If a solution containing 80.040 g of mercury(II) acetate is allowed to react com

Question

If a solution containing 80.040 g of mercury(II) acetate is allowed to react completely with a solution containing 12.026 g of sodium sulfide, how many grams of solid precipitate will be formed?

Assuming complete precipitation, how many moles of each ion remain in solution? If an ion is no longer in solution, enter a zero (0) for the number of moles.

If a solution containing 80.040 g of mercury(II) acetate is allowed to react completely with a solution containing 12.026 g of sodium sulfide, how many grams of solid precipitate will be formed? Number How many grams of the reactant in excess will remain after the reaction? Number Assuming complete precipitation, how many moles of each ion remain in solution? If an ion is no longer in solution, enter a zero (0) for the number of moles. Number Number mol mol Hg" = C,H,o Number Number Na mol mol

Explanation / Answer

The reaction of mercury acetate and sodium sulfide is

Hg(Ac)2 ( Mercury acetate ) + Na2S (sodium sulfide)--> HgS ( Mercuric sulfide), (s)+ 2NaAc (sodium acetate)

1 mole of mercury aceate reacts with 1 mole of Na2S to give 1 mole of Hgs, the solid product

molar mass of mercury acetate Hg(C2H3O2)2=319 g/mole, molar mass of Na2S= 78 g/mole

moles =mass/molar mass , moles : Hg(Ac)2= 80.040/319=0.251, moles of Na2S= 12.026/78=0.154

theoretical molar ratio of Hg(Ac)2: Na2S =1:1, actual ratio of Hg(Ac)2: Na2S= 0.251:0.154= 0.251/0.154:0.154/0.154= 1.63:1

hence Hg(Ac)2 is excess and Na2S is limiting.

0.154 moles of Na2S gives 0.154 moles of HgS and also fomrs 2*0.154=0.308 moles of NaAc.

molar mass of HgS= 233 g/mole, mass of HgS formed= moles* molar mass=0.154*233=35.9 gm

moles of Hg(AC)2 consumed = 0.154, moles of Hg(AC)2 remaining = 0.251-0.154=0.097 moles

mass of Hg(AC)2 remaining =0.097*319 gm =30.9 gm

moles of NaAc= 0.308, moles of Hg(AC)2= 0.097 moles

solution contains NaAC and Hg(AC)2 after reaction

moles off Na+ =0.308 , moles of AC-= moles of AC in NaAC+ moles of Ac in Hg(AC)2 =0.308+2*0.097=0.502

moles of Hg+2=0.097,since Na2S completelly reacts and forms solid products, moles of S2- =0

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