Question

IF 17.2 KILOGRAMS OF AL2O3, 60.4 KILOGRAMS OF NAOH AND 60.4 KILOGRAMS OF HF REACT COMPLETELY HOW MANY KILOGRAMS OF CRYOLITE WILL BE PRODUCED?

Answer 1

Al₂O₃ + 6 NaOH + 12 HF → 2 Na₃AlF₆ + 9 H₂O
                                            Cryolite
(17.5 kg Al₂O₃) / (101.9614 g Al₂O₃/mol) x (2/1) = 0.34326 kmol Na₃AlF₆
(60.4 kg NaOH) / (39.9971 g NaOH/mol) x (2/6) = 0.50336 kmol Na₃AlF₆
(60.4 kg HF) / (20.0064 g HF/mol) x (2/12) = 0.50317 kmol Na₃AlF₆

Since Al₂O₃ produces the least amount of product, Al₂O₃ is the limiting reactant. NaOH and HF are in excess.

(0.34326 kmol Na₃AlF₆) x (209.9413 g/mol) = 72.06 kg Cryolite produced