Question

What mass of precipitate will form if excess Pb(ClO3)2 solution is mixed with 0.300 L of 0.130 M Nal? Assume the reaction goes to completion.

Answer 1

Balanced reaction:

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

Assuming PbI2 is insoluble in water, (Ksp is not given hence we may not calculate molar solubility and so we need to assume PbI2 is insoluble).

number of mole of NaI used = 0.300 L * 0.130 mole/L = 0.039 moles

Here NaI is limiting reagent and Pb(ClO3)2 is in excess.

For every 2 moles of NaI we get 1 mole of PbI2.

So for 0.039 moles of NaI we should get = 0.039 / 2 = 0.0195 moles of PbI2

0.0195 moles of PbI2 = 0.0195moles*461.01g/mol = 8.99 g of PbI2

Thus 8.99 g of PbI2 will be precipitated.