Question

The balanced equation for the reaction of aqueous $Pb{(Cl{O_3})_2}$ with aqueous $NaI$ is
$Pb{(Cl{O_3})_2}(aq) + 2NaI(aq) \to Pb{I_2}(s) + 2NaCl{O_3}(aq)$
What mass of precipitate will form if $1.50L$ of concentrated $Pb{(Cl{O_3})_2}$ is mixed with $0.800L$ of $0.290M\;NaI?$ Assume the reaction goes to completion.

Answer 1

Hint: Lead chlorate will react with sodium iodide to form lead iodide which is an insoluble salt and hence it will precipitate out of the solution and in the given question two values of molarity of $NaI$ is given then we might need to find the number of moles.

Complete step by step solution: Let us first write the chemical equation for the reaction where lead chlorate reacts with sodium iodide to form lead iodide which is an insoluble salt and hence it precipitates out of the solution.
$Pb{(Cl{O_3})_2}(aq) + 2NaI(aq) \to Pb{I_2}(s) + 2NaCl{O_3}(aq)$
We can see that $2$ moles of sodium iodide reacts with $1$ mole of lead chlorate so the reaction will always consume twice as many moles of sodium iodide so it will be considered a limiting reagent and hence will be completely neutralised by the reaction. Therefore we can use the formula for molarity to calculate the number of moles for sodium iodide, we are given with volume of sodium iodide as $0.800L$ and molarity $0.290M$ :
$molarity = \dfrac{{moles\;of\;solute}}{{volume\;of\;solvent}}$
$
  moles = molarity \times volume \\
 \Rightarrow moles = 0.29 \times 0.8 \\
  \Rightarrow moles = 0.232\;moles \\
 $
We can use these moles to calculate the number of moles of lead iodide, as two moles of sodium iodide reacts with one mole of lead iodide so,
$
  {n_{Pb{I_2}}} = \dfrac{{{n_{NaI}}}}{2} \\
  \Rightarrow {n_{Pb{I_2}}} = \dfrac{{0.232}}{2} \\
 \Rightarrow {n_{Pb{I_2}}} = 0.116\;moles \\
 $
Now with the help of these moles, we can easily calculate the mass of the precipitate of lead iodide using the formula of moles:
$
  mole = \dfrac{{mass}}{{molecular\;mass}} \\
\Rightarrow mass = moles \times molecular\;mass \\
\Rightarrow mass = 0.116 \times 461 \\
 \Rightarrow mass = 53.476g \\
 $
Hence, the correct answer is $53.476g$.

Note: Limiting reagents are those reagents which are present in lesser quantity and are completely consumed in the reaction. It can also decide the amount of product formed as well as the amount of the reactants consumed.