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Chapter 4: Problem 81

An essentially \(100 \%\) yield is necessary for a chemical reaction used to analyze a compound, but it is almost never expected for a reaction that is used to synthesize a compound. Explain this difference.

Short Answer

Expert verified
While analyzing a compound necessitates a \(100 \%\) yield to account for all components of the compound, synthesizing a compound often results in less than a \(100 \%\) yield due to factors such as side reactions, loss during purification and limitations in reaction efficiency.

Step by step solution

01

Understand the difference between Analyzing and Synthesizing Compounds

Analyzing a compound involves breaking it down to understand its structure or composition. Synthesizing a compound, on the other hand, involves creating it from simpler chemicals.
02

Define Yield in the context of Chemical Reactions

In a chemical reaction, the 'yield' refers to the amount of product that is formed. A \(100 \%\) yield indicates that the entire starting material was converted into the desired product.
03

Explain why a \(100 \%\) yield is necessary for Analysis

When analyzing a compound, one needs to account for all the components of the compound. Thus, a \(100 \%\) yield is expected because the entire compound should be broken down and accounted for, meaning all of the starting material must be accounted for in the analysis. Any less than a \(100 \%\) yield would suggest incomplete analysis.
04

Explain why a \(100 \%\) yield is almost never expected for Synthesis

In contrast, when synthesizing a compound, it is often not possible to convert all of the starting material into the desired product due to various factors such as side reactions, loss during purification and limitations in reaction efficiency. As such, less than a \(100 \%\) yield is often expected in synthesis.

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Most popular questions from this chapter

Aluminum metal and iron(III) oxide react to give aluminum oxide and iron metal. What is the maximum mass of iron that can be obtained from a reaction mixture containing \(2.5 \mathrm{g}\) of aluminum and \(9.5 \mathrm{g}\) of iron(III) oxide. What mass of the excess reactant remains?

Without per forming detailed calculations, which of the following yields the same mass of \(\mathrm{CO}_{2}(\mathrm{g})\) per gram of compound as does ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) when burned in excess oxygen? (a) \(\mathrm{H}_{2} \mathrm{CO}\); (b) \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH} ;\) (c) \(\mathrm{HOCH}_{2} \mathrm{CHOHCH}_{2} \mathrm{OH}\) (d) \(\mathrm{CH}_{3} \mathrm{OCH}_{3} ;\) (e) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\)

Consider the reaction below. \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{CaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (a) How many grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) are required to react completely with \(415 \mathrm{mL}\) of \(0.477 \mathrm{M} \mathrm{HCl} ?\) (b) How many kilograms of \(\mathrm{Ca}(\mathrm{OH})_{2}\) are required to react with 324 L of a HCl solution that is 24.28\% HCl by mass, and has a density of \(1.12 \mathrm{g} / \mathrm{mL} ?\)

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)? $$2 \mathrm{HCl}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \longrightarrow \mathrm{BaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$

The reaction of potassium superoxide, \(\mathrm{KO}_{2}\), is used in life- support systems to replace \(\mathrm{CO}_{2}(\mathrm{g})\) in expired air with \(\mathrm{O}_{2}(\mathrm{g}) .\) The unbalanced chemical equation for the reaction is given below. $$\mathrm{KO}_{2}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \longrightarrow \mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g})$$ (a) How many moles of \(\mathrm{O}_{2}(\mathrm{g})\) are produced by the reaction of \(156 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g})\) with excess \(\mathrm{KO}_{2}(\mathrm{s}) ?\) (b) How many grams of \(\mathrm{KO}_{2}(\mathrm{s})\) are consumed per \(100.0 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g})\) removed from expired air? (c) How many \(\mathrm{O}_{2}\) molecules are produced per milligram of \(\mathrm{KO}_{2}\) consumed?
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