When `AgNO_(3)`(aq)` reacts with excess of iodine, we get:

Options

A. `AgIO_(3)`
B. `HIO_(3)`
C. `AgO`
D. `HI`

Answer

To solve the question regarding the reaction of silver nitrate (AgNO₃) with excess iodine (I₂), we can follow these steps:

### Step 1: Write the Reactants
The reactants in this reaction are:
- Silver nitrate (AgNO₃) in aqueous solution
- Iodine (I₂) in excess

### Step 2: Write the Chemical Reaction
When silver nitrate reacts with iodine, the products formed are:
- Silver iodate (AgIO₃)
- Silver iodide (AgI)
- Nitric acid (HNO₃)

The unbalanced reaction can be written as:
\[ \text{AgNO}_3 (aq) + \text{I}_2 + \text{H}_2\text{O} \rightarrow \text{AgIO}_3 + \text{AgI} + \text{HNO}_3 \]

### Step 3: Balance the Reaction
To balance the reaction, we need to ensure that the number of atoms of each element is the same on both sides. The balanced equation is:
\[ 6 \text{AgNO}_3 + 3 \text{I}_2 + 3 \text{H}_2\text{O} \rightarrow 1 \text{AgIO}_3 + 5 \text{AgI} + 6 \text{HNO}_3 \]

### Step 4: Identify the Products
From the balanced equation, we can see that the products formed are:
1. Silver iodate (AgIO₃)
2. Silver iodide (AgI)
3. Nitric acid (HNO₃)

### Step 5: Select the Correct Option
Among the products formed, AgIO₃ is listed in the options provided in the question. Therefore, the correct answer is:
- **AgIO₃**

### Final Answer
When AgNO₃ (aq) reacts with excess iodine, we get **AgIO₃** along with AgI and HNO₃.

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