question archive the balanced equation for the reaction of aquaeous Pb(ClO3)2 with aqueous NaI is shown below: Pb(ClO3)2(aq)+2NaI(aq) ---------> PbI2(s) + 2NaClO3(aq) what maass of precipitate will form if 1

the balanced equation for the reaction of aquaeous Pb(ClO3)2 with aqueous NaI is shown below: Pb(ClO3)2(aq)+2NaI(aq) ---------> PbI2(s) + 2NaClO3(aq) what maass of precipitate will form if 1

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the balanced equation for the reaction of aquaeous Pb(ClO3)2 with aqueous NaI is shown below:

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

what maass of precipitate will form if 1.50L of concentrated Pb(ClO3)2 is mixed with .300L of .220 M NaI?

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Answer:

moles NaI = V x M = (0.300 L)(0.220 M) = 0.0660 mol NaI

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

moles PbI2 formed = (0.0660 mol NaI)(1 mole PbI2)/(2 moles NaI) = 0.0330 mol PbI2 formed.

mass PbI2 = (0.0330 mol)(461.01 g)/(1 mole) = 15.2 g PbI2

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