Combustion of benzene? chemical equation?

When a hydrocarbon burns in air, what reactant besides the hydrocarbon is involved in the reaction? Write a balanced chemical equation for the combustion of benzene, C6H6(l), in air. (Use the lowest possible coefficients. Include states-of-matter under SATP conditions in your answer.)


Almost all 'burn' reactions involve oxygen; it's by far the most reactive substance in air.

Hydrocarbon combustions always involve

[some hydrocarbon] + oxygen --> carbon dioxide + steam.

C6H6 + O2 --> CO2 + H2O

Balance carbon, six on each side:

C6H6 + O2 --> 6CO2 + H2O

Balance hydrogen, six on each side:

C6H6 + O2 --> 6CO2 + 3H2O

Now, we have fifteen oxygens on the right and O2 on the left.

Two ways to deal with that. We can use a fraction:

C6H6 + (15/2)O2 --> 6CO2 + 3H2O

Or, if you prefer to have whole number coefficients, double everything

to get rid of the fraction:

2C6H6 + 15O2 --> 12CO2 + 6H2O

With the SATP states thrown in...

C6H6(l) + (15/2)O2(g) --> 6CO2(g) + 3H2O(l)