Question

If a solution containing 76.732 g of mercury(II) nitrate is allowed to react completely with a solution containing 12.026 g of sodium sulfide, how many grams of solid precipitate will be formed? How many grams of the reactant in excess will remain after the reaction?

Answer 1

Balanced equation:
Na₂S(aq) + Hg(NO₃)₂(aq) = HgS(s) + 2 NaNO₃(aq)
Reaction type: double replacement

precipitate = HgS (because all nitrates are soluble)

76.732 g of mercury(II) nitrate = 76.732 /324.599 = 0.236389 Moles

12.026 g of sodium sulfide = 12.026 / 78.044 = 0.154091 Moles

Limiting reagent is sodium sulfide which can produce 0.15409 moles of HgS

0.15409 moles of HgS = 0.15409 x 232.655 = 35.85 gm

Hence 35.85 gm of  solid precipitate will be formed