Question

Balance the equation by oxidation number method. C6H6 + O2 –› CO2 + H2O(acidic)

Answer 1

1. Write the O.N no of each atom in the skeleton equation

C6^-1 H6^+1 +O2^0 = C^+4O2^-2+H2^+1 O2^-2

2. Identify the atoms which undergo change O.N .

C6^-1H6 + O2^0 = C^+4 +H2O2^-2

3. Calculate the total increase n decrease in O.N w.r.t reactant atoms

C6^-1 ---- C^+4. ( here there is increase in oxidation number i.e, -1 to +4)

O2^0 ---- H2O^-2. ( here there is decrease in oxidation number i.e, 0 to -2)

4. Equate the increase and decrease in oxidation number on the reactant side after taking out a common factor of 2

2C6H6 + 15O2 = CO2 + H2O

5. Balance the number of carbon and oxygen atoms on both sides of the equation

2C6H6 + 15O2 = 12CO2 + 6H2O

The H atoms are already balanced in the above equation..

Hope so this helps u !!

Answer 2

$2C6H6 + 15O2 = 12CO2 + 6H2O$ is the balanced equation

Explanation:

1. Write the Oxidation number of each atom in the skeleton equation

$C_{6} ^{-1}H_{6}^{+1} +O_{2} ----- C_{4} ^{+2}O_{2}^{-2} +H_{2}^{+1}O_{2}^{-2}$

2. Identify the atoms which undergo change Oxidation number.

$C_{6} ^{-1}H_{6}^{+1} +O_{2} ----- C_{4} ^{+2}O_{2}^{-2} +H_{2}^{+1}O_{2}^{-2}$

3. Calculate the total increase n decrease in Oxidation number with respect to  the reactant atoms

$C_{6}^{-1} ---- C^{+4}$( here there is increase in oxidation number i.e, -1 to +4)

$O2^0 ---- H_{2} O^{-2}$ here there is decrease in oxidation number i.e, 0 to -2)

4. Equate the increase and decrease in oxidation number on the reactant side after taking out a common factor of 2

$2C6H6 + 15O2 = CO2 + H2O$

5. Balance the number of carbon and oxygen atoms on both sides of the equation

$2C6H6 + 15O2 = 12CO2 + 6H2O$

The H atoms are already balanced in the above equation..

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