Question

What is the theoretical yield (in g of precipitate) when 16.3 mL of a 0.628 M solution of iron(III) chloride is combined with 16.5 mL of a 0.642 M solution of lead(II) nitrate?

Answer 1

Answer: 2.78 grams

Step-by-step explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

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

`\text{no of moles}of FeCl_3={0.628M}* {0.0163 L}=0.010moles`

`\text{no of moles}of Pb(NO_3)_2={0.642M}* 0.0165L}=0.010moles`

`2FeCl_3(aq)+3Pb(NO_3)_2(aq0\rightarrow 3PbCl_2(s)+2Fe(NO_3)_3(aq)`

According to stoichiometry:

3 moles of
`Pb(NO_3)_2` reacts with 2 moles of
`FeCl_3`

Thus 0.010 moles of
`Pb(NO_3)_2` will give=
`(2)/(3)* 0.010=0.006moles` of
`FeCl_3`

`Pb(NO_3)_2` is a limiting reagent as it limits the formation of products and
`FeCl_3` is an excess reagent.

As 3 moles of
`Pb(NO_3)_2` give = 3 moles of
`PbCl_2`

Thus 0.010 moles of
`Pb(NO_3)_2` will give=
`(3)/(3)* 0.010=0.010moles` of
`PbCl_2`

mass of
`PbCl_2=moles* {\text {molar mass}}=0.010* 278 = 2.78grams`

Thus the theoretical yield (in g of precipitate) when 16.3 mL of a 0.628 M solution of iron(III) chloride is combined with 16.5 mL of a 0.642 M solution of lead(II) nitrate is 2.78 grams