Question

DUE TOMORROW!!! 15 POINTS

Answer 1

Answer:

A. write balanced chemical equation (including states), for this process.

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)