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)

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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)
Transcribed Image Text: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)
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