Question

If a solution containing 118.44 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?

How many grams of the reactant in excess will remain after the reaction?

Answer 1

Hg(ClO3)2 + Na2S ==> HgS + 2NaClO3

mols Hg(ClO3)2 = grams/molar mass = ?
mols Na2S = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols Hg(ClO3)2 to mols HgS.
Do the same to convert mols Na2S to mols HgS.
It is likely these two values will not be the same which means one of them must not be right; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Then g ppt = mols HgS (the smaller value) x molar mass HgS = ?

Then to find the amount of the other reagent in excess you must determine the amount of the other reagent used. To do that use the coefficients again, convert mols of the limiting reagent (LR) to mols of the other reagent and multiply that by the molar mass of the other reagent. That will give you the grams of the other reagent used. Subtract from the initial amount to find how much is left.