Question

The metathesis reaction between mercury(I) acetate and potassium iodide is shown below

Hg2(C2H302)2(aq)+2KI(aq) arrow 2 Kc2H3O2(aq)+ Hg2I2(s)

how many grams of solid mercury (I) iodide can be produced by the reaction of 120.0 ml of 0.450 M mercury (I) acetate solution with 150.0 ml of 0.800 M potassium iodide solution

Answer 1

volume , V = 120.0 mL
= 0.12 L


number of mol,
n = Molarity * Volume
= 0.45*0.12
= 5.4*10^-2 mol
volume , V = 150.0 mL
= 0.15 L


number of mol,
n = Molarity * Volume
= 0.8*0.15
= 0.12 mol
Balanced chemical equation is:
Hg2(C2H3O2)2 + 2 KI ---> Hg2I2 + 2 KC2H3O2


1 mol of Hg2(C2H3O2)2 reacts with 2 mol of KI
for 0.054 mol of Hg2(C2H3O2)2, 0.108 mol of KI is required
But we have 0.12 mol of KI

so, Hg2(C2H3O2)2 is limiting reagent
we will use Hg2(C2H3O2)2 in further calculation


Molar mass of Hg2I2,
MM = 2*MM(Hg) + 2*MM(I)
= 2*200.6 + 2*126.9
= 655 g/mol

According to balanced equation
mol of Hg2I2 formed = (1/1)* moles of Hg2(C2H3O2)2
= (1/1)*0.054
= 0.054 mol


mass of Hg2I2 = number of mol * molar mass
= 5.4*10^-2*6.55*10^2
= 35.37 g

Answer: 35.4 g