Respuesta :
Answer: The moles of precipitate (lead (II) iodide) produced is [tex]1.57\times 10^{-5}[/tex] moles
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex] .....(1)
- For lead (II) nitrate:
Molarity of lead (II) nitrate solution = 2.70 M
Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol[/tex]
- For NaI:
Molarity of NaI solution = 0.00157 M
Volume of solution = 20.0 mL = 0.020 L
Putting values in equation 1, we get:
[tex]0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol[/tex]
For the given chemical reaction:
[tex]Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)[/tex]
By Stoichiometry of the reaction:
2 moles of NaI reacts with 1 mole of lead (II) nitrate
So, [tex]3.14\times 10^{-5}[/tex] moles of NaI will react with = [tex]\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol[/tex] of lead (II) nitrate
As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.
Thus, NaI is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
2 moles of NaI produces 1 mole of lead (II) iodide
So, [tex]3.14\times 10^{-5}[/tex] moles of NaI will produce = [tex]\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles[/tex] of lead (II) iodide
Hence, the moles of precipitate (lead (II) iodide) produced is [tex]1.57\times 10^{-5}[/tex] moles
