The balanced equation for the reaction of aqueous pb(clo3)2 with aqueous nai is pb(clo3)2(aq)+2nai(aq)⟶pbi2(s)+2nac lo3(aq) what mass of precipitate will form if 1.50 l of concentrated pb(clo3)2 is mixed with 0.200 l of 0.210 m nai? assume the reaction goes to completion. mass of precipitate:

Answer : The mass of precipitate produced will be, 9.681 grams.

Explanation : Given,

Molarity of NaI = 0.210 M

Volume of solution = 0.2 L

Molar mass of = 461.01 g/mole

First we have to calculate the moles of .

Now we have to calculate the moles of .

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 2 moles of react to give 1 mole of

So, 0.042 moles of react to give moles of

Now we have to calculate the mass of .

Therefore, the mass of precipitate produced will be, 9.681 grams.

a electronegativity decreases beacuse the size of nucleus is bigger.

c) should be the answer.