Cryolite is a compound needed for the Hall-Heroult process for producing aluminum. Cryolite is produced by the following reaction: 6 HF + Al(OH) ₃ + 3 NaOH → Na₃AlF₆ + 6 H₂O
How many grams of cryolite are produced if the reaction has a 67.3% yield and a limiting reagent of 35.4 grams of NaOH?

Question

Quizplus · Accepted Answer

First, calculate the moles of NaOH: 35.4 g / 40.00 g/mol = 0.885 moles. According to the balanced equation, 3 moles of NaOH produce 1 mole of Na3AlF6. Thus, 0.885 moles of NaOH produce 0.295 moles of Na3AlF6. The molar mass of Na3AlF6 is approximately 210 g/mol, so the theoretical yield is 0.295 moles * 210 g/mol = 61.95 g. With a 67.3% yield, the actual yield is 61.95 g * 0.673 = 41.7 g.

Cryolite Is a Compound Needed for the Hall-Heroult Process for Producing