Question

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?

Answer 1

Under a `2*g` mass................

Explanation:

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......