Question

If a solution containing 81.921 g of mercury(II) perchlorate is allowed to react completely with a solution containing 13.180 g of sodium sulfide, how many grams of solid precipitate will form? precipitate: How many grams of the reactant in excess will remain after the reaction? excess reactant:

Answer 1

We know that, mercury (II) perchlorate reacts with sodium sulphide in equimolar quantity, as per the following equation,-

Hg(ClO₄)₂   + Na₂S = HgS + 2NaClO₄

1 mole Hg(ClO₄)₂ = 399.49 gm.

1 mole Na₂S = 78.04 gm.

1 mole HgS = 232.66 gm.

2 mole NaClO₄ = 2 x 122.44 gm = 244.88 gm

So, as per the above equation, 78.04 gm. Na₂S reacts with 399.49 gm.  Hg(ClO₄)_2.

Therefore, 13.18 gm Na₂S reacts with {(399.49 / 78.04) x 13.18} gm. Hg(ClO₄)₂ = 67.468 gm Hg(ClO₄)₂

So, Na2S will be fully consumed and there will be excess Hg(ClO₄)₂ and the amount of excess Hg(ClO₄)₂ = (81.921 - 67.468) gm = 14.453 gm

Now, 78.04 gm. Na₂S produces 232.66 gm. HgS

So, 13.18 gm Na₂S produces {(232.66 / 78.18) x 13.18} gm. HgS = 39.223 gm HgS, which will be precipitated out from the solution.

So, amount of solid precipitate = 39.223 gm HgS

and amount of excess reactant in solution = 14.453 gm Hg(ClO₄)₂