What mass of #"lead chloride"# would result if a #21.5mL# volume of #"lead nitrate"# at #0.0806molL^-1# concentration were mixed with a #18.00mL# volume of #"potassium chloride"# at #0.683molL^-1# concentration?

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Answer 1 · 2017-06-20T03:08:58

Under a #2*g# mass................

We need (i) a stoichiometric equation.....

#Pb(NO_3)_2(aq) + 2KCl(aq) rarr PbCl_2(s)darr+2KNO_3(aq)#

Lead chloride is one of the few water-insoluble halides. It crashes out of water solution with alacrity.

And (ii) we need equivalent quantities of each reagent......

#"Moles of lead nitrate"=21.5xx10^-3*Lxx0.806*mol*L^-1=0.0173*mol#

#"Moles of KCl"=18.00xx10^-3*Lxx0.683*mol*L^-1=0.0123*mol#.

And thus (clearly) chloride anion is the limiting reagent. Stoichiometry predicts that half an equiv of plumbous chloride will precipitate.

#"Mass of"# #PbCl_2# #=# #(0.0123*mol)/2xx278.10*g*mol^-1#

#=1.71*g#

It is unusual that the chloride ion is in stoichiometric deficiency in that you would think we want to salt out the lead content......

What mass of #"lead chloride"# would result if a #21.5*mL# volume of #"lead nitrate"# at #0.0806*mol*L^-1# concentration were mixed with a #18.00*mL# volume of #"potassium chloride"# at #0.683*mol*L^-1# concentration?