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(a) Define the terms theoretical yield, actual yield, and percent yield. (b) Why is the actual yield in a reaction almost always less than the theoretical yield? (c) Can a reaction ever have \(110 \%\) actual yield?

Short Answer

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(a) Theoretical yield is the maximum amount of product that can be obtained from a chemical reaction under ideal conditions. Actual yield is the amount of product actually obtained from a chemical reaction. Percent yield = \(\frac{Actual\,yield}{Theoretical\,yield}\) × 100 (b) Actual yield is usually less than theoretical yield due to factors such as side reactions, incomplete reactions, losses during purification, and presence of impurities. (c) A reaction cannot have a 110% actual yield. The percent yield must be lower than or equal to 100% because actual yield cannot be higher than the theoretical yield.

Step by step solution

01

(a) Definition of theoretical yield, actual yield, and percent yield

Theoretical yield is the maximum amount of product that can be obtained from a chemical reaction under ideal conditions. It is calculated using the stoichiometry of the balanced chemical equation. Actual yield is the amount of product that is actually obtained from a chemical reaction after the completion of the experiment. It is usually less than the theoretical yield due to various factors and limitations. Percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage. It is a measure of the effectiveness of a reaction in producing the desired product. Percent yield = \(\frac{Actual\,yield}{Theoretical\,yield}\) × 100
02

(b) Reason for actual yield being less than theoretical yield

The actual yield is almost always less than the theoretical yield because of various factors that affect the reaction, such as: 1. Side reactions: In addition to the main reaction, there may be other reactions taking place that result in the formation of unwanted products, reducing the amount of desired product. 2. Incomplete reaction: In some cases, not all reactants may react completely, leaving behind some unreacted starting materials, which results in lower yield. 3. Loss during the purification and isolation steps: Some product may be lost during the process of isolation and purification, which will reduce the actual yield. 4. Presence of impurities or non-reproducible conditions: Sometimes, the actual conditions are not the same as the ones assumed for the calculation of the theoretical yield, which can lead to deviations.
03

(c) Possibility of a reaction having 110% actual yield

A reaction cannot have a 110% actual yield. The percent yield must be lower than or equal to 100% because the actual yield cannot be higher than the theoretical yield. If the actual yield is somehow higher than the theoretical yield, it is likely due to errors in the experiment or measurement inaccuracies, and it does not represent a true yield that exceeds the theoretical limit.

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Most popular questions from this chapter

The fermentation of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) produces ethyl alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and \(\mathrm{CO}_{2}\) $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) $$ (a) How many moles of \(\mathrm{CO}_{2}\) are produced when \(0.400\) mol of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) reactsin this fashion? (b) How many grams of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) are needed to form \(7.50 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} ?\) (c) How many grams of \(\mathrm{CO}_{2}\) form when \(7.50 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) are produced?

An element \(X\) forms an iodide \(\left(\mathrm{XI}_{3}\right)\) and a chloride \(\left(\mathrm{XCl}_{3}\right) .\) The iodide is quantitatively converted to the chloride when it is heated in a stream of chlorine: $$ 2 \mathrm{XI}_{3}+3 \mathrm{Cl}_{2} \longrightarrow 2 \mathrm{XCl}_{3}+3 \mathrm{I}_{2} $$ If \(0.5000 \mathrm{~g}\) of \(\mathrm{XI}_{3}\) is treated, \(0.2360 \mathrm{~g}\) of \(\mathrm{XCl}_{3}\) is obtained. (a) Calculate the atomic weight of the element \(\mathrm{X}\) (b) Identify the element \(\bar{X}\).

A chemical plant uses electrical energy to decompose aqueous solutions of \(\mathrm{NaCl}\) to give \(\mathrm{Cl}_{2}, \mathrm{H}_{2}\), and \(\mathrm{NaOH}\) : \(2 \mathrm{NaCl}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \underset{2 \mathrm{NaOH}(a q)}{\longrightarrow}+\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g)\) If the plant produces \(1.5 \times 10^{6} \mathrm{~kg}\left(1500\right.\) metric tons) of \(\mathrm{Cl}_{2}\) daily, estimate the quantities of \(\mathrm{H}_{2}\) and \(\mathrm{NaOH}\) produced.

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Balance the following equations, and indicate whether they are combination, decomposition, or combustion reactions: (a) \(\mathrm{Al}(\mathrm{s})+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{AlCl}_{3}(\mathrm{~s})\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{Li}(s)+\mathrm{N}_{2}(g) \longrightarrow \mathrm{Li}_{3} \mathrm{~N}(s)\) (d) \(\mathrm{PbCO}_{3}(s) \longrightarrow \mathrm{PbO}(s)+\mathrm{CO}_{2}(g)\) (e) \(\mathrm{C}_{7} \mathrm{H}_{8} \mathrm{O}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

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