1. If a solution containing 48.99 g of mercury (II) perchlorate is allowed to react completely with a solution containing 8.564 g of sodium sulfide, how many grams of solid precipitate will be formed?

2. How many grams of the reactant in excess will remain after the reaction?

1. Chemical eqn:

Hg (ClO4) 2 + Na2S - > HgS + 2NaClO4

mercury perchlorate has a molar mass of 399.5g/mol (1d. p) and for Na2S it is 78.0g/mol (1d. p.)

no. of moles of mercury perchlorate = 48.99:399.5 = 0.12263mol (5s. f.)

no. of moles of Na2S = 8.564:78.0 = 0.10979mol (5s. f.)

so no. of moles of Hg (ClO4) 2 / no. of moles of Na2S = 1/1 according to the eqn

so no. of moles of Hg (ClO4) 2 needed if all Na2S is used up is 1/1*0.10979 = 0.10979 mols

since no. of moles of mercury perchlorate needed < no. of moles of it provided, it is in excess and Na2S is the limiting factor.

since HgS is a solid and NaClO4 is aqueous, solid ppt formed will only be HgS.

no. of moles of HgS / no. of moles of NaClO4 = 1/1

no. of moles of HgS produced = 0.10979mols

molar mass of HgS = 232.7g/mol 1d. p.

grams of solid produced = 232.7*0.10979 = 25.5g (3s. f.)

2. reactant in excess is Hg (ClO4) 2,

no. of excess moles = 0.12263-0.10979 = 0.01284mols

grams of excess reactant = 0.01284*399.5 = 5.13g (3s. f.)