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 94.3% yield and a limiting reagent of 27.8 grams of HF?

Question

Quizplus · Accepted Answer

First, we need to determine the limiting reagent. To do this, we need to convert the mass of HF to moles, and then use stoichiometry to determine how many moles of cryolite are produced.

27.8 g HF x (1 mol HF/20.01 g HF) x (1 mol cryolite/6 mol HF) = 0.23 mol cryolite

Now we can use the stoichiometric ratio between cryolite and HF to determine the theoretical yield:

0.23 mol cryolite x (209.96 g cryolite/1 mol cryolite) = 48.6 g cryolite

Finally, we need to apply the given percentage yield of 94.3%:

48.6 g cryolite x 0.943 = 45.9 g cryolite

Therefore, the answer is C) 45.9 g.

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