Question

What is the percent yield of a reaction that produces 12.5g of the gas freon CF₂Cl₂ from 32.9 g of CCl₄ and excess HF.

\(CC{l_4} + 2HF \to C{F_2}\)

Short answer

The percent yield is 48.3%

Step-by-step solution

Given data

From the given balanced equation, we can say that 1 mol of CCl₄ reacts with 2 moles of hydrofluoric acid in order to produce 1 mol of CF₂Cl₂ and 2 moles of hydrogen chloride. HF is in excess so the limiting reagent will be CCl₄

Determine number of moles

Moles of CCl₄

\(\begin{array}{l} = \frac{{Given\,mass}}{{Molecular\,weight}}\\ = \frac{{32.9\,g}}{{153.8\,g}}\\ = 0.214\,mole\end{array}\)

= 32.9g/153.8g=0.214 moles

Now we convert moles into mass

= 0.214 mol x 120.91 g/mol= 25.8 g

Determine percent yield

The formula for percent yield

\(\begin{array}{l} = \frac{{\Pr oduct\,yield}}{{Theoritical\,yield}} \times 100\\ = \frac{{12.5g}}{{25.8}} \times 100\\ = 48.3\% \end{array}\)