Question

If a solution containing 16.38 g of mercury (II) nitrate is allowed to react completely with a solution containing 51.02 g of sodium sulfate, 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(NO3)2 + Na2SO4 -----------------------> HgSO4 + 2 NaNO3

324.6g              142.0g                                   296.6g

16.38 g               51.02 g

324.6 g Hg(NO3)2 -------------------> 142 g Na2SO4 -

16.38 g Hg(NO3)2 -------------------> 142 x 16.38 / 324.6 = 7.17 g

so Na2SO4 is excess reagent

mass of precipitate is formed = 296.6 x 16.38 / 324.6

                                               = 15.97 g

mass of precipitate is formed = 15.97 g

excess reagent remains = 51.02 - 7.17

                                      = 43.85 g