<u>Answer:</u> The mass of lead iodide produced is 9.22 grams
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:
![\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%7D%7B%5Ctext%7BVolume%20of%20solution%20%28in%20L%29%7D%7D)
Molarity of NaI = 0.200 M
Volume of solution = 0.200 L
Putting values in above equation, we get:
![0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles](https://tex.z-dn.net/?f=0.200M%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20NaI%7D%7D%7B0.200%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20NaI%7D%3D%280.200mol%2FL%5Ctimes%200.200L%29%3D0.04moles)
The chemical equation for the reaction of NaI and lead chlorate follows:
![Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)](https://tex.z-dn.net/?f=Pb%28ClO_3%29_2%28aq.%29%2B2NaI%28aq.%29%5Crightarrow%20PbI_2%28s%29%2B2NaClO_3%28aq.%29)
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 =
of lead iodide
To calculate the number of moles, we use the equation:
Molar mass of lead iodide = 461 g/mol
Moles of lead iodide= 0.02 moles
Putting values in above equation, we get:
![0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g](https://tex.z-dn.net/?f=0.02mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20lead%20iodide%7D%7D%7B461g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20lead%20iodide%7D%3D%280.02mol%5Ctimes%20461g%2Fmol%29%3D9.22g)
Hence, the mass of lead iodide produced is 9.22 grams