Question

How many grams of PbCl2 can be produced from the addition of 42 mL of a 0.86 M Pb(NO3)2 solution to 142 mL of 0.65 M NaCl?

Answer 1

At first, we have to balance the equation between NaCl and Pb(NO₃)₂ to find the stoichiometry and which compound is the one that limits the reaction:

$2NaCl_{(ac)}+Pb(NO_3)_{2(ac)} \to 2NaNO_{3(ac)}+PbCl_{2(s)}$

Then, we calculate how much volume of NaCl solution has to react with the Pb(NO₃)₂:

$42\,mL\,sol.Pb(NO_3)_{2(ac)}[\frac{0.86\,mol\,Pb(NO_3)_{2}}{10^3\,mL\,sol.\,Pb(NO_3)_{2(ac)}}*\frac{2\,mol\,NaCl}{1\,mol\,Pb(NO_3)_{2}}*\frac{10^3\,mL\,sol.\,NaCl}{0.65\,mol\,NaCl}]=...$

$...= 111.14\,mL\,sol.\,NaCl$

This means that Pb(NO₃)₂ is the limiting reagent (because we have more than the 111.14 mL of NaCl solution at the start), thus we use that reagent volume to calculate the amount of PbCl₂ produced:

$42\,mL\,sol.Pb(NO_3)_{2(ac)}[\frac{0.86\,mol\,Pb(NO_3)_{2}}{10^3\,mL\,sol.\,Pb(NO_3)_{2(ac)}}*\frac{1\,mol\,PbCl_2}{1\,mol\,Pb(NO_3)_{2}}*\frac{278.09\,g\,PbCl_2}{1\,mol\,PbCl_2}]=...$

$...=10.045\,g\,PbCl_2$

In conclusion, 10.045 g of PbCl₂ were produced on this reaction where Pb(NO₃)₂ was the limiting reagent.

Reference:

• Chang, R. (2010). Chemistry. McGraw-Hill, New York.