If 12.3 kg of Al2O3(s), 59.4 kg of NaOH(l), and 59.4 kg of HF(g) react completely, how many kilograms of cryolite will be produced?

Question

Al2O3(s)+6NaOH(l)+12HF(g)⟶2Na3AlF6+9H2O(g)

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

Answer 1

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

First, we find the limiting reactant. An easy way to do this is to divide the MOLES of each reactant by the corresponding coefficient in the balanced equation.

For Al₂O₃: 12.3 kg x 1000 g / kg x 1 mol Al₂O₃ / 102 g = 121 mols Al₂O₃ (÷1->121)

For NaOH: 59.4 kg x 1000 g / kg x 1 mol NaOH / 40 g = 1485 mols NaOH (÷6->248)

For HF: 59.4 kg x 1000 g / kg x 1 mol HF / 20.0 g = 2970 mols HF (÷12->249)

Since 121 is the lowest value, Al₂O₃ is LIMITING

Next, we use the moles of the limiting reactant (Al₂O₃) to determine the theoretical yield of cryolite (Na₃AlF₆)

Kilograms of cryolite produced:

121 mols Al₂O₃ x 2 mols Na₃AlF₆ / mol Al₂O₃ x 210 g Na₃AlF₆ /mol = 50,820 g Na₃AlF₆ = 50.8 kg Na₃AlF₆

The excess reactants are NaOH and HF. To find how much of each is left over after the reaction is complete, we find the amount used and subtract that from the amount initially present.

moles NaOH used = 121 mol Al₂O₃ x 6 mol NaOH / mol Al₂O₃ = 726 mols NaOH used

mols NaOH left over = 1485 mols - 726 mols = 759 mols

mass NaOH left over = 759 mols NaOH x 40 g / mol x 1 kg / 1000 g = 30.4 kg NaOH left over

moles HF used = 121 mol Al₂O₃ x 12 mols HF / mol Al₂O₃ = 1452 mols HF used

moles HF left over = 2970 mols - 1452 mols = 1518 mols

mass HF left over = 1518 mols HF x 20.0 g / mol x 1 kg / 1000 g = 30.4 kg HF left over