KI reacts with `H_(2)SO_(4)` producing `I_(2)` and `H_(2)S` the volume of 0.2 M `H_(2)SO_(4)` required to produce 0.1 " mol of "`H_(2)S` is

Options

A. 4 L
B. 2.5 L
C. 3.8 L
D. 5 L

Answer

To solve the problem step by step, we will follow the stoichiometric principles based on the reaction between KI and H₂SO₄.

### Step 1: Write the balanced chemical equation
The reaction between potassium iodide (KI) and sulfuric acid (H₂SO₄) produces iodine (I₂), hydrogen sulfide (H₂S), potassium sulfate (K₂SO₄), and water (H₂O). The balanced equation is:

\[ 8 \text{KI} + 5 \text{H}_2\text{SO}_4 \rightarrow 4 \text{K}_2\text{SO}_4 + 4 \text{I}_2 + 4 \text{H}_2\text{S} + 4 \text{H}_2\text{O} \]

### Step 2: Determine the moles of H₂SO₄ required to produce H₂S
From the balanced equation, we see that 5 moles of H₂SO₄ produce 4 moles of H₂S. To find out how many moles of H₂SO₄ are needed to produce 0.1 moles of H₂S, we set up a proportion:

\[ \frac{5 \text{ moles H}_2\text{SO}_4}{4 \text{ moles H}_2\text{S}} = \frac{x \text{ moles H}_2\text{SO}_4}{0.1 \text{ moles H}_2\text{S}} \]

Cross-multiplying gives:

\[ x = \frac{5 \times 0.1}{4} = 0.125 \text{ moles of H}_2\text{SO}_4 \]

### Step 3: Use the molarity to find the volume of H₂SO₄ solution
We know the molarity (M) of the H₂SO₄ solution is 0.2 M. Molarity is defined as:

\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]

Rearranging the formula to find the volume gives:

\[ \text{Volume} = \frac{\text{moles of solute}}{\text{Molarity}} \]

Substituting the values we found:

\[ \text{Volume} = \frac{0.125 \text{ moles}}{0.2 \text{ M}} = 0.625 \text{ liters} \]

### Conclusion
The volume of 0.2 M H₂SO₄ required to produce 0.1 moles of H₂S is **0.625 liters**.

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