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Question #b7bcb

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Answer 1 · 2015-01-08T09:18:43

The answer is $7.65 \cdot 10^{- 3}$ $7.65*10^(-3)$ $moles$ $"moles"$ of $P b^{2 +}$ $Pb^(2+)$ ions were present in the sample.

The balanced chemical equation is:

$P b C l_{2 (a q)} + Z n_{(s)} \to Z n C l_{2 (a q)} + P b_{(s)}$ $PbCl_(2(aq)) + Zn_((s)) -> ZnCl_(2(aq)) + Pb_((s))$

Notice the $1 : 1$ $1:1$ mole ratio between $Z n$ $Zn$ and $P b C l_{2}$ $PbCl_2$ ; this means that one mole $Z n$ $Zn$ will react with 1 mole of $P b C l_{2}$ $PbCl_2$ .

You know that $7.65 \cdot 10^{- 3}$ $7.65*10^(-3)$ moles of $Z n$ $Zn$ had reacted after one day, which automatically means that the exact number of $P b C l_{2}$ $PbCl_2$ moles had reacted as well. The number of $P b C l_{2}$ $PbCl_2$ moles is equal to the number of moles of $P b^{2 +}$ $Pb^(2+)$ ions, since

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

The complete ionic equation looks like this:

$P b_{(a q)}^{2 +} + 2 C l_{(a q)}^{-} + Z n_{(s)} \to Z n_{(a q)}^{2 +} + 2 C l_{(a q)}^{-} + P b_{(s)}$ $Pb_((aq))^(2+) + 2Cl_((aq))^(-) + Zn_((s)) -> Zn_((aq))^(2+) + 2Cl_((aq))^(-) + Pb_((s))$

The net ionic equation is

$P b_{(a q)}^{2 +} + Z n_{(s)} \to Z n_{(a q)}^{2 +} + P b_{(s)}$ $Pb_((aq))^(2+) + Zn_((s)) -> Zn_((aq))^(2+) + Pb_((s))$

This is a single replacement reaction. Since $Z n$ $Zn$ is more reactive metal than $P b$ $Pb$ , the $Z n$ $Zn$ ions will completely replace the $P b$ $Pb$ ions present in the solution.

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