# Question #b7bcb

Jan 8, 2015

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

The balanced chemical equation is:

$P b C {l}_{2 \left(a q\right)} + Z {n}_{\left(s\right)} \to Z n C {l}_{2 \left(a q\right)} + P {b}_{\left(s\right)}$

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

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

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

The complete ionic equation looks like this:

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

The net ionic equation is

$P {b}_{\left(a q\right)}^{2 +} + Z {n}_{\left(s\right)} \to Z {n}_{\left(a q\right)}^{2 +} + P {b}_{\left(s\right)}$

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