2.76 gm of silver carbonate (at mass of Ag 108) on being heated strongly yield a residue weighing 2.16 gm 2.48 gm 2.32 gm 2.64 gm

The question seems to be asking for the weight of the residue after heating 2.76 gm of silver carbonate.

First, we need to calculate the molecular weight of silver carbonate (Ag2CO3). The atomic mass of silver (Ag) is 107.87, carbon (C) is 12.01, and oxygen (O) is 16.00.

So, the molecular weight of Ag2CO3 = 2*(107.87) + 12.01 + 3*(16.00) = 275.75 gm/mol.

Now, we know that on heating, silver carbonate (Ag2CO3) decomposes to form silver (Ag) and carbon dioxide (CO2). The balanced chemical equation for this reaction is:

2Ag2CO3(s) → 4Ag(s) + 2CO2(g) + O2(g)

From this equation, we can see that 2 moles of Ag2CO3 yield 4 moles of Ag. Therefore, 1 mole of Ag2CO3 will yield 2 moles of Ag.

So, the weight of Ag formed from 2.76 gm of Ag2CO3 = 2 * (2.76 gm * (2 * 107.87 gm/mol) / 275.75 gm/mol) = 2.16 gm.

Therefore, the weight of the residue after heating 2.76 gm of silver carbonate is 2.16 gm.