if 15kg of Al2O3, 53.4kg of NaOH and 53.4kg of HF react completely, how many kg of cryolite will be produced and what is the mass of excess reactants after the reaction

Question

(question #35)

Answer 1

This question has been asked and answered at least twice on Wyzant. here is how to do it even though the amounts are slightly different than in the current problem.

Al₂O₃ (s) + 6NaOH (l) + 12HF (g) = 2Na₃AlF₆ + 9H₂O (g) ... balanced equation

molar mass Al2O3 = 102 g/mol

molar mass NaOH = 40 g/mol

molar mass HF = 20 g/mol

molar mass 2Na₃AlF₆ = 210 g/mol

To find the limiting reactant, and hence the theoretical yield of cryolite, and easy way is to simply divide moles of each reactant by the respective coefficient and see which value is lowest.

For Al2O3: 14.2 kg x 1000 g/kg x 1mol / 102 g = 139 mols (÷1->139)

For NaOH: 60.4 kg x 1000 g/kg x 1 mol / 40 g = 1510 mols (÷6->252)

For HF: 60.4 kg HF x 1000 g/kg x 1 mol / 20 g = 3020 mols (÷12->252)

139 is less than 252 so Al2O3 is LIMITING and will dictate how much cryolite can be made.

How many kg of cryolite will be produced?

139 mols Al2O3 x 2 mols Na₃AlF₆ / mol Al2O3 x 210 g Na₃AlF₆ / mol x 1 kg/1000g = 58.4 kg Na₃AlF₆

Which reactants are in excess?

NaOH and HF are in excess

Total mass of excess reactants left over after the reaction:

Find how much NaOH and HF were used, and subtract that from the starting amount...

mols NaOH used = 139 mol Al2O3 x 6 mol NaOH/mol Al2O3 = 834 mols NaOH used

mass NaOH left = 1510 mols - 834 mols (x40 g/mol) = 27,040 g = 27.0 kg NaOH left over

mols HF used = 139 mol Al2O3 x 12 mol HF/mol Al2O3 = 1668 mols HF used

mols HF left = 3020 mols - 1668 mol (x20 g/mol) = 27,040 g = 27.0 kg HF left over

TOTAL mass left over = 27.0 kg + 27.0 kg = 54.0 kg left ov