Residue Mass Obtained by Heating Silver Carbonate

To determine the mass of the residue obtained upon heating silver carbonate ($\text{Ag}_2\text{CO}_3$), we first need to determine its molar mass and understand its thermal decomposition reaction.

Step 1: Calculate the Molar Mass of Silver Carbonate ($\text{Ag}_2\text{CO}_3$)

Given atomic masses:

• $\text{Ag} = 108\text{ g mol}^{-1}$
• $\text{C} = 12\text{ g mol}^{-1}$
• $\text{O} = 16\text{ g mol}^{-1}$

$\text{Molar mass of Ag}_2\text{CO}_3 = 2(108) + 12 + 3(16) = 216 + 12 + 48 = 276\text{ g mol}^{-1}$

Step 2: Calculate the Moles of $\text{Ag}_2\text{CO}_3$

Given mass of $\text{Ag}_2\text{CO}_3 = 2.76\text{ g}$

$\text{Number of moles of Ag}_2\text{CO}_3 = \frac{\text{Given Mass}}{\text{Molar Mass}} = \frac{2.76\text{ g}}{276\text{ g mol}^{-1}} = 0.01\text{ mol}$

Step 3: Chemical Reaction of Thermal Decomposition

Silver carbonate decomposes thermally to give metallic silver, carbon dioxide gas, and oxygen gas because silver oxide ($\text{Ag}_2\text{O}$) is thermally unstable and decomposes further into metallic silver and oxygen:

$\text{Ag}_2\text{CO}_3(s) \xrightarrow{\Delta} 2\text{Ag}(s) + \text{CO}_2(g) + \frac{1}{2}\text{O}_2(g)$

Carbon dioxide ($\text{CO}_2$) and oxygen ($\text{O}_2$) escape as gases, leaving behind metallic silver ($\text{Ag}$) as the solid residue.

Step 4: Calculate the Mass of the Solid Residue ($\text{Ag}$)

From the stoichiometry of the balanced equation: $1\text{ mole of Ag}_2\text{CO}_3 \text{ produces } 2\text{ moles of Ag}$

Therefore: $\text{Moles of Ag residue} = 2 \times 0.01\text{ mol} = 0.02\text{ mol}$

$\text{Mass of residue (Ag)} = \text{Moles of Ag} \times \text{Molar mass of Ag}$ $\text{Mass of residue} = 0.02\text{ mol} \times 108\text{ g mol}^{-1} = 2.16\text{ g}$

Thus, the mass of the residue obtained is $2.16\text{ g}$.

Correct Option: B ($2.16\text{ g}$)