If 10 g of Ag reacts with 1 g of sulfur, the amount of `Ag_(2)S` formed will be [Atomic weight of Ag=108, S=32]?

Options

A. 7.75 g
B. 0.775 g
C. 11 g
D. 10 g

Explanation

To determine the amount of Ag₂S formed when 10 g of Ag reacts with 1 g of sulfur, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between silver (Ag) and sulfur (S) to form silver sulfide (Ag₂S) can be represented as:
\[ 2 \, \text{Ag} + \text{S} \rightarrow \text{Ag}_2\text{S} \]

### Step 2: Calculate the molar masses
- The molar mass of Ag (silver) = 108 g/mol
- The molar mass of S (sulfur) = 32 g/mol
- The molar mass of Ag₂S = \( 2 \times 108 + 32 = 248 \, \text{g/mol} \)

### Step 3: Determine the number of moles of reactants
- Moles of Ag = \( \frac{10 \, \text{g}}{108 \, \text{g/mol}} \approx 0.0926 \, \text{mol} \)
- Moles of S = \( \frac{1 \, \text{g}}{32 \, \text{g/mol}} = 0.03125 \, \text{mol} \)

### Step 4: Identify the limiting reagent
From the balanced equation, we see that 2 moles of Ag react with 1 mole of S. Therefore, we need to find out how much Ag is required for the available sulfur:
- Required moles of Ag for 0.03125 moles of S = \( 0.03125 \, \text{mol} \times 2 = 0.0625 \, \text{mol} \)

Since we have 0.0926 moles of Ag available, which is more than the 0.0625 moles required, sulfur (S) is the limiting reagent.

### Step 5: Calculate the amount of Ag₂S formed
According to the stoichiometry of the reaction:
- 1 mole of S produces 1 mole of Ag₂S.
- Therefore, 0.03125 moles of S will produce 0.03125 moles of Ag₂S.

### Step 6: Convert moles of Ag₂S to grams
Now we can convert the moles of Ag₂S to grams:
- Mass of Ag₂S = moles × molar mass = \( 0.03125 \, \text{mol} \times 248 \, \text{g/mol} \approx 7.75 \, \text{g} \)

Thus, the amount of Ag₂S formed is approximately **7.75 g**.

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