Answer to Question #334320 in Chemistry for Desalinated

Question

CCl4 + SbF3 = CCl2F2 + SbCl3

How many grams of CCl4 are needed to react with 5.0 mol SbF3?

Answer 1

Solution:

Balanced chemical equation:

3CCl₄ + 2SbF₃ → 3CCl₂F₂ + 2SbCl₃

According to stoichiometry:

3 mol of CCl₄ react with 2 mol of SbF₃

X mol of CCl₄ react with 5.0 mol of SbF₃

Thus,

Moles of CCl₄ = (5.0 mol SbF₃) × (3 mol CCl₄ / 2 mol SbF₃) = 7.5 mol CCl₄

The molar mass of CCl₄ is 153.82 g/mol

Therefore,

Mass CCl₄ = (7.5 mol CCl₄) × (153.82 g CCl₄ / 1 mol CCl₄) = 1153.65 g CCl₄

Mass CCl₄ = 1153.65 g

Answer: 1153.65 grams of CCl₄ are needed