Methylamine, CH3NH2, reacts with O2 to form CO2, N2 and H2O. What amount of O2 (in moles) is required to react completely with 1.00 mol of CH3NH2?(A) 2.25 (B) 2.50 (C) 3.00 (D) 4.50
`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.`