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Chapter 4: 4.13 (page 202)

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

Expert verified

The percent yield is 48.3%

Step by step solution

01

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

02

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

03

Determine percent yield

The formula for percent yield

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

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Outline the steps needed to determine the limiting reactant when 0.50 mol of Cr and 0.75 mol of H3PO4 react according to the following chemical equation.\(2Cr + 2{H_3}P{O_4} \to 2CrP{O_4} + 3{H_2}\). Determine the limiting reactant.

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