# Question 7e241

Apr 12, 2015

Yes, a white precipitate of lead(II) chloride will form.

The overall equation is:

Pb(NO_3)_(2(aq)) + 2NaCl_text((aq)rarr PbCl_(2(s)) +2NaNO_(3(aq))#

Lead(II) chloride is insoluble so the ionic equation is:

$P {b}_{\left(a q\right)}^{2 +} + 2 C {l}_{\left(a q\right)}^{-} \rightarrow P b C {l}_{2 \left(s\right)}$

The other ions take no part in the reaction so are spectators.

The number of moles of $P {b}_{\left(a q\right)}^{2 +}$ = 0.15 x 0.1 = 0.015

The number of moles of $C {l}_{\left(a q\right)}^{-}$ ions added = 0.1 x 0.2 = 0.02

0.015 moles of $P {b}^{2 +}$ require 0.015 x 2 = 0.03 moles $C {l}^{-}$ ions.

So there is insufficient $C {l}^{-}$ ion to react with all the $P {b}^{2 +}$ so there will be an XS of $P {b}^{2 +}$ and 0.02/2 = 0.01 moles of $P b C {l}_{2}$ will form.