Question

If a solution containing 39.544 g of mercury(II) perchlorate is allowed to react completely with a solution containing 13.180 g of sodium dichromate, 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 + Na2Cr2O7 -------> HgCr2O7 + 2NaClO4

no of moles of Hg(ClO4)2   = W/G.M.Wt    = 39.544/399.5   = 0.0989 moles

no of moles of Na2Cr2O7   = W/G.M.Wt   = 13.18/262         = 0.0503 moles

1 mole of Hg(ClO4)2 react with 1 moles of Na2Cr2O7

0.0503 mole of Hg(ClO4)2 react with 0.0503 moles of Na2Cr2O7

limiting reagent is Na2Cr2O7

Hg(ClO4)2 + Na2Cr2O7 -------> HgCr2O7 + 2NaClO4

1 moles of Na2Cr2O7 react with Hg(ClO4)2 to gives1 mole HgCr2O7

0.0503 moles of Na2Cr2O7 react with Hg(ClO4)2 to gives 0.0503 moles of HgCr2O7

mass of HgCr2O7   = no of moles * gram molar mass

                                = 0.0503*416.6   = 20.95 g

remaining no of moles of excess reactant Hg(ClO4)2 = 0.0989-0.0503   = 0.0486 moles

mass of excess reactant   = 0.0486*399.5   = 19.42g