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

1. Calculate the moles of HF: 
27.8 g HF / 20.01 g/mol = 1.39 mol HF

2. Use stoichiometry to determine the moles of cryolite produced. Note that the reaction ratio from HF to cryolite is 6:1. 
1.39 mol HF x 1 mol cryolite / 6 mol HF = 0.232 mol cryolite

3. Calculate the theoretical yield of cryolite assuming 100% yield. 
0.232 mol cryolite x 209.96 g/mol = 48.74 g cryolite

4. Calculate the actual yield using the percent yield. 
Actual yield = 94.3% x 48.74 g = 45.9 g

Therefore, the answer is (C) 45.9 g.

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