How many grams of residue is obtained by heating $2.76$ g of silver carbonate?
(Given: Molar mass of C, O and Ag are $12$, $16$ and $108$ g mol$^{-1}$ respectively)

Molar mass of $Ag_2CO_3$ = $2 \times 108 + 12 + 3 \times 16 = 276$ g mol$^{-1}$

Number of moles of $Ag_2CO_3$ = $\dfrac{2.76}{276} = 0.01$ mol

The decomposition reaction of silver carbonate on heating is:

$Ag_2CO_3(s) \rightarrow 2Ag(s) + CO_2(g) + \dfrac{1}{2}O_2(g)$

From the stoichiometry of the reaction, $1$ mole of $Ag_2CO_3$ gives $2$ moles of Ag as residue.

Number of moles of Ag formed = $2 \times 0.01 = 0.02$ mol

Mass of Ag residue = $0.02 \times 108 = 2.16$ g

Answer: $2.16$ g