91Ó°ÊÓ

The inorganic compound sodium aluminium hexafluoride (Na3AlF6) has the formula Na₃AlF₆. This white solid, discovered in 1799 by Peder Christian Abildgaard as the mineral cryolite, exists naturally and is widely employed in the industrial manufacturing of aluminium metal.

Synthetic cryolite is made using the following balancing equation:

$6HF\left(g\right)+Al(OH{)}_3\left(s\right)+3NaOH\left(aq\right)→{\text{Na}}_{3}Al{F}_6\left(aq\right)+6H_{2}O\left(l\right)$

Let's write down everything we know:

$$\begin{gathered} \text{yield}=95.6\% \\ m\left(Al{\left(OH\right)}_3\right)=365kg \\ V\left(NaOH\right)=1.20m^3 \\ w\left(NaOH\right)=50.0\% \\ d\left(NaOH\right)=1.53gm{L}^{-1} \\ V\left(HF\right)=265m^3 \\ p\left(HF\right)=305kPa \\ T\left(HF\right)=91.5^{°}C \end{gathered}$$

By analysing the intensity of the chemical reaction for each component, we may compute the chemical quantities of each component and determine which one is the limiting reactant:

${\mathrm{ξ}}_i=\frac{n_i}{v_i}$

Therefore, we use ${\mathrm{ξ}}_i=\frac{n_i}{v_i}$ to determine

By analysing the intensity of the chemical reaction for each component, we may compute the chemical quantities of each component and determine which one is the limiting reactant:

$$\begin{gathered} n\left(Al{\left(OH\right)}_3\right)=\frac{m\left(Al{\left(OH\right)}_3\right)}{M\left(Al{\left(OH\right)}_3\right)} \\ \left(\text{Al}{\left(\text{OH}\right)}_3\right)=\frac{365·{10}^3\text{g}}{78\text{gmol}} \\ n\left(\text{Al}{\left(\text{OH}\right)}_3\right)=4679.5\text{mol} \\ \left(\text{Al}{\left(\text{OH}\right)}_3\right)=1 \\ \mathrm{ξ}\left(\text{Al}{\left(\text{OH}\right)}_3\right)=4679.5 \\ n\left(NaOH\right)=\frac{m\left(NaOH\right)}{M\left(NaOH\right)} \\ n\left(NaOH\right)=\frac{V\left(NaOH\right)·d\left(NaOH\right)·w\left(NaOH\right)}{M\left(NaOH\right)} \\ n\left(NaOH\right)=22951.7\text{mol} \end{gathered}$$

Hence, the limiting reactant is $22951.7\text{mol}$

We must apply the ideal gas law to compute the amount of$HF$:

$$\begin{gathered} pV=nRT \\ n=\frac{pV}{RT} \\ n\left(HF\right)=\frac{305·{10}^3\text{Pa}·265{\text{m}}^3}{8.314{\text{JK}}^{-1}{\text{mo}}^{-1}·364.65K} \\ n\left(HF\right)=26660\text{mol} \\ v\left(HF\right)=6 \\ \mathrm{ξ}\left(HF\right)=\frac{26660\text{mol}}{6} \\ =4443\text{mol} \end{gathered}$$

The limiting reactant is HF $4443\text{mol}$.

We can now compute cryolite's mass.

$$\begin{gathered} n\left(N{a}_{3}Al{F}_6\right)=\frac{1}{6}n\left(HF\right)n\left(N{a}_{3}Al{F}_6\right)=4443mol \\ M\left(N{a}_{3}Al{F}_6\right)=209.94gmol \\ m\left(N{a}_{3}Al{F}_6\right)=n\left(N{a}_{3}Al{F}_6\right)·M\left(N{a}_{3}Al{F}_6\right) \\ m\left(N{a}_{3}Al{F}_6\right)=4443mol·209.94gmo{l}^{-1} \end{gathered}$$

$$\begin{gathered} m\left(N{a}_{3}Al{F}_6\right)=932763g \\ =933kg \end{gathered}$$

Of course, we must factor in the$95.6\backslash \%$yield :

$$\begin{gathered} {\left(N{a}_{3}Al{F}_6\right)}_{0.956}=m\left(N{a}_{3}Al{F}_6\right)·\frac{95.6\%}{100\%} \\ m{\left(N{a}_{3}Al{F}_6\right)}_{0.956}=933kg·0.956 \\ m{\left(N{a}_{3}Al{F}_6\right)}_{0.956}=891kg \end{gathered}$$

Therefore, the yield is $891kg$