Question

Write the balanced equation for the reaction of aqueous Pb(ClO3)2 with aqueous NaI. Include phases.

chemical equation:

What mass of precipitate will form if 1.50 L of highly concentrated Pb(ClO3)2 is mixed with 0.900 L 0.180 M NaI? Assume the reaction goes to completion.

mass of precipitate:

Answer 1

Answer:-

The balanced chemical equation is written as follows,

Pb(ClO₃)₂(aq) + 2NaI(aq) ---------> PbI₂(s) + 2NaClO₃(aq)

[NaI] = 0.180 m

volume of NaI = 0.900 L

So, Moles of NaI added = 0.180 x 0.900

= 0.162 mol

So, Molar mass of PbI₂ = 207.2 + 2 x 126.9

= 461 g/mol

so, 2 mol NaI = 461 g PbI₂

0.162 mol NaI = 461 x 0.162/2 g PbI₂

= 37.341 g PbI₂