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`2CH_3NH_2+9/2O_2 rarr 2CO_2+N_2+5H_2O`
Here in the reaction `N` will oxidize from -3 to 0 and `O` will reduce from 0 to -2. By interchanging the amount of change of oxidation number the above balanced equation was obtained.
`CH_3NH_2:O_2 = 2:9/2 = 4:9`
Amount of `CH_3NH_2` available `= 1mol`
Amount of `O_2` required `= 1xx9/4= 2.25`
So we need 2.25 moles of `O_2` to react with 1 mol of `CH_3NH_2.`
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