Balancing a complex Chemical equation

[CoinToss]

Homework Statement

Balance the following Chemistry equation, you are allowed to add H{+} and H2O where needed
Given
C₆H₅OCl + NO₃^- --> N₂ + Cl^- + CO₂

2. The attempt at a solution
Balance the carbon first:
C₆H₅OCl + NO₃^- --> N₂ + Cl^- + 6CO₂

Increase Oxygen O in NO3{-} to compensate; also increase N on RHS
C₆H₅OCl + 4NO₃^- --> 2N₂ + Cl^- + 6CO₂

Bring out the H{+} and O{2-} on RHS
C₆H₅OCl + 4NO₃^- --> 2N₂ + Cl^- + 6CO₂ + H2O + 3H⁺

Multiply both sides by 2
2C₆H₅OCl + 8NO₃^- --> 4N₂ + 2Cl^- + 12CO₂ + 2H₂O + 6H⁺

Compensate for 6H{+} on RHS by increasing NO3 by 1 to form H2O on RHS
2C₆H₅OCl + 9NO₃^- --> 4N₂ + 2Cl^- + 12CO₂ + 5H₂O + N^3-

At this point I would assume that H{+} Is required on the LHS to give N2 and H2O on RHS
2C₆H₅OCl + 10NO₃^- + 2H⁺ --> 5N₂ + 2Cl^- + 12CO₂ + 6H₂O

Which is looking pretty good so far-- except the charges on either side are not balanced. And this would be the point where I am stuck. Was there supposed to be a different approach to this? I appreciate your time. Thank you.

 

[Borek]

This is a redox reaction and these are usually difficult to balance by inspection. Try either one of the redox methods (half reactions or oxidation numbers), or use algebraic approach. All these are described here: balancing and stoichiometry.

 

[SteamKing]

Perhaps the NO3 radical comes from an acid which is added to the other organic compound.

 

[CoinToss]

Excellent, I think this would be very helpful. Thank you very much.

@SteamKing: HNO₃ could very well be the said acid but I gather that it isn't of much interest at the moment?

 

[Borek]

NO₃^- is not a radical, just an ion, product of the nitric acid dissociation.

 

[SteamKing]

And maybe some of the H+ ions come from said nitric acid dissociation?

 

[Borek]

Yes. And it is also possible to balance the reaction using HNO₃ as a reactant and HCl as a product.