An ore contains 1.34% of the mineral argentite, `Ag_(2)S`, by mass. How many gram of this ore would have to be processed in order to obtain 1.00 g of pure solid silver, Ag?

Options

A. 74.6 g
B. 85.7 g
C. 107.9 g
D. 134 g

Answer

To solve the problem, we need to determine how many grams of ore containing 1.34% argentite (Ag₂S) are required to extract 1.00 g of pure silver (Ag). Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the molar mass of argentite (Ag₂S)
- The molar mass of silver (Ag) is approximately 108 g/mol.
- The molar mass of sulfur (S) is approximately 32 g/mol.
- Therefore, the molar mass of Ag₂S is calculated as follows:
\[ \text{Molar mass of Ag₂S} = (2 \times 108) + 32 = 216 + 32 = 248 \text{ g/mol} \]

### Step 2: Determine the mass of silver in one mole of Ag₂S
- In one mole of Ag₂S, there are 2 moles of silver:
\[ \text{Mass of silver in Ag₂S} = 2 \times 108 = 216 \text{ g} \]

### Step 3: Calculate the amount of Ag₂S needed to obtain 1 g of silver
- We need to find out how much Ag₂S is required to obtain 1 g of silver. Using the ratio:
\[ \text{Mass of Ag₂S required} = \left( \frac{248 \text{ g of Ag₂S}}{216 \text{ g of Ag}} \right) \times 1 \text{ g of Ag} \]
- Calculating this gives:
\[ \text{Mass of Ag₂S required} = \frac{248}{216} \approx 1.148 \text{ g of Ag₂S} \]

### Step 4: Calculate the total mass of ore needed to obtain the required amount of Ag₂S
- Since the ore contains 1.34% of Ag₂S, we can set up the following proportion:
\[ 1.34\% \text{ of ore} = 1.148 \text{ g of Ag₂S} \]
- To find 100% of the ore, we use the formula:
\[ \text{Mass of ore} = \left( \frac{100}{1.34} \right) \times 1.148 \]
- Calculating this gives:
\[ \text{Mass of ore} \approx \frac{100}{1.34} \times 1.148 \approx 85.7 \text{ g} \]

### Conclusion
To obtain 1.00 g of pure solid silver, approximately **85.7 grams** of the ore must be processed.

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