. This equation shows the stoichiometry of the reaction, meaning the ratio of reactants and products.
Note that this is a simplified equation and does not account for any other possible byproducts or side reactions that may occur during the combustion process.