If a solution containing 118.08 g of mercury(II) chlorate is allowed to react completely with a solution containing 16.642 g of sodium sulfide, how many grams of solid precipitate will form

49.544 g.

Explanation:

The balanced equation of reaction is given below;

Hg(ClO3)2 (aq) + Na2S (aq) > 2 NaClO3 (aq) + HgS (s).

So, the parameters given in the question are; Mass of Hg(ClO3) = 118.08 and the mass of Na2S = 16.642 g.

Therefore, the first thing we are going to be looking at is the reactant which is the limiting reagent from the number of moles

(1). For Hg(ClO3), the molar mass = 367.5 g/mol. Therefore, the number of moles, n = mass/ molar mass.

Number of moles = 118.08/ 367.5.

Number of moles = 0.321 moles.

(2). For Na2S, the molar mass = 78.05 g /mol.

The number of moles = 16.643 / 78.05.

The number of moles= 0.213.

Therefore, the limiting reagent = Na2S.

This means that the excess reagent is Hg(ClO3).

==> In the balanced equation of reaction above, the solid precipitate = HgS.

Hence, the mass of HgS formed = 0.213 × 232.6 g/mol. = 49.544 g.

the answer is c

tell me if wrong