The ratio of the molar amounts of `H_(2)S` needed to precipitate the metal ions from 20 mL each 1 M Cd `(NO_(3))_(2)` and 0.5 M `CuSO_(4)` is

Options

A. `1:1`
B. `2:1`
C. `1:2`
D. indefinite

Explanation

To solve the problem, we need to find the ratio of the molar amounts of H₂S required to precipitate metal ions from two solutions: 20 mL of 1 M Cd(NO₃)₂ and 20 mL of 0.5 M CuSO₄.

### Step-by-step Solution:

1. **Calculate the number of moles of Cd(NO₃)₂:**
- Given: Volume = 20 mL = 0.020 L (since 1 L = 1000 mL)
- Molarity (M) = 1 M
- Number of moles (n) = Molarity × Volume
\[ n_{\text{Cd(NO₃)₂}} = 1 \, \text{M} \times 0.020 \, \text{L} = 0.020 \, \text{moles} \]

2. **Determine the moles of H₂S required for Cd(NO₃)₂:**
- The reaction is:
\[ \text{Cd(NO₃)₂} + \text{H₂S} \rightarrow \text{CdS} + 2 \text{HNO₃} \]
- From the reaction, 1 mole of Cd(NO₃)₂ reacts with 1 mole of H₂S. Therefore, for 0.020 moles of Cd(NO₃)₂, we need:
\[ n_{\text{H₂S (for Cd)}} = 0.020 \, \text{moles} \]

3. **Calculate the number of moles of CuSO₄:**
- Given: Volume = 20 mL = 0.020 L
- Molarity (M) = 0.5 M
- Number of moles (n) = Molarity × Volume
\[ n_{\text{CuSO₄}} = 0.5 \, \text{M} \times 0.020 \, \text{L} = 0.010 \, \text{moles} \]

4. **Determine the moles of H₂S required for CuSO₄:**
- The reaction is:
\[ \text{CuSO₄} + \text{H₂S} \rightarrow \text{CuS} + \text{H₂SO₄} \]
- From the reaction, 1 mole of CuSO₄ reacts with 1 mole of H₂S. Therefore, for 0.010 moles of CuSO₄, we need:
\[ n_{\text{H₂S (for Cu)}} = 0.010 \, \text{moles} \]

5. **Calculate the ratio of moles of H₂S required:**
- Moles of H₂S required for Cd(NO₃)₂: 0.020 moles
- Moles of H₂S required for CuSO₄: 0.010 moles
- Ratio of moles of H₂S for Cd(NO₃)₂ to CuSO₄:
\[ \text{Ratio} = \frac{n_{\text{H₂S (for Cd)}}}{n_{\text{H₂S (for Cu)}}} = \frac{0.020}{0.010} = 2:1 \]

### Final Answer:
The ratio of the molar amounts of H₂S needed to precipitate the metal ions from Cd(NO₃)₂ and CuSO₄ is **2:1**.

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