Question

If a solution containing 84.670 g of mercury(II) nitrate is allowed to react completely with a solution containing 12.026 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(NO3)2 +   Na2S ----------------> HgS (s) + 2 NaNO3

324.6g            78.04g                       232.65g

84.670g            12.026g

here limiting reagent is Na2S . so the product based on that

mass of precipitate HgS formed = 12.026 x 232.65 / 78.04

                                                  = 35.85 g

mass of precipitate HgS formed    = 35.85 g

mercury(II) nitrate consumed = 50.02

mercury(II) nitrate excess will remain after the reaction = 84.670 - 50.02

                                                                                        = 34.649 g