Question

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

What mass of precipitate will form if 1.50 L of highly concentrated Pb(ClO3)2 is mixed with 0.700 L of 0.110 M NaI? Assume the reaction goes to completion.

Answer 1

The balanced equation is

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

Number of moles of NaI = molarity * volume of solution in L

Number of moles of NaI = 0.110 * 0.700 = 0.0770 mol

From the balanced equation we can say that

2 mole of NaI produces 1 mole of PbI2 so

0.0770 mole of NaI will produce

= 0.0770 mole of NaI *(1 mole of PbI2 / 2 mole of NaI)

= 0.0385 mole of PbI2

mass of 1 mole of PbI2 = 461.01 g so

the mass of 0.0385 mole of PbI2 = 17.7 g

Therefore, the mass of PbI2 produced would be 17.7 g