The minimum quantity of `H_(2)`S needed to precipitate 64.5 g of `Cu^(2+)` will be nearly.

Options

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

Explanation

To solve the problem of determining the minimum quantity of H₂S needed to precipitate 64.5 g of Cu²⁺, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between copper(II) ions (Cu²⁺) and hydrogen sulfide (H₂S) can be represented as:
\[ \text{Cu}^{2+} + \text{H}_2\text{S} \rightarrow \text{CuS} + 2\text{H}^+ \]

### Step 2: Calculate the molar mass of Cu²⁺ and H₂S
- The molar mass of copper (Cu) is approximately 63.5 g/mol.
- The molar mass of hydrogen sulfide (H₂S) is calculated as follows:
- Hydrogen (H) = 1 g/mol × 2 = 2 g/mol
- Sulfur (S) = 32 g/mol
- Total molar mass of H₂S = 2 g/mol + 32 g/mol = 34 g/mol

### Step 3: Determine the number of moles of Cu²⁺ in 64.5 g
Using the molar mass of Cu (63.5 g/mol):
\[ \text{Moles of Cu}^{2+} = \frac{\text{mass}}{\text{molar mass}} = \frac{64.5 \text{ g}}{63.5 \text{ g/mol}} \approx 1.016 \text{ moles} \]

### Step 4: Use the stoichiometry of the reaction
From the balanced equation, we see that 1 mole of Cu²⁺ reacts with 1 mole of H₂S. Therefore, the moles of H₂S required will also be approximately 1.016 moles.

### Step 5: Calculate the mass of H₂S needed
Using the molar mass of H₂S (34 g/mol):
\[ \text{Mass of H}_2\text{S} = \text{moles} \times \text{molar mass} = 1.016 \text{ moles} \times 34 \text{ g/mol} \approx 34.544 \text{ g} \]

### Step 6: Round to the nearest significant figure
Since the question asks for the minimum quantity, we can round this to approximately 34 g of H₂S.

### Final Answer
The minimum quantity of H₂S needed to precipitate 64.5 g of Cu²⁺ is nearly **34 g**.

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