Question

In this reaction, what roll does the lead (II) nitrate play when 50.0 mL of 0.100M iron (III) chloride are mixed with 50.0 mL of 0.100M lead (II) nitrate?

Answer 1

Answer: The roll of lead (II) nitrate is that it is a limiting reactant of the reaction.

Solution : Given,

Molarity of $FeCl_3$ = 0.1 M

Volume of $FeCl_3$ = 50.0 ml = 0.05 L     (1 L = 1000 ml)

Molarity of $Pb(NO_3)_2$ = 0.1 M

Volume of $Pb(NO_3)_2$ = 50.0 ml = 0.05 L

First we have to calculate the moles of $FeCl_3$ and $Pb(NO_3)_2$.

$\text{ Moles of }FeCl_3=\text{ Molarity}\times \text{ Volume in L}$

$\text{ Moles of }FeCl_3=(0.100M)\times (0.05L)=0.005moles$

$\text{ Moles of }Pb(NO_3)_2=\text{ Molarity}\times \text{ Volume in L}$

$\text{ Moles of }Pb(NO_3)_2=(0.100M)\times (0.05L)=0.005moles$

The balanced chemical reaction is,

$2FeCl_3+3Pb(NO_3)_2\rightarrow 2Fe(NO_3)_3+3PbCl_2$

From the balanced chemical equation, we conclude that

3 moles of lead nitrate react with 2 moles of ferric chloride.

Thus 0.005 moles of lead nitrate react with $=\frac{2}{3}\times 0.005=0.0033$ moles of ferric chloride.

Moles of ferric chloride will be left unreacted = 0.005 - 0.0033 =0.0017 moles

Limiting reagent is the reagent in the reaction which limits the formation of product.

Excess reagent is the reagent in the reaction which is in excess and thus remains unreacted.

Therefore, in the given reaction, lead nitrate is the limiting reagent and ferric chloride is the excess reagent.

Answer 2

Answer: iron (III) chloride is the excess reactant in the reaction.

Explanation:

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