C6H5COOH+O2 = CO2+H2O

Question

Balance the chemical formula

Answer 1

Lets start from: C₆H₅COOH + O₂ = CO₂ + H₂O or re-write as C₇H₆O₂ + O₂ = CO₂ + H₂O

Left side: 7C, 6H, 4(O)

Right side: 1C, 2H, 3(O) 

Balance the carbons on both side: C₇H₆O₂ + O₂ = 7CO₂ + H₂O

Left side: 7C, 6H, 4(O)
Right side: 7C, 2H, 15(O)

Next balance the hydrogen: C₇H₆O₂ + O₂ = 7CO₂ + 3H₂O

Left side: 7C, 6H, 4(O)
Right side: 7C, 6H, 17(O)

What is left is balancing the Oxygen, leave the O in C7H6O2 alone and work on O2:

C7H6O2 + (15/2)O2 = 7CO2 + 3H2O, the equation is balanced as is but we need coefficients to be whole numbers. Multiply both sides by 2 and you get your final answer as: 2C₇H₆O₂ + 15O₂ = 14CO₂ + 6H₂O

Left side: 14C, 12H, 34(O)
Right side: 14C, 12H, 34(O)

 

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Answer 2

You should first condense the formula so that all representative atoms are combined:

 

C7H6O2  + O2   =  CO2 + H2O

 

as can be seen, they are not balanced

Left Side of Equation    Atom    Right Side of Equation

                               7      C         1

                               6      H         2

                               4      O         3    (*note that they can be added together as long as they are in the same side)

 

While the most striking imbalance belongs to Carbon, place the whole numbers before the formula on the right

 

C7H6O2   +    O2    =   7CO2   + H2O

(C=7, H=6, O=4)          (C=7, H=2, O=15)

 

Next is balance out Hydrogen

C7H6O2   +     O2   = 7CO2    +  3H2O

(C=7,H=6,O=4)       = (C=7, H=6, O=17)

 

Since oxygen is still unbalanced, try to find a number that can satisfy both sides of the equation, i.e. you need 15 more atoms of Oxygen since 2 Oxygen atoms are already present in the carboxylic acid; thus you are left to manipulate only the free O2

 

C7H6O2   + (15/2)O2   =   7CO2   + 3H2O

(C=7,H=6,O=17)          =   (C=7,H=6,O=17)

 

and since it requires all whole numbers simply multiply the whole equation by the denominator (which is 2)

giving you:

 

2C7H6O2   +    15O2   =  14CO2  + 6 H2O

(C=14,H=12,O=34)      =  (C=14,H=12,O=34)

 

 

 

 

Answer 3

Let's start with the Carbons.  There are 6 on the left and 1 on the right.  Therefore, we need to put a 6 in front of the CO₂ term to get 6 carbons on the right.

 

C₆H₅COOH + O₂ ---> 6 CO₂ + H₂O

 

Now for the Hydrogens.  There are 6 on the left and 2 on the right.  We need to put a 3 in front of the H₂O term to get 6 hydrogens on the right.

 

C₆H₅COOH + O₂ ---> 6 CO₂ + 3 H₂O

 

Now, on the right, we have 6 carbons, 6 hydrogens and 15 oxygens.  On the left, we have 4 oxygens.  We need to get 15 oxygens on the left.  If we count the 2 oxygens in the C₆H₅COOH as 2 of our 15, then we have 13 oxygens to get from the O₂ term.  Therefore, we need (13/2) O₂ molecules.

 

C₆H₅COOH + (13/2) O₂ ---> 6 CO₂ + 3 H₂O

 

To get rid of the fraction in front of the O₂ term, multiply everything by 2.

 

2 C₆H₅COOH + 13 O₂ ---> 12 CO₂ + 6 H₂O

 

The equation should now be balanced.  We can check:

Right: 12 carbons, 12 hydrogens, 30 oxygens

Left: 12 carbons, 12 hydrogens, 30 oxygens