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In sodalime decarboxylation, sodalime is made of ' \(x\) ' \(\mathrm{NaOH}\) and ' \(y^{\prime} \mathrm{CaO}\), then \(x / y\) is (A) \(1 / 3\) (B) \(1 / 1\) (C) \(3 / 1\) (D) \(5 / 1\)

Short Answer

Expert verified
The short answer is: In sodalime decarboxylation, the ratio of NaOH to CaO is \(3 / 1\).

Step by step solution

01

Understand the composition of sodalime

Sodalime is a combination of sodium hydroxide (NaOH) and calcium oxide (CaO) which is used in decarboxylation reactions. The mixture is made by adding NaOH and CaO in certain proportions. Our task is to find the ratio x / y, where x is the amount of NaOH and y is the amount of CaO in the mixture.
02

Find the ratio of NaOH to CaO in sodalime

In sodalime, 3 parts NaOH reacts with 1 part CaO, forming a new compound Na2Ca(OH)4. This means that for every 3 moles of NaOH, 1 mole of CaO is needed. So, the ratio of NaOH to CaO in sodalime is 3:1.
03

Choose the correct answer

Based on our findings in step 2, the ratio x / y is 3 / 1. Therefore, the correct answer is: (C) \(3 / 1\).

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Key Concepts

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

Sodium Hydroxide and Calcium Oxide Ratio in Sodalime
Understanding the composition of sodalime is crucial for various chemical processes, especially in organic chemistry. Sodalime, which is used in decarboxylation reactions, comprises sodium hydroxide (NaOH) and calcium oxide (CaO). The ratio of NaOH to CaO holds significance as it determines the compound's efficacy in removing carbon dioxide from compounds.

Typically, sodalime is formulated with a 3:1 ratio, meaning three parts of NaOH to one part of CaO. This specific ratio ensures the formation of the compound Na2Ca(OH)4, which plays an important role in the decarboxylation process. It's interesting to note that varying this ratio could affect the performance of sodalime, highlighting the precision required in chemical synthesis.

Moreover, this ratio is important not only for practical laboratory synthesis but also for students preparing for competitive exams like the JEE MAIN and ADVANCED, where such details are often tested.
Organic Chemistry Decarboxylation Reactions
Decarboxylation reactions are a pivotal concept in organic chemistry, where a carboxyl group is eliminated from a molecule, releasing carbon dioxide. This type of reaction is central to understanding various metabolic pathways and synthetic processes.

In organic synthesis, decarboxylation is often employed to shorten carbon chains, which can be pivotal for creating desired modifications in a molecular structure. The sodalime decarboxylation is one such example where a carboxylic acid salt reacts with sodalime, resulting in the loss of a carbon dioxide molecule and the formation of a hydrocarbon.

For students preparing for exams such as the JEE MAIN and ADVANCED, mastering decarboxylation reactions is essential. These exams frequently test the student's understanding of reaction mechanisms and their ability to apply them in problem-solving.
JEE MAIN and ADVANCED Organic Chemistry
The JEE MAIN and ADVANCED exams are highly competitive entrance examinations for aspiring engineering undergraduates in India. In these examinations, organic chemistry represents a challenging yet rewarding section.

Candidates are tested on their depth of understanding and problem-solving skills in organic reactions, including decarboxylation. The applications of the mole concept, understanding reagent behavior, reaction mechanisms, and synthetic pathways are all fair game in these exams.

Students are advised to gain hands-on experience with a wide range of reactions, such as sodalime decarboxylation, to be well-prepared. Knowing the intricacies, like the sodium hydroxide and calcium oxide ratio in sodalime, can be the difference between a correct and an incorrect answer in such competitive exams.
Mole Concept in Chemistry
The mole concept is a fundamental cornerstone in the field of chemistry. It relates the quantity of substance to the number of entities it contains, bridging the gap between the macroscopic and microscopic worlds.

In the context of the sodium hydroxide and calcium oxide ratio in sodalime, the mole concept allows chemists to quantify the exact proportions of each compound required for the desired chemical reaction. By understanding that a 3:1 mole ratio is necessary, chemists can predictably generate sodalime with consistent and effective results.

This concept is also integral to students studying chemistry, particularly those preparing for exams like the JEE MAIN and ADVANCED, where questions often revolve around calculating reactants, products, and their relationships in chemical equations. A clear understanding of the mole concept enables students to tackle complex problems with confidence.

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Most popular questions from this chapter

The following conversion can be accomplished by: $$ \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCOOH} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{2} \mathrm{OH} $$ (A) \(\mathrm{NaBH}_{4}\) (B) \(\mathrm{Al}\left(\mathrm{Me}_{2} \mathrm{CHO}\right)_{3}, \mathrm{Me}_{2} \mathrm{CHOH}\) (C) \(\mathrm{H}_{2}, \mathrm{Ni}\) (D) \(\mathrm{LiAlH}_{4}\)

Choose the correct option(s): $$ \mathrm{NBS} \stackrel{\mathrm{NaOBr}}{\longrightarrow} \mathrm{P} \stackrel{\mathrm{NaNO}_{2} / \mathrm{HCl}}{\longrightarrow} \mathrm{Q} \stackrel{\Delta}{\longrightarrow} \mathrm{R} $$ (A) (P) on strong heating will produce (R) (B) Formation of (Q) involves \(\mathrm{N}_{2}\) as leaving group (C) Formation of (P) is Hofmann bromamide degradation (D) Rearrangment step is rate determining step to form (P)

Total number of reactions in which carboxylic acid is obtained as a product is (a) \(\mathrm{R}-\mathrm{Cl} \frac{\text { (i) } \mathrm{Mg}}{\text { (ii) } \mathrm{CO}_{2}}\) (b) \(\mathrm{R}-\mathrm{Cl} \frac{\text { (i) } \mathrm{KCN}}{\text { (ii) } \mathrm{H}_{3} \mathrm{O}^{+}}\) (c) \(\mathrm{R}-\mathrm{CHO} \frac{\mathrm{CrO}_{3}}{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow}\) (d) \(\mathrm{R}-\mathrm{C}-\mathrm{Cl} \stackrel{\mathrm{H}_{3} \mathrm{O}^{+}}{\longrightarrow}\) (e) \(\mathrm{R}-\mathrm{CH}_{2} \mathrm{OH} \frac{\mathrm{CrO}_{3}}{\mathrm{H}_{2} \mathrm{SO}_{4}}\) (f) \(\mathrm{R}-\mathrm{CHO} \stackrel{\mathrm{Ag}_{2} \mathrm{O}}{\longrightarrow}\) (g) \(\mathrm{R}-\mathrm{NC} \stackrel{\mathrm{H}_{3} \mathrm{O}^{+}}{\longrightarrow}\)

The decreasing order of rate of esterification of the following acids with MeOH is (I) \(\mathrm{Me}-\mathrm{CH}_{2} \mathrm{COOH}\) (II) Me \(_{2} \mathrm{CHCOOH}\) (III) \(\mathrm{Me}_{3} \mathrm{C}-\mathrm{COOH}\) (A) \(\mathrm{I}>\mathrm{II}>\mathrm{III}\) (B) III \(>\mathrm{II}>\mathrm{I}\) (C) II \(>\mathrm{III}>\mathrm{I}\) (D) II > I > III

\((\mathrm{P}) \stackrel{\mathrm{Red} \mathrm{P} / \mathrm{X}_{2}}{\longrightarrow}(\mathrm{Q})\) (P) has M.F. \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{2}\) which gives fruity smell with alcohol and ' \(\mathrm{X}^{\prime}\) is halogen. Choose the incore: option: (A) Product (Q) is \(\beta\)-halocarboxylic acid (B) Reaction is possible when \(\alpha\)-hydrogen is present (C) Reaction is known as HVZ reaction (D) Red \(\mathrm{P}\) is taken in small amount and ' \(\mathrm{X}\) ' can be chlorine or bromine.

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