Balance the following equations by ion electron method:
a. `CuO + NH_(3) rarr Cu+H_(2)O+N_(2)`

Explanation

To balance the equation `CuO + NH3 → Cu + H2O + N2` using the ion-electron method, we will follow these steps:

### Step 1: Identify the half-reactions
We need to separate the reaction into two half-reactions: one for the reduction of copper and one for the oxidation of nitrogen.

1. **Reduction half-reaction**:
- Copper is reduced from CuO to Cu.
- Oxidation state of Cu in CuO is +2 and in Cu is 0.
- The half-reaction can be written as:
\[ CuO + 2e^- \rightarrow Cu + H_2O \]

2. **Oxidation half-reaction**:
- Nitrogen is oxidized from NH3 to N2.
- Oxidation state of N in NH3 is -3 and in N2 is 0.
- The half-reaction can be written as:
\[ 2NH_3 \rightarrow N_2 + 6e^- + 6H^+ \]

### Step 2: Balance the electrons
Now we need to balance the number of electrons transferred in both half-reactions. The reduction half-reaction involves 2 electrons, while the oxidation half-reaction involves 6 electrons. To balance them, we can multiply the reduction half-reaction by 3.

- **Balanced reduction half-reaction**:
\[ 3CuO + 6e^- \rightarrow 3Cu + 3H_2O \]

### Step 3: Combine the half-reactions
Now we can add the balanced half-reactions together:

- **Oxidation half-reaction**:
\[ 2NH_3 \rightarrow N_2 + 6e^- + 6H^+ \]

- **Reduction half-reaction**:
\[ 3CuO + 6e^- \rightarrow 3Cu + 3H_2O \]

Adding these two half-reactions gives:
\[ 3CuO + 2NH_3 \rightarrow 3Cu + 3H_2O + N_2 \]

### Step 4: Final balanced equation
The final balanced equation is:
\[ 3CuO + 2NH_3 \rightarrow 3Cu + 3H_2O + N_2 \]