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Chapter 4: Problem 73

In the reaction of \(277 \mathrm{g} \mathrm{CCl}_{4}\) with an excess of \(\mathrm{HF}\) \(187 \mathrm{g} \mathrm{CCl}_{2} \mathrm{F}_{2}\) is obtained. What are the (a) theoretical, (b) actual, and (c) percent yields of this reaction? $$\mathrm{CCl}_{4}+2 \mathrm{HF} \longrightarrow \mathrm{CCl}_{2} \mathrm{F}_{2}+2 \mathrm{HCl}$$

Short Answer

Expert verified
The theoretical yield is \(217.64\, g\), the actual yield is \(187\, g\), and the percent yield is \(85.9%\).

Step by step solution

01

Calculate Moles of CCl4

First, one needs to calculate the number of moles of carbon tetrachloride (\(CCl_4\)). This can be done by using the formula \(moles = \frac{mass}{molar\ mass}\), where the molar mass of \(CCl_4\) (carbon tetrachloride) is \(153.823 \, g/mol\). Thus, the moles are \(moles = \frac{277\, g}{153.823\, g/mol} = 1.80 \, mol\).
02

Calculate Theoretical Yield

Next, the theoretical yield of \(CCl_2F_2\) must be calculated. Since the reaction of \(1 \, mole\) of \(CCl_4\) produces \(1 \, mole\) of \(CCl_2F_2\), therefore the theoretical yield of \(CCl_2F_2\) is equal to moles of \(CCl_4\), which is \(1.80 \, mol\). To change moles into grams, use the molar mass of \(CCl_2F_2 = 120.913\, g/mol\), so the theoretical yield is \(120.913 \, g/mol \times 1.80 \, mol = 217.64 \, g\) of \(CCl_2F_2\).
03

Calculate Actual Yield

The problem already provides the actual yield, confirming that in the actual reaction, \(187\, g\) of \(CCl_2F_2\) are produced.
04

Calculate Percent Yield

Finally, calculate the percent yield using the formula \[\% yield = \frac{actual\ yield}{theoretical\ yield} \times 100\%\] Given that the actual yield is \(187\, g\) and the theoretical yield is \(217.64\, g\), the percent yield computes as \(\% yield = \frac{187\, g}{217.64\, g} \times 100\% = 85.9\%\).

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Most popular questions from this chapter

A 10.00 mL sample of \(2.05 \mathrm{M} \mathrm{KNO}_{3}\) is diluted to a volume of \(250.0 \mathrm{mL}\). What is the concentration of the diluted solution?

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