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 determine 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 reaction between copper ions (\( Cu^{2+} \)) and hydrogen sulfide (\( H_2S \)) can be represented as:
\[ Cu^{2+} + H_2S \rightarrow CuS + 2H^+ \]
This shows that one mole of \( Cu^{2+} \) reacts with one mole of \( H_2S \) to produce one mole of copper sulfide (\( CuS \)).

### Step 2: Calculate the molar mass of \( Cu^{2+} \)
The molar mass of copper (\( Cu \)) is approximately 63.5 g/mol. Therefore, 1 mole of \( Cu^{2+} \) corresponds to 63.5 g.

### Step 3: Calculate the molar mass of \( H_2S \)
The molar mass of hydrogen sulfide (\( H_2S \)) can be calculated as follows:
- Hydrogen (H) has a molar mass of about 1 g/mol, and there are 2 hydrogen atoms: \( 2 \times 1 = 2 \) g/mol.
- Sulfur (S) has a molar mass of about 32 g/mol.
Thus, the molar mass of \( H_2S \) is:
\[ 2 + 32 = 34 \text{ g/mol} \]

### Step 4: Determine the amount of \( H_2S \) required for 63.5 g of \( Cu^{2+} \)
From the balanced equation, we know that 1 mole of \( Cu^{2+} \) reacts with 1 mole of \( H_2S \). Therefore, if we have 63.5 g of \( Cu^{2+} \) (which is 1 mole), we will need 1 mole of \( H_2S \) to completely react with it.

Since 1 mole of \( H_2S \) weighs 34 g, we can conclude that:
\[ \text{Minimum quantity of } H_2S = 34 \text{ g} \]

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

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