Question

If a solution containing 78.296 g of mercury(II) acetate 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(CH3COO)2 (aq) + Na2S(aq) -------------> HgS(s) + 2CH3COONa(aq)

no of moles of mercury(II) acetate   = W/G.M.Wt

                                                         = 78.296/318.7   = 0.2457 moles

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

                                                           = 13.18/78   = 0.169moles

1 mole of Hg(CH3COO)2   react with 1 mole Na2S

0.2457 moles of Hg(CH3COO)2 react with 0.2457 moles of Na2S

Na2S is limiting reactant

1 mole of Na2S react with excss of Hg(CH3COO)2 to gives 1 moles of HgS

0.169 moles of Na2S react with excess of Hg(CH3COO)2 to gives 0.169 moles of HgS

mass of HgS   = no of moles * gram molar mass

                      = 0.169*232.66   = 39.32g of HgS

The mass of solid precipitate of HgS = 39.32g

Hg(CH3COO)2 (aq) + Na2S(aq) -------------> HgS(s) + 2CH3COONa(aq)

1 mole of Na2S react with 1 moles of Hg(CH3COO)2

0.169 moles of Na2S react with 0.169 moles of Hg(CH3COO)2 is required

Hg(CH3COO)2 is excess reactant

The no of moles of excess reagent remains after complete the reaction = 0.2457-0.169 = 0.0767moles

The amount of exces reactant remains after complete the reaction = no of moles * gram molar mass

                                                                                                           = 0.0767*318.7 = 24.45g

The mass reactant in excess will remain after the reaction = 24.45g