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 H₂S needed to precipitate all the metal ions from 100 ml of 1M AgNO₃ and 100 ml of 1M CuSO₄, we will follow these steps:

### Step 1: Determine the moles of AgNO₃ and CuSO₄

1. **Calculate the moles of AgNO₃:**
- Volume of AgNO₃ = 100 ml = 0.1 L
- Concentration of AgNO₃ = 1 M
- Moles of AgNO₃ = Volume (L) × Concentration (M) = 0.1 L × 1 M = 0.1 moles

2. **Calculate the moles of CuSO₄:**
- Volume of CuSO₄ = 100 ml = 0.1 L
- Concentration of CuSO₄ = 1 M
- Moles of CuSO₄ = Volume (L) × Concentration (M) = 0.1 L × 1 M = 0.1 moles

### Step 2: Write the reactions and determine H₂S requirements

1. **Reaction with AgNO₃:**
- The balanced chemical reaction is:
\[ 2 \text{AgNO}_3 + \text{H}_2\text{S} \rightarrow \text{Ag}_2\text{S} + 2 \text{HNO}_3 \]
- From the reaction, 2 moles of AgNO₃ require 1 mole of H₂S.
- Therefore, 0.1 moles of AgNO₃ will require:
\[ \text{Moles of H}_2\text{S} = \frac{0.1 \text{ moles AgNO}_3}{2} = 0.05 \text{ moles H}_2\text{S} \]

2. **Reaction with CuSO₄:**
- The balanced chemical reaction is:
\[ \text{CuSO}_4 + \text{H}_2\text{S} \rightarrow \text{CuS} + \text{H}_2\text{SO}_4 \]
- From the reaction, 1 mole of CuSO₄ requires 1 mole of H₂S.
- Therefore, 0.1 moles of CuSO₄ will require:
\[ \text{Moles of H}_2\text{S} = 0.1 \text{ moles CuSO}_4 = 0.1 \text{ moles H}_2\text{S} \]

### Step 3: Calculate the ratio of H₂S required

- **Ratio of H₂S required for AgNO₃ to CuSO₄:**
- H₂S required for AgNO₃ = 0.05 moles
- H₂S required for CuSO₄ = 0.1 moles
- Ratio = H₂S for AgNO₃ : H₂S for CuSO₄ = 0.05 : 0.1 = 1 : 2

### Final Answer:
The ratio of the amounts of H₂S needed to precipitate all the metal ions from 100 ml of 1M AgNO₃ and 100 ml of 1M CuSO₄ is **1 : 2**.

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