Question

If a solution containing 84.096 g of mercury(II) perchlorate is allowed to react completely with a solution containing 13.180 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(ClO4)2 (aq) + Na2S(aq) -------------> HgS(s) + 2NaClO4(aq)

no of moles of Hg(ClO4)2    =   W/G.M.Wt

                                             = 84.096/399.5   =0.21moles

no of moles of Na2S          = W/G.M.Wt

                                             = 13.18/78   = 0.17moles

1 mole of Hg(ClO4)2 react with 1 moles of Na2S

0.21 moles of Hg(ClO4)2 react with 0.21 moles of Na2S

Na2S is limiting reagent

1 mole of Na2S react with Hg(ClO4)2 to gives 1 moles of HgS

0.17 moles of Na2S react with Hg(ClO4)2 to gives = 0.17 moles of HgS

mass of HgS = no of moles * gram molar mass

                      = 0.17*232.65   = 39.55g >>>>answer

0.17 mole of Na2S react with 0.17 moles of Hg(ClO4)2

    Hg(ClO4)2 is excess reagent

no of moles of excess reactant left after complete of the reaction = 0.21-0.17   = 0.04 moles

mass of excess reagent = 0.04*399.5   = 15.98g