Chemistry homework

Question

If a solution containing 107.34 g of mercury(II) nitrate is allowed to react completely with a solution containing 15.488 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

First, write a correctly balanced equation for the reaction taking place:

Hg(NO₃)₂(aq) + Na₂S(aq) ==> HgS(s) + 2NaNO₃(aq) ... balanced equation

Next, we find which reactant is limiting:

moles Hg(NO₃)₂ present = 107.34 g x 1 mol/324.60 g = 0.33068 moles

moles Na₂S present = 15.488 g x 1 mol/78.045 g = 0.19845 moles

Since they react in a 1:1 mole ratio, clearly the Na2S is limiting and will dictate how much HgS will form.

Grams HgS precipitate formed = 0.19845 mol Na₂S x 1 mol HgS/mol Na₂S x 232.66 g/mol = 46.171 g HgS

How much excess reactant is left over? The excess reactant is Hg(NO₃)₂.

moles Hg(NO₃)₂ used up = moles of Na₂S used up = 0.19845 moles

moles Hg(NO₃)₂ left over = 0.33068 moles - 0.19845 moles = 0.13223 moles Hg(NO₃)₂ left over

grams Hg(NO₃)₂ left over = 0.13223 moles x 324.60 g/mol = 42.922 g Hg(NO₃)₂ left over