What will be the volume of `H_(2)S` released at `0^(@)C` and 1 atm pressure when 16.6 g of KI reacts with excess of `H_(2)SO_(4)` according to the equation `KI+H_(2)SO_(4)rarr I_(2)+K_(2)SO_(4)+H_(2)S`+H_(2)O.

Options

A. 2.24 lit
B. 280 ml
C. 224 ml
D. 1.12 lit

Answer

To find the volume of \( H_2S \) released at \( 0^\circ C \) and 1 atm pressure when 16.6 g of KI reacts with excess \( H_2SO_4 \), we can follow these steps:

### Step 1: Calculate the molar mass of KI
- The atomic mass of Potassium (K) = 39 g/mol
- The atomic mass of Iodine (I) = 127 g/mol
- Therefore, the molar mass of KI = \( 39 + 127 = 166 \) g/mol

### Step 2: Calculate the number of moles of KI
- Number of moles of KI = \( \frac{\text{mass of KI}}{\text{molar mass of KI}} \)
- Number of moles of KI = \( \frac{16.6 \, \text{g}}{166 \, \text{g/mol}} = 0.1 \, \text{mol} \)

### Step 3: Use the stoichiometry of the reaction
- The balanced chemical equation is:
\[ 8 \, KI + 5 \, H_2SO_4 \rightarrow 4 \, I_2 + K_2SO_4 + H_2S + 4 \, H_2O \]
- From the equation, 8 moles of KI produce 1 mole of \( H_2S \).
- Therefore, 0.1 moles of KI will produce:
\[ \text{Moles of } H_2S = \frac{0.1 \, \text{mol KI}}{8} = 0.0125 \, \text{mol H}_2S \]

### Step 4: Calculate the volume of \( H_2S \) at STP
- At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters.
- Therefore, the volume of \( H_2S \) produced is:
\[ \text{Volume of } H_2S = 0.0125 \, \text{mol} \times 22.4 \, \text{L/mol} = 0.28 \, \text{L} \]

### Step 5: Convert volume to milliliters
- To convert liters to milliliters, we multiply by 1000:
\[ 0.28 \, \text{L} \times 1000 = 280 \, \text{mL} \]

### Final Answer
The volume of \( H_2S \) released is **280 mL**.

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