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JEE Main 2016 · Chemistry · d- and f-Block Elements and Metallurgy · medium

An ore sample of 40 g contains 62% Ag2S by mass. It is leached with aerated NaCN solution, Ag2S+4NaCN→ 2Na[Ag(CN)2]+Na2S, and the silver is then…

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An ore sample of 40 g40\ \mathrm{g} contains 62%62\% Ag2S\mathrm{Ag_2S} by mass. It is leached with aerated NaCN\mathrm{NaCN} solution, Ag2S+4NaCN→2Na[Ag(CN)2]+Na2S\mathrm{Ag_2S+4NaCN\to 2Na[Ag(CN)_2]+Na_2S}, and the silver is then displaced from the solution by zinc, 2Na[Ag(CN)2]+Zn→Na2[Zn(CN)4]+2Ag\mathrm{2Na[Ag(CN)_2]+Zn\to Na_2[Zn(CN)_4]+2Ag}. With Ag=108\mathrm{Ag}=108, S=32\mathrm{S}=32, Zn=65.4\mathrm{Zn}=65.4, the mass of zinc consumed for complete recovery is
  1. A. 13.1 g13.1\ \mathrm{g}
  2. B. 3.27 g3.27\ \mathrm{g}
  3. C. 6.54 g6.54\ \mathrm{g}
  4. D. 21.6 g21.6\ \mathrm{g}
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Answer: (C) 6.54 g6.54\ \mathrm{g}

Mass of Ag2S=0.62×40=24.8\mathrm{Ag_2S}=0.62\times40=24.8 g; M=248M=248, so 0.1000.100 mol. Each Ag2S\mathrm{Ag_2S} gives 2 mol Na[Ag(CN)2]\mathrm{Na[Ag(CN)_2]}, i.e. 0.2000.200 mol Ag. Zinc displaces silver in the ratio Zn:Ag=1:2\mathrm{Zn:Ag}=1:2 (Zn gives 2 electrons, each Ag+\mathrm{Ag^+} takes 1), so moles of Zn =0.100=0.100 and mass =0.100×65.4=6.54=0.100\times65.4=6.54 g. The 13.113.1 g option takes a 1:11:1 ratio, 3.273.27 g halves the amount, and 21.621.6 g is the mass of silver obtained.
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