Question

In the reaction, $3CuO+2N{H}_3\to N_2+3H_{2}O+3Cu$ the change of $N{H}_3$ to $N_2$ involves:

• A. loss of $6$ electrons per mol of $N_2$
• B. loss of $3$ electrons per mol of $N_2$
• C. gain of $6$ electrons per mol of $N_2$
• D. gain of $3$ electrons per mol of $N_2$

Answer 1

The correct option is A loss of $6$ electrons per mol of $N_2$

Balancing equation

$2N{H}_3+6e^{-}\to N_2+6H^{+}$

So 6 electrons is loss from $N{H}_3$ to form $N_2$