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? A) Lead (II) nitrate increases the amount of precipitate. B) The reactant lead (II) nitrate decreases product yield. C) Lead (II) nitrate is the excess reactant in the reaction. D) The lead (II) nitrate is the reaction's limiting reactant.

Answer 1

Answer: D) The lead (II) nitrate is the reaction's limiting reactant.

Step-by-step explanation:

`\text{no of moles of iron(III) chloride}={\text{Molarity}* {\text{Volume in L}}`

`\text{no of moles of iron(III) chloride}={0.100M}* {0.05L}=5* 10^(-3)`

`\text{no of moles of lead nitrate}={\text{Molarity}* {\text{Volume in L}}`

`\text{no of moles of lead nitrate}={\text{0.100M}* {0.05L}=5* 10^(-3)`

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

As can be seen from the balanced chemical equation, 3 moles of lead nitrate react with 2 moles of ferric chloride.

Thus
`5* 10^(-3)` moles of lead nitrate react with
`=(2)/(3)* {5* 10^(-3)}={3.33* 10^(-3)}` of ferric chloride.

`5* 10^(-3)-{3.33* 10^(-3)}=1.67* 10^(-3)`moles of ferric chloride will be left unreacted.

Limiting reagent is the reagent which limits the formation of product. Excess reagent is one which is in excess and thus remains unreacted.

Thus lead nitrate is the limiting reagent and ferric chloride is the excess reagent.