Question

If a solution containing 45 g of mercury(II) nitrate is allowed to react completely with a solution containing 14.334 g of sodium sulfate. a)How many grams of solid precipitate will be formed? b)How many grams of the reactant in excess will remain after the reaction?

Answer 1

Hg(NO3)2 (aq) + Na2SO4 (aq) ---------------> HgSO4 (s) + 2 NaNO3 (aq)

324.6 g                 142 g                                   296.6 g             

   45 g                    14.334 g                                 ??

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

142 g Na2SO4   ----------------> 296.6 g HgSO4 (s)

14.334 g Na2SO4 ---------------->   ??

mass of solid precipitate = 14.334 x 296.6 / 142 = 29.94 g

mass of solid precipitate = 29.94 g

here excess reagent is Hg(NO3)2.

Hg(NO3)2 (aq) + Na2SO4 (aq) ---------------> HgSO4 (s) + 2 NaNO3 (aq)

324.6 g                 142 g                                   296.6 g             

   45 g                    14.334 g                   

here we need Hg(NO3)2 is = 324.6 x 14.334 / 142 = 32.77 g

but here we have 45 g . so

excess amount of Hg(NO3)2 = 45 - 32.766 g = 12.234 g

excess amount of Hg(NO3)2 = 12.234 g