Question #201015

**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?**

Expert's answer

At first, we have to balance the equation between NaCl and Pb(NO_{3})_{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(NO_{3})_{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(NO_{3})_{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 PbCl_{2} 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 PbCl**_{2}** were produced on this reaction where Pb(NO**_{3}**)**_{2}** was the limiting reagent.**

Reference:

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

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