Question

If a solution containing 42.33 g of mercury(II) chlorate is allowed to react completely with a solution containing 6.256 g of sodium sulfide, how many grams of solid precipitate will form? How many grams of the reactant in excess will remain after the reaction?

Answer 1

Ans :

The reaction is given as :

Hg(ClO4)2 (aq) + Na2S (aq) = HgS (s) + 2NaClO4 (aq)

Number of moles of mercury (II) chlorate = 42.33 / 399.49 = 0.106 mol

Number of moles of sodium sulfide = 6.256 / 78.044 = 0.08 mol

Each mol of mercury (II) chlorate needs one mol of sodium sulfide , so sodium sulfide is the limiting reagent.

0.08 mol of sodium sulfide will make 0.08 mole of H2S solid precipotate

Mass of solid precipitate = 0.08 x 232.655 = 18.65 grams

0.08 mol of sulfide will utilise 0.08 moles of mercury (II) chlorate

So mass utilised = 0.08 x 399.49 = 32.02

Excess mass left = 42.33 g - 32.02 g

= 10.31 g