Question

Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide.
AL2O3(s) + NaOH(l) + HF(g) ---> Na3 AlF6 + H2O(g)

1)Balance the equation.

2)If 15.2 kilograms of Al2O3(s), 57.4 kilograms of NaOH(l), and 57.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced?

3)Which reactants will be in excess?

4)What is the total mass of the excess reactants left over after the reaction is complete?

Answer 1

Answer on Question #40495, Chemistry, Other

**Task:**

Cryolite, $\mathrm{Na_3AlF_6(s)}$ an ore used in the production of aluminium, can be synthesised using aluminium oxide.

$\mathrm{Al_2O_3(s) + NaOH(l) + HF(g) = Na_3AlF_6 + H_2O(g)}$

1) Balance the equation

2) If 15.2 kilograms of $\mathrm{Al_2O_3(s)}$, 57.4 kilograms of $\mathrm{NaOH(l)}$ and 57.4 kilograms of HF gas react completely, how many kilograms of criolite will be produced?

3) Which reactants will be in excess?

4) What is the total mass of the excess reactants left over after the reaction is complete?

**Answer:**

$\mathrm{Al_2O_3(s) + 6NaOH(l) + 12HF(g) = 2Na_3AlF_6 + 9H_2O(g)}$

$$v = \frac{m}{M}$$

where m = mass, grams;

M = molar mass, gram/mol.

$\mathrm{M(Al_2O_3) = 101.96g/mol}$ $\quad \mathrm{M(NaOH) = 39.996g/mol}$ $\quad \mathrm{M(HF) = 20.007g/mol}$

$\mathrm{M(Na_3AlF_6) = 209.95g/mol}$

$\mathrm{v(Al_2O_3) = \frac{15200}{101.96} = 149.08 \text{ moles}}$

$\mathrm{v(NaOH) = \frac{57400}{39.996} = 1435.1 \text{ moles}}$

$\mathrm{v(HF) = \frac{57400}{20.007} = 2869.0 \text{ moles}}$

Let's calculate the amount of $\mathrm{Na_3AlF_6}$, that can be produced from the indicated amount of each reactant:

$\mathrm{v(Na_3AlF_6) = 2 \cdot v(Al_2O_3) = 2 \cdot 149.08 = 298.16 \text{ moles}}$

$\mathrm{v(Na_3AlF_6) = \frac{v(NaOH)}{6} \cdot 2 = \frac{1435.1}{6} \cdot 2 = 478.37 \text{ moles}}$

$\mathrm{v(Na_3AlF_6) = \frac{v(HF)}{12} \cdot 2 = \frac{2869.0}{12} \cdot 2 = 478.17 \text{ moles}}$

As we can see from the previous calculations, the amount of $\mathrm{Al_2O_3}$ is the determining factor.

All other reactants (NaOH, HF) will be in excess.

That is why, the maximum mass of $\mathrm{Na_3AlF_6}$ that can be produced is equal to:

$\mathrm{m(Na_3AlF_6) = v(Na_3AlF_6) \cdot M(Na_3AlF_6)}$

$\mathrm{m(Na_3AlF_6) = 298.16 \cdot 209.95 = 62599 \text{ g} = 62.599 \text{ kg}}$

NaOH and HF will be in excess. Let's calculate this excess:

$$\begin{array}{l} \mathrm{m}(\mathrm{NaOH})_{\mathrm{ex}} = \frac{478.37 - 298.16}{2} \cdot 6 \cdot 39.996 = 21623 \mathrm{g} = 21.623 \mathrm{kg} \\ \mathrm{m}(\mathrm{HF})_{\mathrm{ex}} = \frac{478.17 - 298.16}{2} \cdot 12 \cdot 20.007 = 21609 \mathrm{g} = 21.609 \mathrm{kg} \\ \end{array}$$

The total mass of the excess reactants leftover after the reaction is complete will be:

$$\mathrm{m}(\mathrm{NaOH})_{\mathrm{ex}} + \mathrm{m}(\mathrm{HF})_{\mathrm{ex}} = 21.623 \mathrm{kg} + 21.609 \mathrm{kg} = 43.232 \mathrm{kg}$$