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suppose 14.7 g of iron(iii) chloride (fecl₃) are dissolved in 480. ml o…

Question

suppose 14.7 g of iron(iii) chloride (fecl₃) are dissolved in 480. ml of water. find the composition of the resulting electrolyte solution.
in particular, list the chemical symbols (including any charge) of each dissolved ion in the table below. list only one ion per row.
then, calculate the concentration of each ion in \\( \frac{m e q}{l} \\), and write the concentration in the second column of each row. be sure you round your answers to the correct number of significant digits.

Explanation:

Step1: Write the dissociation equation

$$\text{FeCl}_3(s)\to\text{Fe}^{3 +}(aq)+3\text{Cl}^-(aq)$$

Step2: Calculate the moles of $\text{FeCl}_3$

The molar mass of $\text{FeCl}_3$ is $M=(55.85 + 3\times35.45)\text{ g/mol}=162.2\text{ g/mol}$.
The moles of $\text{FeCl}_3$, $n=\frac{m}{M}=\frac{14.7\text{ g}}{162.2\text{ g/mol}}\approx0.0906\text{ mol}$.

Step3: Calculate the equivalents of each ion

For $\text{Fe}^{3 +}$:
The charge of $\text{Fe}^{3 +}$ is $+ 3$. The number of moles of $\text{Fe}^{3 +}$ is $n_{\text{Fe}^{3+}} = 0.0906\text{ mol}$.
The equivalents of $\text{Fe}^{3 +}$, $Eq_{\text{Fe}^{3+}}=n_{\text{Fe}^{3+}}\times3=0.0906\text{ mol}\times3 = 0.2718\text{ Eq}$.
For $\text{Cl}^-$:
The charge of $\text{Cl}^-$ is $-1$. The number of moles of $\text{Cl}^-$ is $n_{\text{Cl}^-}=3\times0.0906\text{ mol}=0.2718\text{ mol}$.
The equivalents of $\text{Cl}^-$, $Eq_{\text{Cl}^-}=n_{\text{Cl}^-}\times1=0.2718\text{ Eq}$.

Step4: Calculate the volume of the solution

The volume of the solution $V = 480\text{ mL}=0.480\text{ L}$.

Step5: Calculate the concentration in $\frac{\text{mEq}}{\text{L}}$

For $\text{Fe}^{3 +}$:
$C_{\text{Fe}^{3+}}=\frac{Eq_{\text{Fe}^{3+}}\times1000\text{ mEq/Eq}}{V}=\frac{0.2718\text{ Eq}\times1000\text{ mEq/Eq}}{0.480\text{ L}}\approx566\text{ mEq/L}$.
For $\text{Cl}^-$:
$C_{\text{Cl}^-}=\frac{Eq_{\text{Cl}^-}\times1000\text{ mEq/Eq}}{V}=\frac{0.2718\text{ Eq}\times1000\text{ mEq/Eq}}{0.480\text{ L}}\approx566\text{ mEq/L}$.

Answer:

ion$\frac{\text{mEq}}{\text{L}}$
$\text{Cl}^-$$566$