Question

On heating Ag2CO3 decomposes as: 1 Ag2CO3(s)_2Ag(s) + CO2(9) 02(9) 2. If 5.52 g of pure Ag2CO3(s) on heating left a residue of 4.92 g, then % yield of the reaction is (Atomic mass: Ag = 108)​

Answer 1

Given info : On heating Ag₂CO₃ decomposes as : Ag₂CO₃ ⇒ 2Ag + CO₂ + 1/2O₂  , if 5.52 g of pure Ag₂CO₃ on heating left a residue of 4.92g.

To find : the percentage yield of the reaction is ...

Solution : from reaction, it is clear that 1 mole of silver carbonate gives 2 moles of silver.

molecular weight of Ag₂CO₃ = 2(108) + 12 + 3(16) = 276g

weight of silver produced = 2(108) = 216 g

∴ 276 g of silver carbonate gives 216 g of silver.

⇒ 5.52g of silver carbonate gives 216/276 × 5.52 = 4.32g of silver.

∴ % yield of the reaction = found value/given value × 100

    = 4.32g/4.92g × 100

    = 432/4.92 = 87.8 %

therefore the percentage yield of the reaction is 87.8 %.