Write a balanced chemical equation, including states of matter, for the combustion of gaseous benzene, c6h6.

Almost all hydrocarbon '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(l) + O2 (g)--> CO2 (g)+ H2O (g)

Balance carbon, six on each side: 
C6H6(l) + O2 (g)--> 6CO2 (g)+ H2O (g)

Balance hydrogen, six on each side: 
C6H6(l) + O2 (g)--> 6CO2(g) + 3H2O (g)

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

Or, if you prefer to have whole number coefficients, double everything 
to get rid of the fraction: 
2C6H6 (l)+ 15O2 (g)--> 12CO2 (g)+ 6H2O (g)

With the SATP states thrown in... 
C6H6(l) + (15/2)O2(g) --> 6CO2(g) + 3H2O(g)