`H_2SO_4+HI to `__________+___________+__________

Answer

To solve the reaction between sulfuric acid (H₂SO₄) and hydrogen iodide (HI), we will follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- H₂SO₄ (sulfuric acid)
- HI (hydrogen iodide)

### Step 2: Determine the Products
When H₂SO₄ reacts with HI, it produces:
- I₂ (iodine)
- H₂S (hydrogen sulfide)
- H₂O (water)

### Step 3: Write the Unbalanced Reaction
The unbalanced reaction can be written as:
\[ \text{H}_2\text{SO}_4 + \text{HI} \rightarrow \text{I}_2 + \text{H}_2\text{S} + \text{H}_2\text{O} \]

### Step 4: 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 reaction is:
\[ \text{H}_2\text{SO}_4 + 8\text{HI} \rightarrow 4\text{I}_2 + \text{H}_2\text{S} + 4\text{H}_2\text{O} \]

### Step 5: Identify the Oxidation States
- In H₂SO₄, sulfur (S) has an oxidation state of +6.
- In HI, iodine (I) has an oxidation state of -1.
- In the products, iodine (I) in I₂ has an oxidation state of 0, and sulfur (S) in H₂S has an oxidation state of -2.

### Step 6: Classify the Reaction
This reaction is classified as a redox reaction because there is a transfer of electrons:
- Oxidation: Iodine is oxidized from -1 in HI to 0 in I₂.
- Reduction: Sulfur is reduced from +6 in H₂SO₄ to -2 in H₂S.

### Final Answer
The products of the reaction are:
- I₂
- H₂S
- H₂O

So, the complete reaction is:
\[ \text{H}_2\text{SO}_4 + 8\text{HI} \rightarrow 4\text{I}_2 + \text{H}_2\text{S} + 4\text{H}_2\text{O} \]