Answer: The mass of lead iodide produced is 9.22 grams
Explanation:
To calculate the molarity of solution, 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 = 0.200 M
Volume of solution = 0.200 L
Putting values in above equation, we get:
[tex]0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles[/tex]
The chemical equation for the reaction of NaI and lead chlorate 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 reacts produces 1 mole of lead iodide
So, 0.04 moles of NaI will react with = [tex]\frac{1}{2}\times 0.04=0.02mol[/tex] of lead iodide
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of lead iodide = 461 g/mol
Moles of lead iodide= 0.02 moles
Putting values in above equation, we get:
[tex]0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g[/tex]
Hence, the mass of lead iodide produced is 9.22 grams
