Question

Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide.
Balance the equation for the synthesis of cryolite. equation: Al2O3 (s) + NaOH (l) + HF (g) -> Na3AlF6 + H2O (g)
If 11.0 kg of Al2O3(s), 54.4 kg of NaOH(l), and 54.4 kg of HF(g) react completely, how many kilograms of cryolite will be produced?
mass of cryolite produced:___kg Na3AlF6
Which reactants will be in excess? NaOH Al2O3 HF
What is the total mass of the excess reactants left over after the reaction is complete?
total mass of excess reactants:___kg

Answer 1

Ans: Al,03 + 6NaOH) +1246 19) — 2NO3 pifc + 9H20/9) Moles of Al, og given = Cuiven mass = || X 109g = 107.8 moles Molar mass 102 Moles of Naon given = 54.4x10² = 1360 moles 40 Moles of HF given = 54.48103 = 2720 moles 20 for 1 mole of Al₂O₃ ; 6 moles of NaOH and 12 moles of hf required So for 107.8 moles of Al₂O₃ ; NaOH required = 107.886 = 646.8 moles HF required - 107.8X12 = 1293.6 moles But we have NaOH and HF in excess. So Al₂O3 = limiting reagent NOOH, HF = excees reagent Imole of Al2O3 give 2 moles of cryolite. 107.8 moles of $1,03 give ouyolite - 2x107.8 moles = 215.6 moles

Mass of cryolite produced = Moles x Molar mass - 215.68210 = 45276 g Mass of cryolite produced - 45.276 Rg y Reactants in excess = NaOH, HF Moles of Naon present = 1360 Moles of Nooh used = 646.8 Moles of Naon left = 71312 Moss of Nooh left = 28.528 kg Moles of HF present 2720 Moles of HF used : 1293.6 Moles of HF left = 1426.4 Moss of He left = 28.528 kg Mass of excess reactants left = 28.528 +28.528 = 57.056 kg.