Calculate the residue obtained on strongly heating 2.76g of Ag2CO3
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Answered by
472
Ag2CO3 on strong heating decomposes as
2Ag2CO3(s)-4Ag(s)+2CO2(g)+O2(g)
Molar mass of Ag2CO3 =276g
Molar mass of Ag=108g
2 mol of Ag2CO3 gives 4 moles of Ag
552g of Ag2CO3 gives 432g of Ag(residue)
2.76g of Ag2CO3 gives
432/552*2.76=2.16g of Ag as residue
2Ag2CO3(s)-4Ag(s)+2CO2(g)+O2(g)
Molar mass of Ag2CO3 =276g
Molar mass of Ag=108g
2 mol of Ag2CO3 gives 4 moles of Ag
552g of Ag2CO3 gives 432g of Ag(residue)
2.76g of Ag2CO3 gives
432/552*2.76=2.16g of Ag as residue
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Answered by
53
Sol. 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)
Molecular weight= 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.
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