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)