The amount of arsenic pentasulhide that can be obtained when `35.5g` arsenic acid is treated with axess `H_(2)S` in the presence of conc. `HCl` `("assuming" 100% "conversion")` is :

Options

A. 0.50 mol
B. 0.25 mol
C. 0.125 mol
D. 0.333 mol

Explanation

To solve the problem of how much arsenic pentasulfide can be obtained from 35.5 grams of arsenic acid when treated with excess \( H_2S \) in the presence of concentrated \( HCl \), we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction between arsenic acid (\( H_3AsO_4 \)) and hydrogen sulfide (\( H_2S \)) is:
\[ 2 H_3AsO_4 + 5 H_2S \rightarrow As_2S_5 + 8 H_2O \]

### Step 2: Calculate the molar mass of arsenic acid
The molar mass of arsenic acid (\( H_3AsO_4 \)) can be calculated as follows:
- Hydrogen (H): 1 g/mol × 3 = 3 g/mol
- Arsenic (As): 74.92 g/mol × 1 = 74.92 g/mol
- Oxygen (O): 16 g/mol × 4 = 64 g/mol

Adding these together:
\[ \text{Molar mass of } H_3AsO_4 = 3 + 74.92 + 64 = 141.92 \text{ g/mol} \approx 142 \text{ g/mol} \]

### Step 3: Calculate the number of moles of arsenic acid
To find the number of moles of arsenic acid in 35.5 grams, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles of } H_3AsO_4 = \frac{35.5 \text{ g}}{142 \text{ g/mol}} \approx 0.25 \text{ moles} \]

### Step 4: Use stoichiometry to find moles of arsenic pentasulfide produced
From the balanced equation, we see that:
\[ 2 \text{ moles of } H_3AsO_4 \text{ produce } 1 \text{ mole of } As_2S_5 \]
Thus, the number of moles of arsenic pentasulfide produced from 0.25 moles of arsenic acid can be calculated as:
\[ \text{Moles of } As_2S_5 = \frac{1}{2} \times \text{moles of } H_3AsO_4 = \frac{1}{2} \times 0.25 = 0.125 \text{ moles} \]

### Final Answer
The amount of arsenic pentasulfide that can be obtained is **0.125 moles**.

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