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: Write the balanced chemical equation
The unbalanced reaction is:
\[ KI + H_2SO_4 \rightarrow I_2 + K_2SO_4 + H_2S + H_2O \]

To balance the equation, we find that:
\[ 8 KI + 5 H_2SO_4 \rightarrow 4 K_2SO_4 + 4 I_2 + H_2S + 4 H_2O \]

### Step 2: Calculate the number of moles of KI
The molar mass of KI is:
- K: 39 g/mol
- I: 127 g/mol
- Total: \( 39 + 127 = 166 \) g/mol

Now, calculate the number of moles of KI in 16.6 g:
\[ \text{Moles of } KI = \frac{\text{mass}}{\text{molar mass}} = \frac{16.6 \text{ g}}{166 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 3: Use the stoichiometry of the balanced equation
From the balanced equation, we see that:
- 8 moles of KI produce 1 mole of \( H_2S \).

Now, calculate the moles of \( H_2S \) produced from 0.1 moles of KI:
\[ \text{Moles of } H_2S = \frac{0.1 \text{ moles of } KI}{8} = 0.0125 \text{ moles of } H_2S \]

### 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:
\[ \text{Volume of } H_2S = \text{moles of } H_2S \times 22.4 \text{ L/mol} = 0.0125 \text{ moles} \times 22.4 \text{ L/mol} = 0.28 \text{ L} \]

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

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

---