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)

Question

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)

Quantrex Academy · Accepted Answer

Correct answer: B. 2.16 g

Molar mass of Ag_2CO_3=2 × 108 + 12 + 3 × 16 = 276 g mol ^ -1 Number of moles of Ag_2CO_3=(2.76)/(276) = 0.01 mol The decomposition reaction of silver carbonate on heating is: Ag_2CO_3(s) arrow 2Ag(s) + CO_2(g) + (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 × 0.01 = 0.02 mol Mass of Ag residue = 0.02 × 108 = 2.16 g Answer: 2.16 g

How many grams of residue is obtained by heating 2.76 g of silver...

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