# 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?

Jun 20, 2017

Under a $2 \cdot g$ mass................

#### Explanation:

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

$P b {\left(N {O}_{3}\right)}_{2} \left(a q\right) + 2 K C l \left(a q\right) \rightarrow P b C {l}_{2} \left(s\right) \downarrow + 2 K N {O}_{3} \left(a q\right)$

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......

$\text{Moles of lead nitrate} = 21.5 \times {10}^{-} 3 \cdot L \times 0.806 \cdot m o l \cdot {L}^{-} 1 = 0.0173 \cdot m o l$

$\text{Moles of KCl} = 18.00 \times {10}^{-} 3 \cdot L \times 0.683 \cdot m o l \cdot {L}^{-} 1 = 0.0123 \cdot m o l$.

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

$\text{Mass of}$ $P b C {l}_{2}$ $=$ $\frac{0.0123 \cdot m o l}{2} \times 278.10 \cdot g \cdot m o {l}^{-} 1$

$= 1.71 \cdot g$

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