Respuesta :
Answer: The mass of precipitate (lead (II) iodide) that will form is 13.83 grams
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]
Molarity of NaI solution = 0.300 M
Volume of solution = 0.200 L
Putting values in above equation, we get:
[tex]0.300M=\frac{\text{Moles of NaI}}{0.200L}\\\\\text{Moles of NaI}=(0.300mol/L\times 0.200L)=0.06mol[/tex]
The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:
[tex]Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)[/tex]
By Stoichiometry of the reaction:
2 moles of NaI produces 1 mole of lead (II) iodide
So, 0.06 moles of NaI will produce = [tex]\frac{1}{2}\times 0.06=0.03mol[/tex] of lead (II) iodide
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of lead (II) iodide = 0.03 moles
Molar mass of lead (II) iodide = 461.1 g/mol
Putting values in above equation, we get:
[tex]0.03mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.03mol\times 461.1g/mol)=13.83g[/tex]
Hence, the mass of precipitate (lead (II) iodide) that will form is 13.83 grams
