Answer to Question #139038 in General Chemistry for Lauren Johnson

Question

275 mL of 3.0 M Pb(NO3)2 is mixed with 625 mL of 2.5 M NaCl, what is the theoretical yield (in grams) of PbCl2 in this reaction?

Answer 1

1000ml contains 3 moles of Pb(NO3)2

So,275 ml contains 3*275/1000 =0.825 moles of Pb(NO3)2.

Similarly 625 ml contains 2.5*625/1000

=1.56 moles of Nacl.

The reaction is

Pb(NO3)2 +2Nacl= Pbcl2 +Na(NO3)2

So limiting reagent is Pb(NO3)2

From the reaction,

1 mole Pb(NO3)2 react and 1 mole Pbcl2 produced.

0.825 moles Pb(NO3)2 react and 0.825 moles of Pbcl2 produced.

0.825 moles =(0.825*278.1) g

=229.43 g of Pbcl2.