How many moles of `H_(2)SO_(4)` are requried to produce 1 " mol of "`H_(2)S` when KI reacts with `H_(2)SO_(4)` producing `I_(2)` and `H_(2)S`?

Options

A. 5
B. 4
C. 3
D. 2

Explanation

To solve the problem of how many moles of \( H_2SO_4 \) are required to produce 1 mole of \( H_2S \) when potassium iodide reacts with sulfuric acid, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between potassium iodide (\( KI \)) and sulfuric acid (\( H_2SO_4 \)) can be represented as follows:

\[ 8 KI + 5 H_2SO_4 \rightarrow 4 I_2 + 4 H_2S + 4 K_2SO_4 + 4 H_2O \]

### Step 2: Identify the stoichiometric coefficients
From the balanced equation, we can see the stoichiometric coefficients for \( H_2SO_4 \) and \( H_2S \):
- 5 moles of \( H_2SO_4 \) produce 4 moles of \( H_2S \).

### Step 3: Set up the ratio
To find out how many moles of \( H_2SO_4 \) are needed to produce 1 mole of \( H_2S \), we can set up a proportion based on the coefficients from the balanced equation:

\[ \frac{5 \text{ moles of } H_2SO_4}{4 \text{ moles of } H_2S} = \frac{x \text{ moles of } H_2SO_4}{1 \text{ mole of } H_2S} \]

### Step 4: Solve for \( x \)
Cross-multiplying gives us:

\[ 5 \cdot 1 = 4 \cdot x \]

This simplifies to:

\[ 5 = 4x \]

Now, divide both sides by 4:

\[ x = \frac{5}{4} = 1.25 \]

### Step 5: Conclusion
Thus, to produce 1 mole of \( H_2S \), we need 1.25 moles of \( H_2SO_4 \).

### Final Answer
**1.25 moles of \( H_2SO_4 \) are required to produce 1 mole of \( H_2S \).**

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