Question

If a solution containing 35.10 g of mercury(II) chlorate is allowed to react completely with a solution containing 6.256 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(CO3)2(aq) + Na2S(aq) ------> HgS(s) + 2NaClo3(aq)

no of moles of mercury(II) chlorate = weight of substance/ Gram molar mass

                                                       = 35.1/367.5 = 0.095moles

no of moles of Na2S   = 6.256/78 = 0.08 moles

1 mole of Hg(CLO3)2 react with 1 moles of Na2S

So limiting reagent is Na2S

1 mole of Na2S react with mercury(II) chlorate to form 1 mole of HgS

78gm of Na2S react with mercury(II) chlorate to form 232.6gm of HgS

6.256gm of Na2S react with mercury(II) chlorate to form = 232.6*6.256/78 = 18.65 gm of HgS

Remaining no of moles of excess reagent = 0.095-0.08 = 0.015moles

mass of mercury(II) chlorate = no of moles*gram molar mass

                                            = 0.015*367.5 = 5.5125gm