Question

If a solution containing 22.90 g of mercury(II) nitrate is allowed to react completely with a solution containing 7.410 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

324.6 g         142 g                                    296.6        170 g

22.90             7.410                                     ??

here limiting reagent is Na2SO4. so product formed according to that.

142 g of Na2SO4 ----------------- 296.6 HgSO4

7.410 g of Na2SO4 ----------------- ??

mass of HgSO4 = 296.6 x 7.410 / 142

                          = 15.48 g

mass of solid precipitate will be formed = 15.48 g

here Hg(NO3)2 is excess reagent. because

we need only 16.94 g of Hg(NO3)2. but we have 22.90 g so

the excess amount is = 22.90 - 16.94 = 5.96 g