The minimum quantity in gram of `H_(2)S` needed to precipitate 63.5 g of `Cu^(2+)` will ne nearly

Options

A. 63.5g
B. 31.75g
C. 34g
D. 20g

Explanation

To find the minimum quantity of \( H_2S \) needed to precipitate 63.5 g of \( Cu^{2+} \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced reaction for the precipitation of copper(II) sulfide from copper(II) ions and hydrogen sulfide is:
\[ Cu^{2+} + H_2S \rightarrow CuS + 2H^+ \]
This shows that one mole of \( Cu^{2+} \) reacts with one mole of \( H_2S \) to produce one mole of \( CuS \).

### Step 2: Calculate the number of moles of \( Cu^{2+} \)
To find the number of moles of \( Cu^{2+} \), we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of \( Cu \) is approximately 63.5 g/mol. Therefore:
\[ \text{Number of moles of } Cu^{2+} = \frac{63.5 \, \text{g}}{63.5 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 3: Determine the number of moles of \( H_2S \) needed
From the balanced equation, we see that 1 mole of \( H_2S \) is required to precipitate 1 mole of \( Cu^{2+} \). Thus:
\[ \text{Number of moles of } H_2S = 1 \, \text{mol} \]

### Step 4: Calculate the mass of \( H_2S \)
Next, we need to calculate the mass of \( H_2S \) required. The molar mass of \( H_2S \) can be calculated as follows:
- Hydrogen (H) has a molar mass of approximately 1 g/mol, and there are 2 hydrogen atoms: \( 2 \times 1 = 2 \, \text{g/mol} \)
- Sulfur (S) has a molar mass of approximately 32 g/mol.

Thus, the molar mass of \( H_2S \) is:
\[ \text{Molar mass of } H_2S = 2 + 32 = 34 \, \text{g/mol} \]

Now, using the number of moles calculated earlier:
\[ \text{Mass of } H_2S = \text{Number of moles} \times \text{Molar mass} = 1 \, \text{mol} \times 34 \, \text{g/mol} = 34 \, \text{g} \]

### Final Answer
The minimum quantity of \( H_2S \) needed to precipitate 63.5 g of \( Cu^{2+} \) is **34 grams**.

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