Question

If a solution containing 37.525 g of mercury(II) acetate is allowed to react completely with a solution containing 12.026 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(CH3COO)2_(aq) + Na2Cr2O7_(aq) ---------> HgCr2O7_(s) + 2CH3COONa_(aq)   

according to balanced reaction

261.97 g Na2Cr2O7 reacts with 318.68 g Hg(CH3COO)2

12.026 g Na2Cr2O7 reacts with 12.026 x 318.68 / 261.97 = 14.63 g Hg(CH3COO)2

but we have 37.525 g Hg(CH3COO)2. so Hg(CH3COO)2 is exess reagent

Na2Cr2O7 is limiting reagent .

261.97 g Na2Cr2O7 gives 416.58 g HgCr2O7

12.026 g Na2Cr2O7 gives 12.026 x 416.58 / 261.97 = 19.12 g

mass of precipitate formed = 19.12 g

mass of exess reactant (Hg(CH3COO)2) left = 37.525 - 14.63 = 22.895 g