A blackened silver coin weighing 15 g on treatment with HCI yielded 25 ml of `H_2 S` at `12^@ C` and 775 mmHg. What percentage of the original silver tranished?

Options

A. 0.1088
B. 0.0734
C. 0.0365
D. 0.0153

Answer

To solve the problem, we need to determine the percentage of silver that tarnished on the coin. Here’s a step-by-step solution:

### Step 1: Understand the Reaction
The tarnished silver coin contains silver sulfide (Ag₂S). When treated with hydrochloric acid (HCl), it produces hydrogen sulfide (H₂S) and silver chloride (AgCl). The balanced chemical equation for the reaction is:
\[ Ag_2S + 2HCl \rightarrow 2AgCl + H_2S \]

### Step 2: Calculate the Moles of H₂S Produced
We are given the volume of H₂S produced (25 mL) at a temperature of 12°C and a pressure of 775 mmHg. First, we need to convert the volume to liters and the pressure to atmospheres.

- Volume in liters:
\[ V = 25 \, \text{mL} = 25 \times 10^{-3} \, \text{L} \]

- Pressure in atmospheres:
\[ P = \frac{775 \, \text{mmHg}}{760 \, \text{mmHg/atm}} = 1.028 \, \text{atm} \]

- Temperature in Kelvin:
\[ T = 12 + 273 = 285 \, \text{K} \]

### Step 3: Use the Ideal Gas Law to Calculate Moles of H₂S
Using the ideal gas equation \( PV = nRT \), we can solve for the number of moles (n):
\[ n = \frac{PV}{RT} \]
Where \( R = 0.0821 \, \text{L atm/(K mol)} \).

Substituting the values:
\[ n = \frac{(1.028 \, \text{atm}) \times (25 \times 10^{-3} \, \text{L})}{(0.0821 \, \text{L atm/(K mol)}) \times (285 \, \text{K})} \]
Calculating this gives:
\[ n \approx 1.09 \times 10^{-3} \, \text{moles of H₂S} \]

### Step 4: Relate Moles of H₂S to Moles of Silver
From the balanced equation, 1 mole of H₂S corresponds to 2 moles of AgCl, which means it corresponds to 2 moles of Ag. Therefore, the moles of Ag produced are:
\[ \text{Moles of Ag} = 2 \times (1.09 \times 10^{-3}) = 2.18 \times 10^{-3} \, \text{moles of Ag} \]

### Step 5: Calculate the Mass of Silver
Now, we can calculate the mass of silver using its molar mass (107.8 g/mol):
\[ \text{Mass of Ag} = \text{Moles of Ag} \times \text{Molar Mass of Ag} \]
\[ \text{Mass of Ag} = 2.18 \times 10^{-3} \, \text{moles} \times 107.8 \, \text{g/mol} \approx 0.235 \, \text{g} \]

### Step 6: Calculate the Percentage of Silver that Tarnished
To find the percentage of silver that tarnished, we divide the mass of silver that reacted by the original mass of the coin and multiply by 100:
\[ \text{Percentage of Ag tarnished} = \left( \frac{0.235 \, \text{g}}{15 \, \text{g}} \right) \times 100 \approx 1.57\% \]

### Final Answer
The percentage of the original silver that tarnished is approximately **1.57%**.

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