The weight of a residue obtained by heating 2.76 g of silver carbonate is

Options

A. 2.76 g
B. 2.98 g
C. 2.16 g
D. 2.44 g

Explanation

To find the weight of the residue obtained by heating 2.76 g of silver carbonate (Ag₂CO₃), we can follow these steps:

### Step 1: Write the decomposition reaction of silver carbonate.
When silver carbonate is heated, it decomposes according to the following reaction:
\[ \text{2 Ag}_2\text{CO}_3 \rightarrow 4 \text{Ag} + 2 \text{CO}_2 + \text{O}_2 \]

### Step 2: Calculate the molar mass of silver carbonate (Ag₂CO₃).
The molar mass of Ag₂CO₃ can be calculated as follows:
- Molar mass of Ag (Silver) = 107.87 g/mol (approximately 108 g/mol)
- Molar mass of C (Carbon) = 12.01 g/mol
- Molar mass of O (Oxygen) = 16.00 g/mol

Thus, the molar mass of Ag₂CO₃ is:
\[ \text{Molar mass of Ag}_2\text{CO}_3 = 2 \times 108 + 12 + 3 \times 16 = 216 + 12 + 48 = 276 \text{ g/mol} \]

### Step 3: Determine the number of moles of silver carbonate in 2.76 g.
Using the molar mass calculated, we can find the number of moles of Ag₂CO₃ in 2.76 g:
\[ \text{Number of moles of Ag}_2\text{CO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{2.76 \text{ g}}{276 \text{ g/mol}} = 0.01 \text{ moles} \]

### Step 4: Use stoichiometry to find the mass of silver produced.
From the balanced equation, we see that 2 moles of Ag₂CO₃ produce 4 moles of Ag. Therefore, the ratio is:
\[ \frac{4 \text{ moles of Ag}}{2 \text{ moles of Ag}_2\text{CO}_3} = 2 \text{ moles of Ag per mole of Ag}_2\text{CO}_3 \]

Thus, for 0.01 moles of Ag₂CO₃, the moles of Ag produced will be:
\[ \text{Moles of Ag} = 0.01 \text{ moles of Ag}_2\text{CO}_3 \times 2 = 0.02 \text{ moles of Ag} \]

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

### Step 6: Determine the weight of the residue.
The residue after heating silver carbonate consists of the silver produced. Therefore, the weight of the residue obtained by heating 2.76 g of silver carbonate is:
\[ \text{Weight of residue} = 2.16 \text{ g} \]

### Final Answer:
The weight of the residue obtained by heating 2.76 g of silver carbonate is **2.16 g**.

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