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

To solve this problem, we'll need to use stoichiometry. First, we need to determine the limiting reagent. We can do this by using the balanced equation and converting the mass of NaOH to moles:
NaOH + Al(OH)3 + 6HF → Na3AlF6 + 6H2O
35.4 g NaOH x (1 mol NaOH/40 g NaOH) = 0.885 mol NaOH

Next, we'll use the stoichiometry of the balanced equation to find the theoretical yield of cryolite:
0.885 mol NaOH x (1 mol Na3AlF6 / 3 mol NaOH) x (NA3AlF6 = 209.94 g/mol) = 62.4 g Na3AlF6

However, the problem tells us that the reaction has a 67.3% yield, so we need to multiply the theoretical yield by the percent yield to find the actual yield:
62.4 g Na3AlF6 x 0.673 = 41.7 g Na3AlF6

Therefore, the answer is B) 41.7 g.

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