An ore contains `1.24%` of the mineral argentite, `Ag_(2)S` by mass. How many grams of this ore would have to be processed in order to obtain 1.0 g of pure solid silver?

Options

A. 46.3 g
B. 92.6 g
C. 69.45 g
D. 23.15 g

Explanation

To solve the problem step by step, we need to determine how many grams of ore containing 1.24% argentite (Ag₂S) are required to obtain 1.0 g of pure silver (Ag).

### Step 1: Determine the moles of silver needed
To find out how many moles of silver correspond to 1.0 g, we use the molar mass of silver (Ag), which is approximately 108 g/mol.

\[ \text{Moles of Ag} = \frac{\text{mass of Ag}}{\text{molar mass of Ag}} = \frac{1.0 \, \text{g}}{108 \, \text{g/mol}} \approx 0.00926 \, \text{mol} \]

### Step 2: Relate moles of silver to moles of argentite
From the decomposition of argentite (Ag₂S), we know that:

\[ \text{1 mole of Ag₂S produces 2 moles of Ag} \]

Thus, to find the moles of Ag₂S needed to produce 0.00926 moles of Ag, we use the ratio:

\[ \text{Moles of Ag₂S} = \frac{0.00926 \, \text{mol Ag}}{2} \approx 0.00463 \, \text{mol Ag₂S} \]

### Step 3: Calculate the mass of argentite required
Next, we need to calculate the molar mass of argentite (Ag₂S):

\[ \text{Molar mass of Ag₂S} = 2 \times 108 \, \text{g/mol (Ag)} + 32 \, \text{g/mol (S)} = 248 \, \text{g/mol} \]

Now, we can find the mass of argentite required:

\[ \text{Mass of Ag₂S} = \text{moles of Ag₂S} \times \text{molar mass of Ag₂S} = 0.00463 \, \text{mol} \times 248 \, \text{g/mol} \approx 1.15 \, \text{g} \]

### Step 4: Calculate the mass of ore needed
Since the ore contains 1.24% of argentite, we can set up the equation to find the mass of ore (X) needed to obtain 1.15 g of argentite:

\[ \frac{1.24}{100} \times X = 1.15 \, \text{g} \]

Solving for X:

\[ X = \frac{1.15 \, \text{g}}{0.0124} \approx 92.74 \, \text{g} \]

### Conclusion
To obtain 1.0 g of pure silver, approximately **92.74 grams** of the ore must be processed.

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