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)

Step-by-step solution

Molar mass of Ag₂CO₃ = 2 108 + 12 + 3 16 = 276 g mol ⁻¹ Number of moles of Ag₂CO₃ = 2.76 276 = 0.01 mol The decomposition reaction of silver carbonate on heating is: Ag₂CO₃(s) 2Ag(s) + CO₂(g) + 1 2 O₂(g) From the stoichiometry of the reaction, 1 mole of Ag₂CO₃ gives 2 moles of Ag as residue. Number of moles of Ag formed = 2 0.01 = 0.02 mol Mass of Ag residue = 0.02 108 = 2.16 g Answer: 2.16 g