Question

Upon mixing 0.050 L of 0.010 M aqueous solution of mercury (II) nitrate with 0.020 L of 0.10 M aqueous solution of sodium sulfide, a precipitate of mercury (II) sulfide forms.

A) Write a balanced molecular equation of this reaction.

B) Determine the theoretical yield of mercury (II) sulfide formed from the limiting reagent.

C) Calculate the % yield of product when 0.082 grams of mercury (II) sulfide is isolated.

Answer 1

A)

Na2S(aq) + Hg(NO3)2(aq) --> HgS(s) + 2NaNO3(aq) a

sodium sulfide + mercury nitrate ---> mercuric sulfide + sodium nitrate

B)

Given Volume of Mercury nitrate V1 = 0.050 L

Molarity of Mercury nitrate M1 = 0.010 M

No. of moles of Mercury nitrate n1 = V1 * M1 = 0.050 L * 0.010 M = 5 * 10^-4 moles

similarly for sodium sulfide V2 = 0.020 L

M2 = 0.010 M

n2 = V2 * M2 = 0.02 L * 0.1 M = 20 * 10^-4 moles

1 mole of sodium sulfide requires 1 mole of mercury nitrate

so mercury nitrate is limiting reactant as it is present in less quantity and according to stoichometry

theoretical yield will be based on limiting reactant

1 mole of mercury nitrate yields 1 mole of HgS

5 * 10^-4 moles of mercury nitrate yields 5 * 10^-4 mole of HgS

Theoretical yield = 5 * 10^-4 moles of HgS

C) Given mass of HgS = 0.082 g

Molar mass of HgS 232.66 g/mol

No. of moles = Mass / Molar mass = 0.082 g / 232.66 g/mol = 3.52 * 10^-4 moles

so actual yield = 3.52 * 10^-4 moles

% yield = (actual yield / theoretical yield) * 100 % = (3.52 * 10^-4 moles / 5 * 10^-4 moles) * 100 % = 70.49 %

% yield = 70.49 %