Producing Cryolite from Chemical Reactions

Calculating Mass of Cryolite Produced

If 13.2 kg of Al₂O₃ (s), 50.4 kg of NaOH (l), and 50.4 kg of HF (g) react completely, how many kilograms of cryolite will be produced? The mass of cryolite produced will be, 54.38 kg

Solution:

Given,

Mass of Al₂O₃ = 13.2 kg = 13200 g

Mass of NaOH = 50.4 kg = 50400 g

Mass of HF = 50.4 kg = 50400 g

Molar mass of Al₂O₃ = 101.9 g/mole

Molar mass of NaOH = 40 g/mole

Molar mass of HF = 20 g/mole

Molar mass of Na₃AlF₆ = 209.9 g/mole

First we have to calculate the moles of Al₂O₃, NaOH, and HF.

Moles of Al₂O₃ = Mass of Al₂O₃ / Molar mass of Al₂O₃ = 13200g / 101.9g/mole = 129.54 moles

Moles of NaOH = Mass of NaOH / Molar mass of NaOH = 50400g / 40g/mole = 1260 moles

Moles of HF = Mass of HF / Molar mass of HF = 50400g / 20g/mole = 2520 moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

Al₂O₃ + 6NaOH + 12HF → 2Na₃AlF₆ + 9H₂O

The mole ratio of Al₂O₃, NaOH, and HF is 1:6:12

The ratio of given moles of Al₂O₃, NaOH, and HF is 129.54:1260:2520

From this we conclude that NaOH and HF are excess reagents because the given moles are greater than the required moles, and Al₂O₃ is a limiting reagent that limits the formation of product.

Now we have to calculate the moles of Na₃AlF₆

As 1 mole of Al₂O₃ reacts to give 2 moles of Na₃AlF₆

So, 129.54 moles of Al₂O₃ react to give 259.08 moles of Na₃AlF₆

Now we have to calculate the mass of Na₃AlF₆

Mass of Na₃AlF₆ = Moles of Na₃AlF₆ × Molar mass of Na₃AlF₆

Mass of Na₃AlF₆ = (259.08 moles) × (209.9 g/mole) = 54380.892 g = 54.38 kg

Thus, the mass of cryolite produced will be 54.38 kg