The action of `H_(2)SO_(4)` on KI gives `I_(2)` and `H_(2)S` Calculate the volume of `0.2 M H_(2)SO_(4)` to produce `3.4g H_(2)S`

Explanation

To solve the problem of calculating the volume of `0.2 M H₂SO₄` required to produce `3.4 g H₂S`, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between potassium iodide (KI) and sulfuric acid (H₂SO₄) can be represented as follows:

\[ 8 \text{KI} + 5 \text{H₂SO₄} \rightarrow 4 \text{K₂SO₄} + \text{H₂S} + 4 \text{I₂} + 4 \text{H₂O} \]

### Step 2: Calculate the molar mass of H₂S
The molar mass of H₂S (Hydrogen Sulfide) can be calculated as:
- H: 1 g/mol × 2 = 2 g/mol
- S: 32 g/mol × 1 = 32 g/mol

Thus, the molar mass of H₂S is:
\[ 2 + 32 = 34 \text{ g/mol} \]

### Step 3: Calculate the number of moles of H₂S produced
Using the mass of H₂S given (3.4 g), we can calculate the number of moles using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

So,
\[ \text{Number of moles of H₂S} = \frac{3.4 \text{ g}}{34 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 4: Determine the moles of H₂SO₄ required
From the balanced equation, we see that 5 moles of H₂SO₄ produce 1 mole of H₂S. Therefore, to find the moles of H₂SO₄ needed for 0.1 moles of H₂S, we can use the ratio:
\[ \text{Moles of H₂SO₄} = 0.1 \text{ moles H₂S} \times \frac{5 \text{ moles H₂SO₄}}{1 \text{ mole H₂S}} = 0.5 \text{ moles H₂SO₄} \]

### Step 5: Calculate the mass of H₂SO₄ required
Now, we can calculate the mass of H₂SO₄ needed using its molar mass (98 g/mol):
\[ \text{Mass of H₂SO₄} = \text{Moles} \times \text{Molar Mass} = 0.5 \text{ moles} \times 98 \text{ g/mol} = 49 \text{ g} \]

### Step 6: Calculate the volume of H₂SO₄ solution required
Given that the molarity (M) of the H₂SO₄ solution is 0.2 M, we can use the formula:
\[ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]

Rearranging gives:
\[ \text{Volume} = \frac{\text{Number of moles}}{\text{Molarity}} \]

Substituting the values:
\[ \text{Volume} = \frac{0.5 \text{ moles}}{0.2 \text{ M}} = 2.5 \text{ liters} \]

### Final Answer
The volume of `0.2 M H₂SO₄` required to produce `3.4 g H₂S` is **2.5 liters**.

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