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 find the amount of `H2S` required to precipitate `1.69 g` of `BaS` from a `BaCl2` solution, we can follow these steps:

### Step 1: Write the balanced chemical reaction
The reaction between barium chloride (`BaCl2`) and hydrogen sulfide (`H2S`) can be written as:
\[ \text{BaCl}_2 + \text{H}_2\text{S} \rightarrow \text{BaS} + 2\text{HCl} \]

### Step 2: Calculate the moles of `BaS`
To find the moles of `BaS`, we use the formula:
\[ \text{Moles} = \frac{\text{mass}}{\text{molar mass}} \]

First, we need to calculate the molar mass of `BaS`:
- Atomic weight of `Ba` = 137 g/mol
- Atomic weight of `S` = 32 g/mol

Thus, the molar mass of `BaS` is:
\[ 137 + 32 = 169 \text{ g/mol} \]

Now, we can calculate the moles of `BaS`:
\[ \text{Moles of BaS} = \frac{1.69 \text{ g}}{169 \text{ g/mol}} \approx 0.01 \text{ moles} \]

### Step 3: Determine the moles of `H2S` required
From the balanced equation, we see that the stoichiometric coefficients of `BaS` and `H2S` are both 1. This means that 1 mole of `BaS` requires 1 mole of `H2S`. Therefore, the moles of `H2S` required is also:
\[ \text{Moles of H2S} = 0.01 \text{ moles} \]

### Step 4: Calculate the mass of `H2S`
Next, we need to calculate the molar mass of `H2S`:
- Atomic weight of `H` = 1 g/mol (and there are 2 hydrogen atoms)
- Atomic weight of `S` = 32 g/mol

Thus, the molar mass of `H2S` is:
\[ 2 \times 1 + 32 = 34 \text{ g/mol} \]

Now we can calculate the mass of `H2S` required:
\[ \text{Mass of H2S} = \text{Moles of H2S} \times \text{Molar mass of H2S} \]
\[ \text{Mass of H2S} = 0.01 \text{ moles} \times 34 \text{ g/mol} = 0.34 \text{ g} \]

### Final Answer
The amount of `H2S` required to precipitate `1.69 g` of `BaS` is **0.34 g**.

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