# Answer to Question #201015 in General Chemistry for Coen Leonard

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?

1
2021-05-31T05:57:54-0400

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.

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