The amount of `H_2 S` required to precipitate 1.69 g BaS from `BaCl_2` solution is (Atomic weight Ba=137, S = 32 and H = 1) :

Options

A. 3.4 g
B. 1.7 g
C. 0.34 g
D. 0.17 g

Answer

To solve the problem of how much \( H_2S \) is required to precipitate 1.69 g of \( BaS \) from a \( BaCl_2 \) solution, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between barium chloride (\( BaCl_2 \)) and hydrogen sulfide (\( H_2S \)) can be written as:
\[ BaCl_2 + H_2S \rightarrow BaS + 2HCl \]
From this equation, we can see that 1 mole of \( H_2S \) produces 1 mole of \( BaS \).

### Step 2: Calculate the molar mass of \( BaS \)
The molar mass of \( BaS \) can be calculated using the atomic weights given:
- Atomic weight of Barium (Ba) = 137 g/mol
- Atomic weight of Sulfur (S) = 32 g/mol

Thus, the molar mass of \( BaS \) is:
\[ Molar \, mass \, of \, BaS = 137 + 32 = 169 \, g/mol \]

### Step 3: Calculate the number of moles of \( BaS \) in 1.69 g
Using the molar mass of \( BaS \), we can find the number of moles of \( BaS \) in 1.69 g:
\[ \text{Number of moles of } BaS = \frac{\text{mass}}{\text{molar mass}} = \frac{1.69 \, g}{169 \, g/mol} \approx 0.01 \, mol \]

### Step 4: Determine the amount of \( H_2S \) required
From the balanced equation, we know that 1 mole of \( H_2S \) is required to produce 1 mole of \( BaS \). Therefore, the number of moles of \( H_2S \) required is also 0.01 moles.

### Step 5: Calculate the mass of \( H_2S \) required
Now, we need to calculate the mass of \( H_2S \) required using its molar mass:
- Atomic weight of Hydrogen (H) = 1 g/mol
- Therefore, the molar mass of \( H_2S \) is:
\[ Molar \, mass \, of \, H_2S = 2 \times 1 + 32 = 34 \, g/mol \]

Now we can find the mass of \( H_2S \) required:
\[ \text{Mass of } H_2S = \text{Number of moles} \times \text{Molar mass} = 0.01 \, mol \times 34 \, g/mol = 0.34 \, g \]

### Final Answer
The amount of \( H_2S \) required to precipitate 1.69 g of \( BaS \) is **0.34 g**.

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