If a solution containing 18.96 g of mercury (ii) perchlorate is allowed to react completely with a solution containing 6.256 g of sodium sulfate, how many grams of solid precipitate will be formed? how many grams of the reactant in excess will remain after the reaction?

The balanced chemical reaction would be written as:

Hg(ClO4)2 + Na2SO4 → 2 NaClO4 + HgSO4(s)

We are given the amounts of both of the reactants to be used for the reaction. We use these values as the starting point of the calculations. 

18.96 g Hg(ClO4)2 ( 1 mol / 399.4912 g ) = 0.0475 mol  Hg(ClO4)2 
6.256 g Na2SO4 ( 1 mol / 142.04 g ) = 0.0440 mol Na2SO4

We have a 1 is to 1 ratio of the reactants. Thus, the limiting reactant would be Na2SO4. We use this value for calculating the amount of products in the reaction.

0.0440 mol Na2SO4 ( 1 mol HgSO4 / 1 mol Na2SO4 ) ( 296.65 g / mol ) = 13.06 g HgSO4

mass of excess reactant = (0.0475 - 0.0440) mol ( 399.4912 g / 1 mol ) = 172.180 g Hg(ClO4)2