What are the steps in finding the balanced equation for the complete combustion of benzene?

Well, you need for Balanced Writing about Combustion of Benzene.

Since Benzene is an Aromatic Hydrocarbon, YOU ARE ATTENDING ITS MOLECULES CONTAIN CARBON- AND HYDROGEN-ATOMs, no more species.
Now, YOU MAY STATE ITS CHEMICAL PRODUCTs HAVE TO RESULT CARBON DIOXIDE (e.g. CO2) AND WATER (e.g. H2O) LIKE VAPOURs.
So, you begin writing

C6H6(l) + O2(g) ---> CO2(g) + H2O(g)

WHICH CARRY OUT ALL CITED SPECIES.
THERE IS A TROUBLE : Chemical Writing isn't balanced one.

Step 1) YOU MULTIPLY BY 6 CO2 AT RIGHT TERM TO BALANCE Carbon Atoms

C6H6(l) + O2(g) ---> 6 CO2(g) + H2O(g)

Step 2) YOU MULTIPLY BY 3 H2O AT RIGHT TERM TO BALANCE Hydrogen Atoms

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

Step 3) YOU MULTIPLY BY "15/2" O2 AT LEFT TERM TO BALANCE Oxygen Atoms

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

Step 4) YOU MULTIPLY BY 4 ALL THE SPECIES AT TWO TERMs TO GIVE Whole Coefficients

2 C6H6(l) + 15 O2(g) ---> 12 CO2(g) + 6 H2O(g)

WARNING!!
I DISCUSSED COMPLETE COMBUSTION PROCESS DESPITE THIS PHENOMENON RESULTS A DIFFICULT ONE.
BENZENE IS AN "AROMATIC HYDROCARBON" SO IT BURNS DIFFICULTLY.
Usually, Benzene leaves up forming stinking fumes rich by itself since it keep unchanged among Burning Flames.

I hope this helps you.

Benzene is a hydrocarbon, C6H6

Hydrocarbons combust with oxygen to produce CO2 and H2O
C6H6 + O2 ==> CO2 + H2O
Balance by trial and error. But you can help the "trial" part by some snooping around. For example, I look at C6H6 so I know that must produce 6CO2 so I place a 6 coefficient for CO2.
C6H6 + O2 ==> 6CO2 + H2O
Then I see 6H on the left and I know immediately to place a 3 for H2O on the right.
C6H6 + O2 ==> 6CO2 + 3H2O
Now I count O on the right and I see 15. On the left I can get that number ONLY with a fraction for O2. That fraction is 15/2 BUT we usually don't write these equations with fractions so I multiply everything by 2 in order to clear the fraction.. That gives 2C6H6 + 15O2 ==> 12CO2 + 6H2O

To find the balanced equation for the complete combustion of benzene, you need to follow these steps:

Step 1: Write the unbalanced equation.
Start by writing the unbalanced equation for the combustion of benzene. In this case, the equation is:

C6H6 + O2 -> CO2 + H2O

Step 2: Balance the carbon atoms.
Count the number of carbon atoms on both sides of the equation. In this case, there are 6 carbon atoms on the left side and only 1 on the right side. To balance the carbon atoms, put a coefficient of 6 in front of the CO2 on the right side:

C6H6 + O2 -> 6CO2 + H2O

Now there are 6 carbon atoms on both sides.

Step 3: Balance the hydrogen atoms.
Next, count the hydrogen atoms on both sides. There are 6 hydrogen atoms in the benzene (C6H6) and only 2 hydrogen atoms in the water (H2O) on the right side. To balance the hydrogen atoms, put a coefficient of 3 in front of the H2O on the right side:

C6H6 + O2 -> 6CO2 + 3H2O

Now there are 6 carbon atoms and 12 hydrogen atoms on both sides.

Step 4: Balance the oxygen atoms.
Finally, balance the oxygen atoms. The only source of oxygen in this equation is O2 on the left side. Count the number of oxygen atoms on both sides. There are 2 oxygen atoms in O2 on the left side and 2 x 6 = 12 oxygen atoms in CO2 on the right side, and 3 x 1 = 3 oxygen atoms in H2O on the right side. In total, there are 12 + 3 = 15 oxygen atoms on the right side. To balance the oxygen atoms, put a coefficient of 15/2 = 7.5 in front of the O2 on the left side. However, since we cannot have fractional coefficients, multiply the entire equation by 2 to get rid of the decimal:

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

Now there are 12 carbon atoms, 24 hydrogen atoms, and 30 oxygen atoms on both sides, which means the equation is balanced.

So, the balanced equation for the complete combustion of benzene is:
2C6H6 + 15O2 -> 12CO2 + 6H2O