Fa solution containing 85.14 g of mercury(ii) nitrate is allowed to react completely with a solution containing 14.334 g of sodium sulfide, how many grams of solid precipitate will be formed? mass: 29.69 g how many grams of the reactant in excess will remain after the reaction? mass: g assuming complete precipitation, how many moles of each ion remain in solution? if an ion is no longer in solution, enter a zero (0) for the number of moles. hg2+ : 0 mol no–3 : 0 mol na+ : 0 mol s2− : 0 mol

Hg(NO₃)₂(ac) + Na₂S(ac) → HgS(s) + 2Na⁺ + 2NO₃⁻

Now we calculate the moles of each reagent -using the molecular weights-, in order to determine the limiting reactant:

Because the stoichiometric ratio between the reactants is 1:1, we compare the number of moles of each one upfront.

moles Hg(NO₃)₂ > moles Na₂S

Thus Na₂S is the limiting reagent.

So in order to find the mass of solid precipitate, we must calculate it using the moles of Na₂S:

The mass of the solid precipitate is 42.760 g.

Mass of Hg(NO₃)₂ remaining =

The mass of the remaning reactant in excess is 25.49 g.

Hg⁺²: 0 mol

NO₃⁻:

Na⁺:

S²⁻: 0 mol

hypostyle hall;

wanswer:

explanation: