Write the balanced chemical equation for the following reaction:- Nitrrogen is obtained in the reaction of aqueous ammonia with potassium permanganate

Answer

To write the balanced chemical equation for the reaction of aqueous ammonia with potassium permanganate, we will follow these steps:

### Step 1: Write the unbalanced equation
The reactants in this reaction are potassium permanganate (KMnO4) and ammonia (NH3). The products formed are manganese dioxide (MnO2), potassium hydroxide (KOH), water (H2O), and nitrogen gas (N2). The unbalanced equation can be written as:
\[ \text{KMnO}_4 + \text{NH}_3 \rightarrow \text{MnO}_2 + \text{KOH} + \text{H}_2\text{O} + \text{N}_2 \]

### Step 2: Assign coefficients to each compound
We will assign coefficients to each compound in the equation:
- Let the coefficient of KMnO4 be \( A \)
- Let the coefficient of NH3 be \( B \)
- Let the coefficient of MnO2 be \( C \)
- Let the coefficient of KOH be \( D \)
- Let the coefficient of H2O be \( E \)
- Let the coefficient of N2 be \( F \)

The equation can now be represented as:
\[ A \text{KMnO}_4 + B \text{NH}_3 \rightarrow C \text{MnO}_2 + D \text{KOH} + E \text{H}_2\text{O} + F \text{N}_2 \]

### Step 3: Write the balance equations
Next, we will write the balance equations for each element:
1. **Potassium (K)**: \( A = D \)
2. **Manganese (Mn)**: \( A = C \)
3. **Oxygen (O)**: \( 4A = 2C + D + E \)
4. **Nitrogen (N)**: \( B = 2F \)
5. **Hydrogen (H)**: \( 3B = D + 2E \)

### Step 4: Assume a value for one variable
To simplify the balancing process, we can assume a value for one of the coefficients. Let's assume \( A = 1 \):
- Then \( D = 1 \) (from K balance)
- \( C = 1 \) (from Mn balance)

### Step 5: Substitute and solve for remaining coefficients
Now substituting \( A = 1 \) into the oxygen balance equation:
\[ 4(1) = 2(1) + 1 + E \]
\[ 4 = 2 + 1 + E \]
\[ E = 1 \]

Now substituting \( A = 1 \) into the nitrogen balance:
\[ B = 2F \]
We will need to find \( B \) and \( F \) using the hydrogen balance:
\[ 3B = 1 + 2(1) \]
\[ 3B = 3 \]
\[ B = 1 \]

Now substituting \( B = 1 \) into \( B = 2F \):
\[ 1 = 2F \]
\[ F = \frac{1}{2} \]

### Step 6: Write the balanced equation
Now we have all coefficients:
- \( A = 1 \)
- \( B = 1 \)
- \( C = 1 \)
- \( D = 1 \)
- \( E = 1 \)
- \( F = \frac{1}{2} \)

Substituting these coefficients into the equation gives:
\[ 1 \text{KMnO}_4 + 1 \text{NH}_3 \rightarrow 1 \text{MnO}_2 + 1 \text{KOH} + 1 \text{H}_2\text{O} + \frac{1}{2} \text{N}_2 \]

### Step 7: Eliminate the fraction
To eliminate the fraction, we multiply the entire equation by 2:
\[ 2 \text{KMnO}_4 + 2 \text{NH}_3 \rightarrow 2 \text{MnO}_2 + 2 \text{KOH} + 2 \text{H}_2\text{O} + 1 \text{N}_2 \]

### Final Balanced Equation
The balanced chemical equation is:
\[ 2 \text{KMnO}_4 + 2 \text{NH}_3 \rightarrow 2 \text{MnO}_2 + 2 \text{KOH} + 2 \text{H}_2\text{O} + \text{N}_2 \]

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