Question #7e241

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Question #7e241

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Answer 1 · 2015-04-12T11:01:57

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

The overall equation is:

$P b {(N O_{3})}_{2 (a q)} + 2 N a C l_{(aq} \to P b C l_{2 (s)} + 2 N a N O_{3 (a q)}$ $Pb(NO_3)_(2(aq)) + 2NaCl_text((aq)rarr PbCl_(2(s)) +2NaNO_(3(aq))$

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

$P b_{(a q)}^{2 +} + 2 C l_{(a q)}^{-} \to P b C l_{2 (s)}$ $Pb_((aq))^(2+)+2Cl_((aq))^(-)rarrPbCl_(2(s))$

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

The number of moles of $P b_{(a q)}^{2 +}$ $Pb_((aq))^(2+)$ = 0.15 x 0.1 = 0.015

The number of moles of $C l_{(a q)}^{-}$ $Cl_((aq))^-$ ions added = 0.1 x 0.2 = 0.02

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

So there is insufficient $C l^{-}$ $Cl^-$ ion to react with all the $P b^{2 +}$ $Pb^(2+)$ so there will be an XS of $P b^{2 +}$ $Pb^(2+)$ and 0.02/2 = 0.01 moles of $P b C l_{2}$ $PbCl_2$ will form.

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Question #7e241

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