Question

Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide.

Balance the equation.

Al2O3(s) + NaOH(l) + HF(g) -- Na3AlF6 + H2O(g)

If 16.4 kilograms of Al2O3(s), 51.4 kilograms of NaOH(l), and 51.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced?

Which reactants will be in excess?

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

Answer 1

Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide.

Solution :-

Balanced equation is as follows

Al2O3 + 6 NaOH + 12 HF --- > 2 Na3AlF6 + 9 H2O

If 16.4 kilograms of Al2O3(s), 51.4 kilograms of NaOH(l), and 51.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced?

Lets calculate the moles of each reactant

(16.4 kg * 1000g/1kg)*(1mol /101.96 g) = 160.85 mol Al2O3

(51.4 kg * 1000 g / 1kg)*(1 mol / 40 g) =1285 mol NaOH

(51.4 kg*1000 g /1kg)*(1mol / 20.01 g) = 2568.7 mol HF

Now lets calculate the moles of the NaOH and HF needed to react with Al2O3

160.85 mol Al2O3 * 6 mol NaOH / 1 mol Al2O3 = 965.1 mol NaOH

Moles of HF needed

160.85 mol Al2O3 * 12 mol HF / 1 mol Al2O3 = 1930.2 mol HF

So the Al2O3 is limiting reactant because the moles of NaOH and HF needed are less than moles of NaOH and HF available

So lets calculate the theoretical mass of the product that can be formed

(160.85 mol Al2O3 *2 mol Na3AlF6 / 1 mol Al2O3)*(209.94 g / 1 mol Na3AlF6) = 67525 g

67525 g * 1 kg / 1000 g = 67.5 kg Na3AlF6

So the mass of the cryolite that can be formed = 67.5 kg

Which reactants will be in excess?

Reactant that remain in excess are HF and NaOH

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

Solution :-

Lets find the mass of each reactant remain in excess

Mass of NaOH = (1285 mol – 965.1 mol ) * 40.0 g per mol = 12796 g = 12.796 kg

Mass of HF = (2568.7 mol – 1930.2 mol ) * 20.01 g per mol = 12776 g = 12.776 kg

Total mass of excess reactant = 12.796 kg + 12.776 kg = 25.572 kg