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(viii) CuO+NH_3 to Cu+N_2+H_2O |
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Answer» Solution :`CuO+NH_3 to Cu+N_2+H_2O`. `Cu+2e^(-) to Cu` (Reduction) ……………(1) `N^(-3) to N_(2)^(0)+3e^(-)` (oxidation) ……………(2) MULTIPLY equation (2) by 2 to balance nitrogen atom `2N^(-3) to N_(2)^(0)+6e^(-)` ……………(3) Multiply Equation (1) by 3 to BALACNE the number of electrons. `3Cu^(2+) +6e^(-)to 3Cu` ……………(4) ADD equation (3) and (4) . `3Cu^(2+)+2N^(-3) to 3Cu+N_(2)` Over all balanced equation `3CuO +2NH_(3) to 3Cu+N_(2)+H_(2)O` Balance O atom by adding `H_2O` on the SIDE falling short of it. |
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