Answer:
A. write balanced chemical equation (including states), for this process.
Step-by-step explanation:
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)