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