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Chapter 4: Problem 78

Azobenzene, an intermediate in the manufacture of dyes, can be prepared from nitrobenzene by reaction with triethylene glycol in the presence of \(\mathrm{Zn}\) and KOH. In one reaction, 0.10 L of nitrobenzene \((d=1.20 \mathrm{g} / \mathrm{mL})\) and \(0.30 \mathrm{L}\) of triethylene glycol \((d=1.12 \mathrm{g} / \mathrm{mL})\) yields \(55 \mathrm{g}\) azobenzene. What are the (a) theoretical yield, (b) actual yield, and (c) percent yield of this reaction? $$ 2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}+4 \mathrm{C}_{6} \mathrm{H}_{14} \mathrm{O}_{4} \frac{\mathrm{Zn}}{\mathrm{KOH}} $$ nitrobenzene triethylene glycol $$ \left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{N}\right)_{2}+4 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{4}+4 \mathrm{H}_{2} \mathrm{O} $$ azobenzene

Short Answer

Expert verified
Theoretical yield is 89g, actual yield is 55g, and the percent yield of the reaction is 61.8%.

Step by step solution

01

Calculate theoretical yield

First, convert the volume of each reactant to mass using the provided densities, and then convert each to moles using the molecular weights. So, for nitrobenzene: \( \text{mass} = \text{density} × \text{volume} = (1.20g/mL) × (100mL) = 120g \). The molar mass of nitrobenzene \( \text{C}_6\text{H}_5\text{NO}_2 \) is approximately 123.12g/mol, so \( 120g ÷ 123.12g/mol = 0.97 mol \). For triethylene glycol, the calculations will be similar. According to the chemical equation, it takes 2 mol of nitrobenzene to produce 1 mol of azobenzene, this is approximately \( 0.97 mol ÷ 2 = 0.49 mol \). Therefore, the theoretical yield is \( 0.49 mol × 182g/mol = 89g. \)
02

Identify actual yield

The actual yield of the reaction is given in the problem as 55g.
03

Calculate percent yield

The percent yield is calculated by dividing the actual yield by the theoretical yield, and multiplying the result by 100. So, the percent yield is \( (55g ÷ 89g) x 100 = 61.8% \).

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Most popular questions from this chapter

Under appropriate conditions, copper sulfate, potassium chromate, and water react to form a product containing \(\mathrm{Cu}^{2+},\) \(\mathrm{CrO}_{4}{^2}{^-},\) and \(\mathrm{OH}^{-}\) ions. Analysis of the compound yields \(48.7 \% \mathrm{Cu}^{2+}, 35.6 \% \mathrm{CrO}_{4}{^2}{-},\) and \(15.7 \% \mathrm{OH}^{-}\). (a) Determine the empirical formula of the compound. (b) Write a plausible equation for the reaction.

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