Find the amount of residue obtained in grams when 2.76g of silver carbonate is strongly heated. (Molar mass of Ag=108g/mol) Ag2CO3Δ−→2Ag+CO2+12O2
A
2.16g
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B
2.48g
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C
2.32g
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D
2.64g
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Solution
The correct option is A2.16g Ag2CO3heat−−→2Ag+CO2↑+12O2↑ Number of moles of Ag2CO3=given massmolar mass=2.76276=0.01mol
According to stoichiometry, 1 mol of Ag2CO3 decomposes to produce 2 mol of Ag 0.01 mol of Ag2CO3 decomposes to produce 21×0.01=0.02 mol of Ag Mass of 0.02 mol of Ag=0.02×108=2.16g of Ag