How many grams of `H_(2) S` will react with 6.32 g of `KMnO_(4)` to produce `K_(2) SO_(4) and MnO_(2)` ?

Explanation

To solve the problem of how many grams of `H₂S` will react with 6.32 g of `KMnO₄` to produce `K₂SO₄` and `MnO₂`, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical reaction for the reaction between `H₂S` and `KMnO₄` can be represented as:

\[ 5 \, H_2S + 2 \, KMnO_4 \rightarrow K_2SO_4 + 2 \, MnO_2 + 3 \, H_2O \]

### Step 2: Determine the molar mass of `KMnO₄`
The molar mass of `KMnO₄` can be calculated as follows:
- K: 39 g/mol
- Mn: 55 g/mol
- O: 16 g/mol × 4 = 64 g/mol

Adding these together:
\[ 39 + 55 + 64 = 158 \, g/mol \]

### Step 3: Calculate the number of moles of `KMnO₄`
Using the mass of `KMnO₄` given (6.32 g), we can calculate the number of moles:

\[ \text{Moles of } KMnO_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{6.32 \, g}{158 \, g/mol} \approx 0.0400 \, mol \]

### Step 4: Use stoichiometry to find moles of `H₂S`
From the balanced equation, we see that 2 moles of `KMnO₄` react with 5 moles of `H₂S`. Therefore, we can set up a proportion to find the moles of `H₂S`:

\[ \frac{5 \, \text{moles } H_2S}{2 \, \text{moles } KMnO_4} = \frac{x \, \text{moles } H_2S}{0.0400 \, \text{moles } KMnO_4} \]

Cross-multiplying gives:

\[ x = 0.0400 \, \text{moles } KMnO_4 \times \frac{5 \, \text{moles } H_2S}{2 \, \text{moles } KMnO_4} = 0.1000 \, \text{moles } H_2S \]

### Step 5: Calculate the mass of `H₂S`
Now we can calculate the mass of `H₂S` using its molar mass. The molar mass of `H₂S` is:

- H: 1 g/mol × 2 = 2 g/mol
- S: 32 g/mol

Thus, the molar mass of `H₂S` is:

\[ 2 + 32 = 34 \, g/mol \]

Now, we can find the mass of `H₂S`:

\[ \text{Mass of } H_2S = \text{moles} \times \text{molar mass} = 0.1000 \, mol \times 34 \, g/mol = 3.40 \, g \]

### Final Answer
The mass of `H₂S` that will react with 6.32 g of `KMnO₄` is **3.40 grams**.

---