Question

The balanced equation for the reaction of aqueous Pb(ClO3)2 with aqueous NaI is Pb(ClO3)2(aq)+2NaI(aq)⟶PbI2(s)+2NaClO3(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 1

Answer : The mass of $PbI_2$ precipitate produced will be, 9.681 grams.

Explanation : Given,

Molarity of NaI = 0.210 M

Volume of solution = 0.2 L

Molar mass of $PbI_2$ = 461.01 g/mole

First we have to calculate the moles of $NaI$.

$\text{Moles of }NaI=\text{Molarity of }NaI\times \text{Volume of solution}=0.210M\times 0.2L=0.042moles$

Now we have to calculate the moles of $PbI_2$.

The balanced chemical reaction is,

$Pb(ClO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaClO_3(aq)$

From the balanced reaction we conclude that

As, 2 moles of $NaI$ react to give 1 mole of $PbI_2$

So, 0.042 moles of $NaI$ react to give $\frac{0.042}{2}=0.021$ moles of $PbI_2$

Now we have to calculate the mass of $PbI_2$.

$\text{Mass of }PbI_2=\text{Moles of }PbI_2\times \text{Molar mass of }PbI_2$

$\text{Mass of }PbI_2=(0.021mole)\times (461.01g/mole)=9.681g$

Therefore, the mass of $PbI_2$ precipitate produced will be, 9.681 grams.