Answer to Question #104283 in General Chemistry for Samantha

Question

If a solution containing 41.98g of mercury(II) perchlorate is allowed to react completely with a solution containing 6.256g of sodium sulfide, how many grams of solid precipitate will form?

Answer 1

Hg(ClO4)2 + Na2S = HgS + 2NaClO4

n(Hg(ClO4)2)= 41.98g/399.49g/mol = 0.105 mol

n(Na2S) = 6.256g/78g/mol = 0.08 mol

Hg(ClO4)2 in the excess, Na2S reacts completely.

n(HgS) = 0.08 mol

m(HgS) = 0.08mol*232.66g/mol = 18.6 g - mass of the solid precipitate.