If a solution containing 107.10 g of mercury (II) perchlorate is allowed to react completely with a solution containing 16.642 g of sodium sulfide, how many grams of solid precipitate will be formed?

Question

Gabrielle Orr

Chemistry

6 October, 23:44

If a solution containing 107.10 g of mercury (II) perchlorate is allowed to react completely with a solution containing 16.642 g of sodium sulfide, how many grams of solid precipitate will be formed?

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June Weber · Answer 1

When mercury (II) perchlorate is allowed to react with soldium sulfide, the resulting products are mercury (II) sulfide and sodium perchlorate. The equation is Hg (ClO4) 2 + Na2S = 2NaClO4 + HgS. We must need to find first which of the two reactants is the limiting one. The limiting reactant is sodium sulfide from the given. The solid precipitate between the two products is HgS. Hence the amount of solid precipitate produced is 49.63 grams HgS.