1- Na3AlF6 can be made by the reaction below:Al203 (s) + 6 NaOH (aq) + 12HF (6) → 2 Na:AlF, (s) + 9 H20 (I)In an experiment; 6.55 g Al;0, and excess HF were dissolved in 1.75 L of 0.15 M NaOH.If 20 g Na3AIF6 was obtained,a Which one is the limiting reagent?b What is the actual yield?|What is the theoretical yield?d What is the percent yield for this experiment?(Mw (Na3AlF6): 209.94 g/mol; (Al>0:): 101.96 g/mol; (NaOH): 39.997 g/mol)

Moles of Al2O3 = Mass/Molar mass = 6.55g/(101.96gmol-1) = 0.0642 moles Moles of NaOH = Molarity ×…

Answered: 1- Na3AlF6 can be made by the reaction…

Chemistry & Chemical Reactivity

10th Edition

ISBN:9781337399074

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter21: The Chemistry Of The Main Group Elements

Section: Chapter Questions

Problem 120IL

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