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Chapter 26: Problem 80

A \(10.6 \mathrm{g}\) sample of benzaldehyde was allowed to react with \(5.9 \mathrm{g} \mathrm{KMnO}_{4}\) in an excess of \(\mathrm{KOH}(\mathrm{aq}) .\) After filtration of the \(\mathrm{MnO}_{2}(\mathrm{s})\) and acidification of the solution, \(6.1 \mathrm{g}\) of benzoic acid was isolated. What was the percent yield of the reaction? [ Hint: Write half-equations for the oxidation and reduction half-reactions.

Short Answer

Expert verified
The percent yield of the reaction is approximately 49.96%.

Step by step solution

01

Write the balanced chemical equation

We need to establish the balanced chemical equation for the reaction. Apply the principles of balancing equations separately to the oxidation and reduction half-reactions. The reduction half-reaction is \(5 \mathrm{MnO}_{4}^{-} + 6 \mathrm{H}^{+} + 3 e^{-} \Rightarrow 5 \mathrm{MnO}_{2} + 3 \mathrm{H}_{2} \mathrm{O}\) and the oxidation half-reaction is \(\mathrm{C_{6}H_{5}CHO + H_{2}O \Rightarrow C_{6}H_{5}COOH + 2 \mathrm{H}^{+} + 2 e^{-}\). Add these equations to get the overall balanced reaction.
02

Calculate the theoretical yield

Using the balanced equation, calculate the theoretical yield of benzoic acid. Determine how many moles of benzaldehyde, \(C_{6}H_{5}CHO\), are reacting by dividing the masses by the respective molar masses. The number of moles of benzaldehyde is \(\frac{10.6 \mathrm{g}}{106.12 \mathrm{g/mol}} = 0.10 \mathrm{mol}\). According to the balanced equation, one molecule of benzaldehyde yields one molecule of benzoic acid, so the theoretical yield of benzoic acid is also 0.10 mol. Convert this to grams using the molar mass of benzoic acid. Theoretical yield = \(0.10 \mathrm{mol} \times 122.12 \mathrm{g/mol} = 12.21 \mathrm{g}\)
03

Determine percent yield

Applying the formula for percent yield which is \(\frac{actual \ yield}{theoretical \ yield} \times 100\%\), we compute the percent yield as \(\frac{6.1 \mathrm{g}}{12.21 \mathrm{g}} \times 100\% \approx 49.96\% \)

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Most popular questions from this chapter

Supply condensed or structural formulas for the following substances. (a) 1,5 -cyclooctadiene (an intermediate in the manufacture of resins) (b) 3,7,11 -trimethyl- 2,6,10 -dodecatriene- 1 -ol (farnesol \(-\) odor of lily of the valley) [Hint: Dodeca means \(12 .]\) (c) 2,6 -dimethyl- 5 -hepten- 1 -al (used in the manufacture of perfume).

Supply condensed structural formulas for the following substances. (a) 2,4,6 -trinitrotoluene (TNT-an explosive) (b) methyl salicylate (oil of wintergreen) [Hint: Salicylic acid is \(o \text { -hydroxybenzoic acid. }]\) (c) 2 -hydroxy-1,2,3-propanetricarboxylic acid (citric \(\left.\operatorname{acid}, \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\right)\).

Draw the \(E\) and \(Z\) isomers of (a) 3 -methyl- 3 -hexene; \((b)\) 3-fluoro-2-methyl-3-hexene.

Supply a structural formula for each of the following compounds. (a) 1,3,5 -trimethylbenzene; (b) \(p\) -nitrophenol; (c) 3 -amino- 2,5 -dichlorobenzoic acid (a plant-growth regulator).

Use the half-reaction method to balance the following redox equations. (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}+\mathrm{Fe}+\mathrm{H}^{+} \longrightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}+\mathrm{Fe}^{3+}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{H}^{+} \longrightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}+\mathrm{Cr}^{3+}+\mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{MnO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \longrightarrow\) \(\mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{OH}+\mathrm{MnO}_{2}+\mathrm{OH}^{-}\).
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