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How do I balance this equation? C6H5COOH (l) -> H2O (l) + CO2 (g)Liquid benzoic acid...
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We are looking at the combustion of benzoic acid to water and CO2. Combustion always occurs in the presence of oxygen from the atmosphere.
C6H5CO2H +O2 --> CO2 + H2O
Every carbon from benzoic acid is converted to CO2 so we can put a 7 in front of the CO2.
C6H5CO2H + O2 --> 7CO2 + H2O
Every hydrogen from benzoic acid is converted to H2O so we can put a 3 in front of the water.
C6H5CO2H + O2 --> 7CO2 + 3H2O
So we now have 7 carbons on the left and right and 6 hydrogens on the left and right. Now we need to balance the oxygen. We have 17 oxygens on the right side so we need 17 oxygens on the left side. We already have 2 oxygens on the left from benzoic acid so the remaining 15 oxygens must come from the O2. That means that we need a 7.5 in front of the O2.
C6H5CO2H + 7.5O2 --> 7CO2 + 3H2O
If we only want whole number integers as coefficients, so multiply everything by 2. This gives the final balanced chemical equation as:
2C6H5CO2H + 15O2 --> 14CO2 + 6H2O
Posted by ncchemist on October 22, 2012 at 4:12 AM (Answer #1)
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