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Balance the following equation by oxidation number methodC_(6)H_(6)+O_(2)rarrCO_(2)+H_(2)O |
Answer» Solution :![]() (ii) BALANCE the changes in O.N. by multiplying the oxidant and reductant by suitable numbers `2C_(6)H_(6)+15O_(2)rarrCO_(2)+H_(2)O` (iii) Balance the equation atomically (except O and H). `2C_(6)H_(6)+15O_(2)rarr 12CO_(2) + H_(2)O` (iv) Balance O atoms by ADDING one `H_(2)O` MOLECULE to the RHS for making the number of molecules of `H_(2)O` to be 6. `2C_(6)H_(6)+15O_(2)rarr12CO_(2)+6H_(2)O` |
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