Question

A solution contains Ag+ and Hg2+ ions. The addition of 0.100 L of 1.44 M NaI solution is just enough to precipitate all the ions as AgI and HgI2. The total mass of the precipitate is 33.3 g. Find the mass of AgI in the precipitate.

(I already tried 11.3g and it didn't work)

Answer 1

Moles of NaI added = 0.1*1.44= 0.144 moles

Molecular weight o NaI= 150

Mass of NaI added = 0.144*150= 21.6 gms

Ag+ + NaI---> AgI + Na (1)

similarly Hg⁺² + 2NaI----> HgI2+2Na+ (2)

let x =moles of AgI for the first reaction 0.144-x= moles of NaI used for HgI2 reactino

1 mole of NaI produces 1 mole of AgI ( from reaction (1)

x moles of NaI produces x moles of AgI= x*235 gms ( Molecular weight of AgI= 235)= 235x

(0.144-x ) moles of NaI produces (0.144-x)/2 moles of HgI2 ( from reaction (2))= (0.144-x)*454/2} gms=227*(0.144-x)

Total mass =33.3 gms

235x+ 227*(0.144 -x)= 33.33

235x-227x+227*0.144x= 33.3

8x= 33.3-32.69

8x= 0.61 x =0.61/8=0.07625 moles of NaI

AgI produced =0.07625*235=17.91 gms