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Question: The balanced equation for the reaction of aqueous \(Pb{(Cl{O_3})_2}\) with aqueous \(NaI\) is \(Pb...

The balanced equation for the reaction of aqueous Pb(ClO3)2Pb{(Cl{O_3})_2} with aqueous NaINaI is
Pb(ClO3)2(aq)+2NaI(aq)→PbI2(s)+2NaClO3(aq)Pb{(Cl{O_3})_2}(aq) + 2NaI(aq) \to Pb{I_2}(s) + 2NaCl{O_3}(aq)
What mass of precipitate will form if 1.50L1.50L of concentrated Pb(ClO3)2Pb{(Cl{O_3})_2} is mixed with 0.800L0.800L of 0.290M  NaI?0.290M\;NaI? Assume the reaction goes to completion.

Explanation

Solution

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 NaINaI 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(ClO3)2(aq)+2NaI(aq)→PbI2(s)+2NaClO3(aq)Pb{(Cl{O_3})_2}(aq) + 2NaI(aq) \to Pb{I_2}(s) + 2NaCl{O_3}(aq)
We can see that 22 moles of sodium iodide reacts with 11 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 0.800L0.800L and molarity 0.290M0.290M :
molarity=moles  of  solutevolume  of  solventmolarity = \dfrac{{moles\;of\;solute}}{{volume\;of\;solvent}}
moles=molarity×volume ⇒moles=0.29×0.8 ⇒moles=0.232  moles  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,
nPbI2=nNaI2 ⇒nPbI2=0.2322 ⇒nPbI2=0.116  moles  {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=massmolecular  mass ⇒mass=moles×molecular  mass ⇒mass=0.116×461 ⇒mass=53.476g  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 53.476g53.476g.

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.