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.

Answer 2

The mass of the precipitate (i.e PbI₂) obtained from the reaction between aqueous Pb(ClO₃)₂ and aqueous NaI is 9.681 g

How to determine the mole of NaI

• Volume of NaI = 0.2 L
• Molarity of NaI = 0.21 M

• Mole of NaI =?

Mole = Molarity x Volume

Mole of NaI = 0.21 × 0.2

Mole of NaI = 0.042 mole

How to determine the mole of PbI₂ produced

Balanced equation

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

From the balanced equation above,

2 moles of NaI reacted to produce 1 mole of PbI₂

Therefore,

0.042 mole of NaI will react to produce = 0.042 / 2 = 0.021 mole of PbI₂

How to determine the mass of PbI₂

• Mole of PbI₂ = 0.021 mole
• Molar mass of PbI₂ = 207 + (127×2) = 461 g/mol

• Mass of PbI₂ =?

Mass = mole × molar mass

Mass of PbI₂ = 0.021 × 461

Mass of PbI₂ = 9.681 g

Learn more about stoichiometry:

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