Question

Freon – 12, CCl2F2, is prepared from CCl4 by reaction with HF. The other product of this reaction is HCl. Outline the steps needed to determine the percent yield of a reaction that produces 12.5 g of CCl2F2 from 32.9 g of CCl4. Freon – 12 has been banned and is no longer used as a refrigerant because it catalyzes the decomposition of ozone and has a very long lifetime in the atmosphere. Determine the percent yield.

Answer 1

CCl₄ + 2HF → CCl₂F₂ + 2HCl

m(CCl₄) = 32.9 g

m’(CCl₂F₂) = 12.5 g

M(CCl₄) = 153.82 g/mol

n(CCl₄) $= \frac{32.9}{153.82} = 0.2138 \;mol$

n(CCl₂F₂) = n(CCl₄) = 0.2138 mol

M(CCl₂F₂) = 120.91 g/mol

m(CCl₂F₂) $= 0.2138 \times 120.91 = 25.85 \;g$

Proportion:

25.85 g – 100 %

12.5 g – x

$x = \frac{12.5 \times 100}{28.85} = 43.32$ %

The percent yield = 43.32 %.