Question

Ag2CO3, a white powder, forms solid grey silver, CO2 and O2 when heated in a test tube. What conclusions can you draw regarding these 4 substances – Ag2CO3, Ag, CO2, O2.

Answer 1

Solution 1 :

Unlike other metal carbonates that usually decomposes into metal oxides liberating carbon dioxide, silver carbonate on heating decomposes into elemental silver liberating mixture of carbon dioxide and oxygen gas as -

Ag2CO3 (s) 2Ag (s) + CO2 (g) + 1/2 O2 (g) MW = 276g 2 × 108 = 216 g Hence, 2.76 g of Ag2CO3 on heating will give :

216/276× 2.76 = 2.16 g Ag as residue.

Solution 2 :

The rxn is as follows Ag2CO3=Ag + Co2 + O2

No need to balance the equation just apply POAC.

2×moles of Ag2Co3= 1× Mole of Ag and moles

= Weight in gram/molr mass put the values and u will get the same answer.

POAC is very simple just conserve the no of atoms of that particular element on both the sides and get the answer.

Solution 3 :

Here, First write the equation..Ag2CO3 => 2Ag + CO2 + 1/2 O2263.6g 203.6g (standard molecular wt)

But, Ag2CO3=2.76gSo, 263.6g Ag2CO3 - 203.6g 2Ag 2.76g Ag2CO3. ?

So, 2.76 × 203.6/263.6 =2.13g ~2.16g

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Hope it helped :)

Answer 2

SOLUTION

MassofAg2O3=2.76gMolarMassofAg2O3=276g/molMolarMassofAg=108g/molMoleofAg2O3=2.76276=0.01moleAg2O3−→Δ2Ag+CO2+12O21moleofAg2O3=2molesofAg⟹Massof0.02moleofAg=0.02×108g=2.16gMassofresidue=MassofAg2O3−MassofAg=(2.76−2.16)g=0.6g