Balance the equation that represents a reaction between sulfuric acid and hydroiodic acid. Sulfuric acid and hydroiodic acid react to produce dihydrogen sulfide gas, iodine, and steam. (5 points) ?H2SO4 + ?HI → ?H2S + ?I2 + ?H2O

Answer 1

To balance the chemical equation for the reaction between sulfuric acid (H₂SO₄) and hydroiodic acid (HI) producing dihydrogen sulfide (H₂S), iodine (I₂), and steam (H₂O), we can follow these steps:

Unbalanced equation: \[ \text{H}_2\text{SO}_4 + \text{HI} \rightarrow \text{H}_2\text{S} + \text{I}_2 + \text{H}_2\text{O} \]

Balancing the equation:

1. Count the number of each type of atom on both sides of the equation:

   • Left side:
     • H: 2 (from H₂SO₄) + 1 (from HI) = 3
     • S: 1 (from H₂SO₄)
     • O: 4 (from H₂SO₄)
     • I: 1 (from HI)
   • Right side:
     • H: 2 (from H₂S) + 2 (from H₂O) = 4
     • S: 1 (from H₂S)
     • I: 2 (from I₂)
     • O: 1 (from H₂O)

2. Adjust coefficients to balance the equation. Start with iodine (I) because it is present in fewer species:

   • We need 2 HI for every I₂ produced, so we put a coefficient of 2 in front of HI.

   \[ \text{H}_2\text{SO}_4 + 2 \text{HI} \rightarrow \text{H}_2\text{S} + \text{I}_2 + \text{H}_2\text{O} \]

3. Recount atoms after adding the coefficient:

   • Left side:
     • H: 2 (from H₂SO₄) + 2 (from 2HI) = 4
     • S: 1
     • O: 4
     • I: 2
   • Right side:
     • H: 2 (from H₂S) + 2 (from H₂O) = 4
     • S: 1
     • O: 1
     • I: 2

4. Count the oxygen atoms. There are 4 on the left side and 1 on the right side. We need to adjust H₂O. Since we have to account for the 4 oxygens, we can put a coefficient of 3 before H₂O.

5. The equation now looks like this: \[ \text{H}_2\text{SO}_4 + 2 \text{HI} \rightarrow \text{H}_2\text{S} + \text{I}_2 + 3 \text{H}_2\text{O} \]

6. Finally, check again the balance:

   • Left side:
     • H: 2 (H₂SO₄) + 2 (2HI) = 4, which equals 6 (2 from H₂S + 6 from 3H₂O)
     • S: 1
     • O: 4 (4 total)
     • I: 2

The balanced equation would therefore be: \[ \text{H}_2\text{SO}_4 + 2 \text{HI} \rightarrow \text{H}_2\text{S} + \text{I}_2 + 2 \text{H}_2\text{O} \]

To ensure both sides balance, let's finalize our coefficients: The final balanced equation is: \[ \text{H}_2\text{SO}_4 + 4 \text{HI} \rightarrow \text{H}_2\text{S} + 2 \text{I}_2 + 2 \text{H}_2\text{O} \]

So, the final coefficients are:

• 1 for H₂SO₄
• 4 for HI
• 1 for H₂S
• 2 for I₂
• 2 for H₂O