Question

Cryolite$\left({\mathrm{Na}}_3{\mathrm{AlF}}_6\right)$is used in the production of aluminum from its ores. It is made by the reaction

$\mathbf{6}\mathbf{NaOH}\mathbf{+}{\mathbf{Al}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{+}\mathbf{12}\mathbf{HF}\to \mathbf{2}{\mathbf{Na}}_{\mathbf{3}}{\mathbf{AlF}}_{\mathbf{6}}\mathbf{+}\mathbf{9}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

Calculate the mass ofcryolite that can be prepared by the complete reaction of 28.7g${\mathbf{Al}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$.

Short answer

The mass of cryolite is 1182.1 g.

Step-by-step solution

Calculating molar masses

We know that the molar masses of Al, Na, F, and O are26.98g, 22.99g,19.00g, and 16.00grespectively.

$$\begin{gathered} \mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathbf{}\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{}\mathit{o}\mathit{f}{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\mathbf{\text{=}}\left(\text{2×26.98 + 3×16.00}\right)\mathbf{\text{g}} \\ \mathbf{\text{=}}\left(\text{53.96 + 48.00}\right)\mathbf{\text{g}} \\ \mathbf{\text{= 101.96g}} \end{gathered}$$

Calculating mass of cryolite produced in complete reaction

As per the reaction:

${\text{6NaOH + Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{+ 12HF}\to {\text{2Na}}_{\text{3}}{\text{AlF}}_{\text{6}}{\text{+ 9H}}_{\text{2}}\text{O}$

.$\mathbf{1}\mathbf{mol}\mathbf{.}{\mathbf{Al}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{produces}\mathbf{2}\mathbf{mol}\mathbf{.}{\mathbf{Na}}_{\mathbf{3}}{\mathbf{AlF}}_{\mathbf{6}}$

101.96g ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$produces $2\times 209.95$g

${\text{28.7g.Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{produces 'x' amount of cryolite}$

whereby x is

$$\begin{gathered} =\frac{\text{2×09.98g}}{\text{101.96g}}\times \text{28.7g}\times \text{10} \\ \text{= 1181.94g} \end{gathered}$$

Therefore,the mass of cryolite produced from the complete reaction is$\text{1181.94}$ g.