When `2.76 g` of silver carbonate is strongly heated, it yields a residue weighing

Options

A. 2.16 g
B. 2.48 g
C. 2.32 g
D. 2.64 g

Explanation

To solve the problem of determining the weight of the residue obtained when 2.76 g of silver carbonate (Ag2CO3) is strongly heated, we can follow these steps:

### Step 1: Write the decomposition reaction
When silver carbonate is heated, it decomposes into silver (Ag), carbon dioxide (CO2), and oxygen (O2). The balanced chemical equation for this reaction is:

\[ 2 \text{Ag}_2\text{CO}_3 (s) \rightarrow 4 \text{Ag} (s) + 2 \text{CO}_2 (g) + \text{O}_2 (g) \]

### Step 2: Calculate the molar mass of silver carbonate (Ag2CO3)
The molar mass of silver carbonate can be calculated as follows:
- Molar mass of Ag = 108 g/mol
- Molar mass of C = 12 g/mol
- Molar mass of O = 16 g/mol

Thus, the molar mass of Ag2CO3 is:
\[ 2 \times 108 + 12 + 3 \times 16 = 216 + 12 + 48 = 276 \text{ g/mol} \]

### Step 3: Determine the number of moles of Ag2CO3 in 2.76 g
To find the number of moles of Ag2CO3, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
So,
\[ \text{Number of moles of Ag2CO3} = \frac{2.76 \text{ g}}{276 \text{ g/mol}} = 0.01 \text{ moles} \]

### Step 4: Calculate the moles of silver (Ag) produced
From the balanced equation, 2 moles of Ag2CO3 produce 4 moles of Ag. Therefore, 1 mole of Ag2CO3 produces 2 moles of Ag. Thus, the moles of Ag produced from 0.01 moles of Ag2CO3 is:
\[ \text{Moles of Ag} = 0.01 \text{ moles Ag2CO3} \times \frac{4 \text{ moles Ag}}{2 \text{ moles Ag2CO3}} = 0.02 \text{ moles Ag} \]

### Step 5: Calculate the mass of silver (Ag) produced
Now, we can calculate the mass of silver produced using its molar mass:
\[ \text{Mass of Ag} = \text{Number of moles} \times \text{Molar mass} \]
\[ \text{Mass of Ag} = 0.02 \text{ moles} \times 108 \text{ g/mol} = 2.16 \text{ g} \]

### Conclusion
The residue obtained after heating 2.76 g of silver carbonate is **2.16 g of silver (Ag)**.

---