Question

the weight residue obtained by heating 2.7 g of silver carbonate is ?

Answer 1

Hey there !

Solution :

2Ag₂CO₃ + Heat ( Δ ) → 4 Ag + 2Co₂ + O₂

According to stoichiometry, we get that 2 moles of Silver carbonate on heating gives 4 moles of Silver as residue.

So we are asked to find out how many grams of silver will the Silver Carbonate give if we heated 2.7 g of Silver Carbonate.

Moles of Silver carbonate in 2.7 g can be calculated as :

Number of moles = Given Mass / Molecular Mass

Molecular Mass of Silver carbonate is 276 g

Number of moles = 2.7 g / 276 g

Number of moles = 0.009 which can be approximated to 0.01 moles. So on substituting in the stoichiometry we get,

2 Moles of Silver Carbonate → 4 moles of Silver

0.01 Moles of Silver Carbonate → ?

Let the unknown quantity be ' x '

On cross multiplying we get,

2 * x = 4 * 0.01

=> 2x = 0.04

=> x = 0.04 / 2

=> x = 0.02

Hence we get 0.02 moles of Silver as residue. So the amount of silver as residue obtained in grams can be given as : 0.02 * 108 = 2.16 g.

Hence the amount of silver obtained is 2.16 g.

Hope my answer helped !