What mass of "lead chloride"lead chloride would result if a 21.5*mL21.5mL volume of "lead nitrate"lead nitrate at 0.0806*mol*L^-10.0806molL1 concentration were mixed with a 18.00*mL18.00mL volume of "potassium chloride"potassium chloride at 0.683*mol*L^-10.683molL1 concentration?

1 Answer
Jun 20, 2017

Under a 2*g2g mass................

Explanation:

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

Pb(NO_3)_2(aq) + 2KCl(aq) rarr PbCl_2(s)darr+2KNO_3(aq)Pb(NO3)2(aq)+2KCl(aq)PbCl2(s)+2KNO3(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*molMoles of lead nitrate=21.5×103L×0.806molL1=0.0173mol

"Moles of KCl"=18.00xx10^-3*Lxx0.683*mol*L^-1=0.0123*molMoles of KCl=18.00×103L×0.683molL1=0.0123mol.

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

"Mass of"Mass of PbCl_2PbCl2 == (0.0123*mol)/2xx278.10*g*mol^-10.0123mol2×278.10gmol1

=1.71*g=1.71g

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