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.
