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
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.