To solve the problem of determining the percentage of silver that tarnished from a blackened silver coin weighing 15 g, we can follow these steps:
### Step 1: Understand the Reaction
The tarnished silver coin likely contains silver sulfide (Ag₂S), which reacts with hydrochloric acid (HCl) to produce silver chloride (AgCl) and hydrogen sulfide (H₂S). The balanced reaction is:
\[ \text{Ag}_2\text{S} + 2 \text{HCl} \rightarrow 2 \text{AgCl} + \text{H}_2\text{S} \]
### Step 2: Calculate 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. We need to convert these values to appropriate units and use the ideal gas equation \( PV = nRT \) to find the number of moles of H₂S.
1. **Convert pressure from mmHg to atm**:
\[
P = \frac{775 \, \text{mmHg}}{760 \, \text{mmHg/atm}} = 1.0184 \, \text{atm}
\]
2. **Convert volume from mL to L**:
\[
V = 25 \, \text{mL} = 0.025 \, \text{L}
\]
3. **Convert temperature from Celsius to Kelvin**:
\[
T = 12 + 273 = 285 \, \text{K}
\]
4. **Use the ideal gas equation to find moles of H₂S**:
\[
n = \frac{PV}{RT} = \frac{(1.0184 \, \text{atm})(0.025 \, \text{L})}{(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1})(285 \, \text{K})}
\]
\[
n \approx 0.0010895 \, \text{mol}
\]
### Step 3: Relate Moles of H₂S to Moles of Silver
From the balanced equation, 1 mole of Ag₂S produces 1 mole of H₂S. Therefore, the moles of silver (Ag) produced from the reaction can be calculated as follows:
- Since 1 mole of Ag₂S produces 2 moles of AgCl, the moles of silver (Ag) will be:
\[
\text{Moles of Ag} = 2 \times n_{\text{H}_2\text{S}} = 2 \times 0.0010895 \approx 0.002178 \, \text{mol}
\]
### Step 4: Calculate the Mass of Silver
To find the mass of silver that tarnished, we use the molar mass of silver (Ag), which is approximately 108 g/mol:
\[
\text{Mass of Ag} = \text{Moles of Ag} \times \text{Molar Mass of Ag} = 0.002178 \, \text{mol} \times 108 \, \text{g/mol} \approx 0.2352 \, \text{g}
\]
### Step 5: Calculate the Percentage of Silver Tarnished
Now, we can find the percentage of silver that tarnished from the original mass of the coin:
\[
\text{Percentage of tarnished silver} = \left( \frac{\text{Mass of Ag}}{\text{Original mass of coin}} \right) \times 100 = \left( \frac{0.2352 \, \text{g}}{15 \, \text{g}} \right) \times 100 \approx 1.568\%
\]
### Final Answer
The percentage of the original silver that tarnished is approximately **1.57%**.
