If a solution containing 107.68 g of mercury(II) perchlorate is allowed to react completely with a solution containing 17.796 g of sodium sulfide, how many grams of solid precipitate will be formed?

Answer 1

This is a limiting reagent (LR) problem and you know that when amounts are given for BOTH reactants.
Hg(ClO4)2 + Na2S ==> HgS + 2NaClO4

1a. mols Hg(ClO4)2 = grams/molar mass = ?
b. mols Na2S = g/molar mass = ?

2a. Using the coefficients in the balanced equation, convert mols Hg(ClO4)2 to mols HgI.
b. Do the same and convert mols Na2S to mols HgS.

3. The mols HgS formed in 2a ia likely not to agree with mols HgS in 2b. The correct answer is ALWAYS the smaller number and the reagent responsible for that is the LR.

4. Using the smaller number convert mols HgS to g by
g = mols x molar mass. The solid ppt is HgS.