The minium quantity of `H_(2)S` needed to precipitate 63.5 g of `Cu^(2+)` will be nearly.

Options

A. 63.5 g
B. 31.75 g
C. 34 g
D. 2.0 g

Answer

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

### Step 1: Determine the molar mass of \( Cu^{2+} \)
The molar mass of copper (Cu) is 63.5 g/mol. Since \( Cu^{2+} \) is simply copper with a +2 charge, its molar mass remains the same.

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

### Step 3: Determine the stoichiometry of the reaction
The balanced chemical equation for the precipitation of copper sulfide (\( CuS \)) from \( Cu^{2+} \) and \( H_2S \) is:
\[ Cu^{2+} + H_2S \rightarrow CuS + 2H^+ \]
From the equation, we see that 1 mole of \( Cu^{2+} \) reacts with 1 mole of \( H_2S \).

### Step 4: Calculate the number of moles of \( H_2S \) required
Since we need 1 mole of \( H_2S \) to react with 1 mole of \( Cu^{2+} \), we will need:
\[ \text{Number of moles of } H_2S = 1 \, \text{mol} \]

### Step 5: Determine the molar mass of \( H_2S \)
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.
- Sulfur (S) has a molar mass of approximately 32 g/mol.

Thus, the molar mass of \( H_2S \) is:
\[ \text{Molar mass of } H_2S = (2 \times 1) + 32 = 34 \, \text{g/mol} \]

### Step 6: Calculate the mass of \( H_2S \) required
To find the mass of \( H_2S \) needed, we use the formula:
\[ \text{Mass (g)} = \text{Number of moles} \times \text{Molar mass (g/mol)} \]
Substituting the values:
\[ \text{Mass of } H_2S = 1 \, \text{mol} \times 34 \, \text{g/mol} = 34 \, \text{g} \]

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

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