Question

A water-soluble salt formed by Hg2+ and chlorate 105.27 g was dissolved in a solution containing 16.642 g of sodium sulfide, which gave rise to an insoluble mercury salt.

How much insoluble mercury salt is formed?

How much of the excess reagent remains?

Answer 1

Hg(ClO3)2 + Na2S ==> HgS + 2NaClO3
mols Hg(ClO3)2 = g/molar mass = 105.27/367.49 = 0.28646
mols Na2S = 16.642/78.045 = 0.21324
Since the reactants are 1:1 then 0.21324 is the smaller moles so Na2S will be the limiting reagent (LR) and Hg(ClO3)2 will be the excess reagent (ER).
mols HgS formed = 0.21324 x (1 mol HgS/1 mol Na2S) = 0.21324 mols HgS formed. Grams HgS = mols HgS x molar mass HgS = ?
How much Hg(ClO3)2 is used by the 0.21324 mols Na2S. That's
mols Hg(ClO3)2 used = mols Na2S x (1 mol HgClO3)2/1 mol Na2S) = 0.21324 mols Na2S x (1/1) = 0.21324
0.28646 = mols Hg(ClO3)2 to start
-0.21324 = mols used
----------------------------
0.07322 mols Hg(ClO3)2 remaining.
grams remaining = mols remaining x molar mass = ?
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