Write balanced equation for the following :
Phosphorus is treated with concentrated nitric acid.

Answer

To write the balanced equation for the reaction of phosphorus (P₄) with concentrated nitric acid (HNO₃), follow these steps:

### Step 1: Write the unbalanced equation
The initial reaction can be represented as:
\[ \text{P}_4 + \text{HNO}_3 \rightarrow \text{H}_3\text{PO}_4 + \text{NO}_2 + \text{H}_2\text{O} \]

### Step 2: Identify the products
When phosphorus reacts with concentrated nitric acid, the products formed are phosphoric acid (H₃PO₄), nitrogen dioxide (NO₂), and water (H₂O).

### Step 3: Balance the equation
1. **Balance phosphorus (P)**:
- There are 4 phosphorus atoms in P₄, so we need 4 H₃PO₄ on the product side.
\[ \text{P}_4 + \text{HNO}_3 \rightarrow 4\text{H}_3\text{PO}_4 + \text{NO}_2 + \text{H}_2\text{O} \]

2. **Balance hydrogen (H)**:
- There are 12 hydrogen atoms in 4 H₃PO₄, so we need 12 HNO₃ to provide enough hydrogen.
\[ \text{P}_4 + 12\text{HNO}_3 \rightarrow 4\text{H}_3\text{PO}_4 + \text{NO}_2 + \text{H}_2\text{O} \]

3. **Balance nitrogen (N)**:
- 12 HNO₃ gives 12 nitrogen atoms, and we need to balance it with NO₂. Each NO₂ contains one nitrogen atom, so we will need 12 NO₂.
\[ \text{P}_4 + 12\text{HNO}_3 \rightarrow 4\text{H}_3\text{PO}_4 + 12\text{NO}_2 + \text{H}_2\text{O} \]

4. **Balance oxygen (O)**:
- Count the total number of oxygen atoms.
- Left side: 12 HNO₃ contributes 36 O atoms.
- Right side: 4 H₃PO₄ contributes 16 O, 12 NO₂ contributes 24 O, and 1 H₂O contributes 1 O, totaling 41 O.
- To balance the oxygen, we need to adjust the water.
\[ \text{P}_4 + 12\text{HNO}_3 \rightarrow 4\text{H}_3\text{PO}_4 + 12\text{NO}_2 + 2\text{H}_2\text{O} \]

5. **Final check**:
- Count all atoms to ensure they are balanced:
- P: 4 on both sides
- H: 12 on both sides
- N: 12 on both sides
- O: 36 on both sides (12 from HNO₃ and 24 from products)

### Final Balanced Equation
The balanced equation for the reaction of phosphorus with concentrated nitric acid is:
\[ \text{P}_4 + 12\text{HNO}_3 \rightarrow 4\text{H}_3\text{PO}_4 + 12\text{NO}_2 + 2\text{H}_2\text{O} \]