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Chapter 17: Problem 22

Sodium is heated in air at \(300^{\circ} \mathrm{C}\) to form \(\mathrm{X} . \mathrm{X}\) absorbs \(\mathrm{CO}_{2}\) and forms \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{Y}\). Which of the following is \(\mathrm{Y}\) ? (a) \(\mathrm{O}_{2}\) (b) \(\mathrm{O}_{3}\) (c) \(\mathrm{H}_{2}\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}\)

Short Answer

Expert verified
The compound \( \mathrm{Y} \) formed is \( \mathrm{O}_2 \).

Step by step solution

01

Understanding the Reaction of Sodium with Air

When sodium is heated in air, it reacts primarily with oxygen to form sodium oxide, \ \( \mathrm{Na}_2\mathrm{O} \). This is the compound referred to as \( \mathrm{X} \).
02

Reaction of Sodium Oxide with Carbon Dioxide

The sodium oxide \( \mathrm{Na}_2\mathrm{O} \) then reacts with carbon dioxide \( \mathrm{CO}_2 \) to form sodium carbonate \( \mathrm{Na}_2\mathrm{CO}_3 \). The reaction is: \ \( \mathrm{Na}_2\mathrm{O} + \mathrm{CO}_2 \rightarrow \mathrm{Na}_2\mathrm{CO}_3 \). In this reaction, the product \( \mathrm{Y} \) along with \( \mathrm{Na}_2\mathrm{CO}_3 \) needs to be determined.
03

Analyzing Possible Compounds for \( \mathrm{Y} \)

Since no leftover or additional gases are produced other than \( \mathrm{Na}_2\mathrm{CO}_3 \), the compound \( \mathrm{Y} \) should be neutral or result from the stoichiometry of the equation. Comparing the given options: none of them are directly formed except for neutral residual oxygen.
04

Determining the Nature of \( \mathrm{Y} \) Given the Equation Context

In the reaction of \( \mathrm{Na}_2\mathrm{O} \) and \( \mathrm{CO}_2 \), no extra product is visible at simple stoichiometry levels besides theoretical completeness. Hence, \( \mathrm{Y} \) is most likely extraneous or a residual by-product related to extra oxygen not involved directly but remaining in a form such as \( \mathrm{O}_2 \).
05

Confirming \( \mathrm{Y} \) from Given Options

Choosing from the options, the answer should logically be related to an oxygen-related molecule since \( \mathrm{O}_2 \) could be residual from the initial air and heating reaction, making the possible choice (a) \( \mathrm{O}_2 \) correct considering the compound nature and reaction condition.

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Key Concepts

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

Sodium Reactions
Sodium is a highly reactive metal that readily undergoes reactions with various elements, especially when exposed to air. In its pure form, sodium is stored in oil to prevent such reactions from occurring inadvertently. When heated at high temperatures, such as 300掳C, sodium will interact with oxygen in the air. This interaction is primarily due to sodium's high reactivity and the abundant availability of oxygen in the air.
  • When heated in air, sodium reacts to form sodium oxide (\( \mathrm{Na}_2\mathrm{O} \)).
  • This reaction showcases sodium's nature to form stable oxide layers by combining with oxygen.
  • Sodium oxide is a significant compound formed due to heating, indicative of sodium's affinity for oxygen at elevated temperatures.
Understanding sodium reactions with air is crucial in industrial and laboratory settings, ensuring proper handling and safety protocols.
Heating Metals
Heating metals is a common practice in chemistry to induce reactions that are not otherwise visible under room temperature conditions. Each metal has its specific reactivity profile when exposed to high temperatures, and sodium is particularly interesting because of its vigorous nature.
  • Upon heating, sodium reacts exothermically, meaning it releases energy as it forms sodium oxide.
  • The reaction's outcome is significantly temperature-dependent, with 300掳C being optimal for sodium to produce \( \mathrm{Na}_2\mathrm{O} \).
  • This process serves as a demonstration of sodium鈥檚 energetic behavior and the importance of controlled heating environments.
Proper understanding of these reactions highlights the necessity of safety measures and containment when dealing with reactive metals like sodium.
Sodium Oxide Reactions
Sodium oxide (\( \mathrm{Na}_2\mathrm{O} \)), formed when sodium is heated in air, is itself a reactive compound. Its interaction with other chemicals makes it practical for various chemical processes, such as the production of sodium carbonate. When \( \mathrm{Na}_2\mathrm{O} \) engages with carbon dioxide, a notable reaction occurs that provides insight into inorganic synthesis pathways.
  • The reaction formula is \( \mathrm{Na}_2\mathrm{O} + \mathrm{CO}_2 \rightarrow \mathrm{Na}_2\mathrm{CO}_3 \).
  • This reaction demonstrates how a basic oxide like sodium oxide can capture carbon dioxide to form a carbonate.
  • Sodium carbonate (\( \mathrm{Na}_2\mathrm{CO}_3 \)) is a product of such reactions, with applications in detergents, glass production, and more.
Awareness of sodium oxide's reactive properties ensures useful applications in chemical and manufacturing arenas.
Carbon Dioxide Reaction
Carbon dioxide (\( \mathrm{CO}_2 \)) is a common gaseous compound in various chemical and environmental processes. Its ability to react with sodium oxide highlights its utility beyond its role as a greenhouse gas. This reaction offers an example of how carbon dioxide can be transformed into another chemically valuable product, sodium carbonate.
  • This transformation is a demonstration of \( \mathrm{CO}_2 \)'s reactivity, leading to the formation of stable carbonate compounds.
  • Sodium carbonate formed through this reaction is seen in everyday products, making the reaction industrially significant.
  • Managing \( \mathrm{CO}_2 \) through chemical reactions not only aids in manufacturing but also opens pathways for reducing atmospheric carbon.
Understanding these reactions helps in developing strategies for carbon capture and utilization, a focus of modern environmental chemistry.

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Most popular questions from this chapter

Match the following \begin{tabular}{ll} \hline List-I & List-II \\ \hline (a) Aqueous solution of (p) Anhydrone \(\mathrm{NaOCl}\) & (q) Javelle water \\ (b) \(\mathrm{KClO}_{3}\) & (r) Euchlorine \\ (c) \(\mathrm{Mg}\left(\mathrm{ClO}_{4}\right)_{2}\) & (s) Berthelot's salt \\ (d) \(\mathrm{Cl}_{2}\) and \(\mathrm{ClO}_{2}\) mixture & \end{tabular}

The molecular formula of potash alum is (a) \(\mathrm{K}_{2} \mathrm{Al}_{2} \mathrm{~S}_{2} \mathrm{H}_{48} \mathrm{O}_{39}\) (b) \(\mathrm{KAl}_{2} \mathrm{~S}_{2} \mathrm{H}_{48} \mathrm{O}_{40}\) (c) \(\mathrm{K}_{2} \mathrm{AlS}_{2} \mathrm{H}_{48} \mathrm{O}_{40}\) (d) \(\mathrm{K}_{2} \mathrm{Al}_{2} \mathrm{~S}_{2} \mathrm{H}_{48} \mathrm{O}_{40}\)

\(1.04 \mathrm{~g}\) of bleaching powder was made into a paste with water and then the volume was made upto 200 ml. \(25 \mathrm{ml}\) of this solution was found to oxidize \(13.4\) \(\mathrm{ml}\) of a standard solution of arsenic containing \(4.3 \mathrm{~g}\) \(\mathrm{Al}_{2} \mathrm{O}_{3}\) per litre. Find the percentage available chlorine in the sample. (a) \(31.7 \%\) (b) \(36.7 \%\) (c) \(41.7 \%\) (d) \(63.6 \%\)

Which of the following changes occur when an excess of \(\mathrm{CO}_{2}\) gas is passed into a clear solution of lime water? (a) a white precipitate containing both \(\mathrm{CaCO}_{3}\) and \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\) is formed (b) initially a white precipitate of \(\mathrm{CaCO}_{3}\) is formed which changes into soluble \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\) on passing \(\mathrm{CO}_{2}\) gas (c) a white precipitate of \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\) is formed (d) a white precipitate of \(\mathrm{CaCO}_{3}\) is formed

Fluorosis disease is caused due to the reaction of \(\ldots \ldots\). with excess of fluoride in the body. (a) \(\mathrm{Mg}\) (b) \(\mathrm{Fe}\) (c) \(\mathrm{K}\) (d) \(\mathrm{Ca}\)
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