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""lead chloride" would result if a 21.5*mL21.5*mL volume of "lead nitrate""lead nitrate" at 0.0806*mol*L^-10.0806*mol*L^-1 concentration were mixed with a 18.00*mL18.00*mL volume of "potassium chloride""potassium chloride" at 0.683*mol*L^-10.683*mol*L^-1 concentration?