Ratio of the amounts of `H_(2)S` needed to precipitate all the metal ions from 100 ml of `1M AgNO_(3)` and 100 ml of 1M `CuSO_(4)` will be :

Options

A. `1:1`
B. `1:2`
C. `2:1`
D. `3:4`

Explanation

To solve the problem of finding the ratio of the amounts of \( H_2S \) needed to precipitate all the metal ions from 100 ml of \( 1M \, AgNO_3 \) and 100 ml of \( 1M \, CuSO_4 \), we can follow these steps:

### Step 1: Calculate the moles of \( AgNO_3 \) and \( CuSO_4 \)

1. **For \( AgNO_3 \)**:
- Molarity (M) = 1 M
- Volume (V) = 100 ml = 0.1 L
- Moles of \( AgNO_3 \) = Molarity × Volume = \( 1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.1 \, \text{moles} \)

2. **For \( CuSO_4 \)**:
- Molarity (M) = 1 M
- Volume (V) = 100 ml = 0.1 L
- Moles of \( CuSO_4 \) = Molarity × Volume = \( 1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.1 \, \text{moles} \)

### Step 2: Write the balanced chemical equations

1. **For the reaction of \( AgNO_3 \) with \( H_2S \)**:
\[ 2AgNO_3 + H_2S \rightarrow Ag_2S + 2HNO_3 \]
- From the balanced equation, 2 moles of \( AgNO_3 \) react with 1 mole of \( H_2S \).

2. **For the reaction of \( CuSO_4 \) with \( H_2S \)**:
\[ CuSO_4 + H_2S \rightarrow CuS + H_2SO_4 \]
- From the balanced equation, 1 mole of \( CuSO_4 \) reacts with 1 mole of \( H_2S \).

### Step 3: Calculate the amount of \( H_2S \) required for each reaction

1. **For \( AgNO_3 \)**:
- Moles of \( H_2S \) required = \( \frac{0.1 \, \text{moles of } AgNO_3}{2} = 0.05 \, \text{moles of } H_2S \)

2. **For \( CuSO_4 \)**:
- Moles of \( H_2S \) required = \( 0.1 \, \text{moles of } CuSO_4 \) (1:1 ratio)

### Step 4: Find the ratio of \( H_2S \) required for \( AgNO_3 \) to \( CuSO_4 \)

- Ratio of \( H_2S \) for \( AgNO_3 \) to \( CuSO_4 \):
\[ \text{Ratio} = \frac{0.05 \, \text{moles of } H_2S}{0.1 \, \text{moles of } H_2S} = \frac{1}{2} \]

### Final Answer
The ratio of the amounts of \( H_2S \) needed to precipitate all the metal ions from \( AgNO_3 \) to \( CuSO_4 \) is \( 1:2 \).

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