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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Theoretical Yield
The theoretical yield is a term used in chemistry to describe the maximum amount of product that can be created from a given amount of reactants. It is calculated under the assumption that the reaction goes to completion without any losses. This means all the reactants are fully converted into the products as described by the balanced chemical equation.

To calculate the theoretical yield, you need two main things:
  • The balanced chemical equation for the reaction, which tells you the ratio of reactants to products.
  • The amount of reactants that you have, often measured in grams or moles.
Once you have these, convert the mass of reactants into moles, find the limiting reactant (the reactant that runs out first), and use the stoichiometry of the reaction to find the moles of the desired product. Finally, convert these moles back into grams. This calculated mass is the theoretical yield of the reaction.

In our example with azobenzene, we calculated the theoretical yield to be 89 grams, using 0.97 moles of nitrobenzene and considering the stoichiometry of the reaction.
Actual Yield
In contrast to the theoretical yield, the actual yield is the amount of product actually produced when the chemical reaction is carried out in a laboratory or industrial setting. It is often less than the theoretical yield due to various factors like incomplete reactions, side reactions, or loss of product during recovery.

For any given reaction, the actual yield can be measured by weighing the amount of product formed and purified. It is crucial from a practical perspective because it reflects real-world efficiency.

In the problem above, the actual yield of azobenzene is given as 55 grams. This means 55 grams of azobenzene were successfully recovered after the reaction under the specified conditions.
Stoichiometry
Stoichiometry is the part of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. By relying on stoichiometry, chemists can predict how much of each substance is needed or produced, which is essential for both laboratory work and industrial processes.

A balanced chemical equation gives the stoichiometric coefficients, which are numbers that indicate the proportions of reactants and products. These coefficients allow you to determine the amount of product from a given amount of reactants and vice versa.

In our reaction to produce azobenzene, the stoichiometry is defined by the balanced equation:
  • 2 moles of nitrobenzene react with 4 moles of triethylene glycol to produce 1 mole of azobenzene.
Using stoichiometry, we calculated that with 0.97 moles of nitrobenzene, assuming it was the limiting reagent, the maximum amount of azobenzene that could be formed would be 0.49 moles, which equates to a theoretical yield of 89 grams.
Chemical Reactions
Chemical reactions involve the breaking and forming of chemical bonds to transform reactants into products. These transformations often accompany a change in energy and can be influenced by a variety of factors. Each reaction is carefully represented by a chemical equation, which balances the number of atoms for each element on both sides of the reaction.

Reactions can be categorized into various types such as synthesis, decomposition, single replacement, and double replacement, among others. In this context, the reaction involves the conversion of nitrobenzene into azobenzene in the presence of triethylene glycol, Zn, and KOH.

Understanding the mechanics of chemical reactions is crucial for deciphering how reactants interact to form products, what conditions influence these processes, and how to control them for optimal yield.

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Most popular questions from this chapter

Hydrogen gas, \(\mathrm{H}_{2}(\mathrm{g}),\) is passed over \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})\) at \(400^{\circ} \mathrm{C} .\) Water vapor is formed together with a black residue-a compound consisting of \(72.3 \% \mathrm{Fe}\) and \(27.7 \%\) O. Write a balanced equation for this reaction.

A \(25.00 \mathrm{mL}\) sample of \(\mathrm{HCl}(\mathrm{aq})\) was added to a \(0.1000 \mathrm{g}\) sample of \(\mathrm{CaCO}_{3}\). All the \(\mathrm{CaCO}_{3}\) reacted, leaving some excess HCl(aq). \(\mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) $$ The excess HCl(aq) required 43.82 mL of 0.01185 M \(\mathrm{Ba}(\mathrm{OH})_{2}\) to complete the following reaction. What was the molarity of the original HCl(aq)? $$2 \mathrm{HCl}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \longrightarrow \mathrm{BaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$

A drop \((0.05 \mathrm{mL})\) of \(12.0 \mathrm{M} \mathrm{HCl}\) is spread over a sheet of thin aluminum foil. Assume that all the acid reacts with, and thus dissolves through, the foil. What will be the area, in \(\mathrm{cm}^{2}\), of the cylindrical hole produced? (Density of \(\mathrm{Al}=2.70 \mathrm{g} / \mathrm{cm}^{3} ;\) foil thickness \(=0.10 \mathrm{mm} .)\) \(2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{g})\)

A 99.8 mL sample of a solution that is \(120 \%\) KI by mass \((d=1.093 \mathrm{g} / \mathrm{mL})\) is added to \(96.7 \mathrm{mL}\) of another solution that is \(14.0 \% \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) by mass \((d=1.134 \mathrm{g} / \mathrm{mL})\) How many grams of \(\mathrm{PbI}_{2}\) should form? \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{KI}(\mathrm{aq}) \longrightarrow \mathrm{PbI}_{2}(\mathrm{s})+2 \mathrm{KNO}_{3}(\mathrm{aq})\)

Chalkboard chalk is made from calcium carbonate and calcium sulfate, with minor impurities such as \(\mathrm{SiO}_{2} .\) Only the \(\mathrm{CaCO}_{3}\) reacts with dilute \(\mathrm{HCl}(\mathrm{aq})\) What is the mass percent \(\mathrm{CaCO}_{3}\) in a piece of chalk if a 3.28 -g sample yields \(0.981 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g}) ?\) $$\begin{aligned} \mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) & \longrightarrow \mathrm{CaCl}_{2}(\mathrm{aq}) +\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) \end{aligned}$$

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