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Chapter 4: Problem 142

In the reaction of \(2.00 \mathrm{mol} \mathrm{CCl}_{4}\) with an excess of \(\mathrm{HF}\), \(1.70 \mathrm{mol} \mathrm{CCl}_{2} \mathrm{F}_{2}\) is obtained. $$ \mathrm{CCl}_{4}+2 \mathrm{HF} \longrightarrow \mathrm{CCl}_{2} \mathrm{F}_{2}+2 \mathrm{HCl} $$ (a) The theoretical yield is \(1.70 \mathrm{mol} \mathrm{CCl}_{2} \mathrm{F}_{2}\) (b) The theoretical yield is \(1.00 \mathrm{mol} \mathrm{CCl}_{2} \mathrm{F}_{2}\) (c) The theoretical yield depends on how large an excess of HF is used. (d) The percent yield is \(85 \%\)

Short Answer

Expert verified
(a) The theoretical yield is 2.00 mol \( \mathrm{CCl}_{2}\mathrm{F}_{2} \) (d) The percent yield is \( 85 \% \)

Step by step solution

01

Theoretical Yield Calculation

The theoretical yield is determined from the stochiometric coefficients of reactants and products in the balanced chemical reaction. In this reaction: \( \mathrm{CCl}_4 + 2\mathrm{HF} \rightarrow \mathrm{CCl}_2\mathrm{F}_2 + 2\mathrm{HCl} \) , for every one mole of \( \mathrm{CCl}_4 \) reacted, one mole of \( \mathrm{CCl}_2\mathrm{F}_2 \) is produced. Hence, the theoretical yield of \( \mathrm{CCl}_2\mathrm{F}_2 \) when 2.00 mol of \( \mathrm{CCl}_4 \) are reacted is 2.00 mol.
02

Percent yield Calculation

Percent yield is the ratio of the actual yield to the theoretical yield expressed as a percentage. It is calculated using the formula: \( \text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100 \) Here, the actual yield is 1.70 mol and the theoretical yield is 2.00 mol. Thus, the percent yield is \( \frac{1.70\, \text{mol}}{2.00\, \text{mol}} \times 100 = 85 \% \)

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