If a solution containing 102.78 gof mercury(II) nitrate is allowed to react completely with a solution containing 17.796 g of sodium sulfide, how many grams of solid precipitate will form?

53 g of HgS will be formed

Explanation:

We determine the main reaction:

Hg(NO₃)₂(aq) + Na₂S (aq) → HgS (s)↓ + 2NaNO₃(aq)

We determine the limiting reactant (we need to convert the mass to moles, firstly)

102.78 g / 324.59 g/mol = 0.316 moles of nitrate

17.796 g / 78.06 = 0.228 moles of sulfide

Ratio is 1:1, so the limiting reactant is the Na₂S. For 0.316 moles of nitrate, I need the same amount of sulfide and I only have 0.228 moles.

Ratio is again 1:1. For 0.228 moles of sodium sulfide I can produce the same amount of mercury(II) sulfide

We convert the moles to mass → 232.65 g/mol . 0.228 mol = 53 g of HgS will be formed

it is stable as it is in group 0 which have full outer shells of electrons