Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Write the chemical equation for the reaction between (a) sodium and oxygen; (b) lithium and nitrogen; (c) sodium and water; (d) potassium superaxide and water.

Short Answer

Expert verified
(a) \(4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O}\); (b) \(6\text{Li} + \text{N}_2 \rightarrow 2\text{Li}_3\text{N}\); (c) \(2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2\); (d) \(4\text{KO}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{KOH} + 3\text{O}_2 + 2\text{H}_2\text{O}_2\).

Step by step solution

01

Sodium and Oxygen Reaction

Determine the reactants and products. Sodium (\text{Na}) reacts with oxygen (\text{O}_2) to form sodium oxide (\text{Na}_2\text{O}). Balance the chemical equation by ensuring the same number of each type of atom on both sides of the equation. The balanced chemical equation is: \[4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O}\]
02

Lithium and Nitrogen Reaction

Identify the reactants and products. Lithium (\text{Li}) reacts with nitrogen (\text{N}_2) to form lithium nitride (\text{Li}_3\text{N}). Balance the chemical equation by ensuring that there are equal numbers of each type of atom on both sides of the equation. The balanced equation is: \[6\text{Li} + \text{N}_2 \rightarrow 2\text{Li}_3\text{N}\]
03

Sodium and Water Reaction

Determine the reactants and products. Sodium (\text{Na}) reacts with water (\text{H}_2\text{O}) to produce sodium hydroxide (\text{NaOH}) and hydrogen gas (\text{H}_2). Balance the chemical equation to maintain the law of conservation of mass. The balanced chemical equation is: \[2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2\]
04

Potassium Superoxide and Water Reaction

Identify reactants and products. Potassium superoxide (\text{KO}_2) reacts with water (\text{H}_2\text{O}) to form potassium hydroxide (\text{KOH}), oxygen (\text{O}_2), and hydrogen peroxide (\text{H}_2\text{O}_2). To balance the equation, ensure equal numbers of each type of atom on both sides. The balanced equation is: \[4\text{KO}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{KOH} + 3\text{O}_2 + 2\text{H}_2\text{O}_2\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Balancing Chemical Equations
The delicate art of balancing chemical equations lies at the heart of chemistry, ensuring that the law of conservation of mass is upheld. Simply put, what goes into a chemical reaction must equal what comes out. For example, when sodium (\text{Na}) combines with oxygen (\text{O}_2), we get sodium oxide (\text{Na}_2\text{O}). To balance this reaction, you need to ensure that the number of sodium and oxygen atoms is the same on both sides. The balanced representation of this reaction is: \[\begin{equation}4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O}ewline\end{equation}\] In a nutshell, balancing chemical equations involves tweaking the coefficients (the numbers before the chemical formulas) to get the same number of each type of atom on both sides of the reaction arrow.
Reaction Stoichiometry
Reaction stoichiometry can be thought of as the recipe for a chemical reaction, detailing the exact proportions of reactants needed to generate desired products. It tells us not just what substances react, but how much of each substance is involved. For instance, in the reaction between lithium (\text{Li}) and nitrogen (\text{N}_2) to yield lithium nitride (\text{Li}_3\text{N}), stoichiometry comes into play to balance the equation as: \[\begin{equation}6\text{Li} + \text{N}_2 \rightarrow 2\text{Li}_3\text{N}ewline\end{equation}\] Reaction stoichiometry is crucial for understanding the quantities involved in a reaction, which is important in both laboratory work and industrial processes. If we don't know the correct proportions, we can't produce compounds efficiently or predict the outcome of reactions.
Alkali Metal Reactions
Alkali metals, such as sodium (\text{Na}) and potassium (\text{K}), are known for their vigorous reactions with other substances, including water. When sodium reacts with water, it forms sodium hydroxide (\text{NaOH}) and hydrogen gas (\text{H}_2), as shown in the equation:\[\begin{equation}2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2ewline\end{equation}\] These reactions are highly exothermic, releasing a significant amount of heat. They're a perfect illustration of the reactivity of alkali metals, and understanding these reactions is fundamental for further studies in inorganic chemistry and safety procedures in chemical handling.
Inorganic Chemistry
Inorganic chemistry is the branch of chemistry concerned with the properties and behaviors of inorganic compounds, which are compounds not containing carbon atoms. This field covers a vast range of substances, including metals, minerals, and organometallic compounds. The reaction between potassium superoxide (\text{KO}_2) and water is an interesting example from inorganic chemistry. This reaction forms potassium hydroxide (\text{KOH}), oxygen (\text{O}_2), and hydrogen peroxide (\text{H}_2\text{O}_2), as captured in the balanced equation:\[\begin{equation}4\text{KO}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{KOH} + 3\text{O}_2 + 2\text{H}_2\text{O}_2ewline\end{equation}\] Such reactions highlight the diversity of inorganic compounds and their transformations, as well as the importance of understanding their behavior for applications in various fields, from environmental to industrial.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Predict the products of each of the following reactions and then balance each equation: (a) \(\mathrm{Mg}(\mathrm{s})+\mathrm{Br}_{2}(\mathrm{l}) \rightarrow\) (b) \(\mathrm{BaO}(s)+\mathrm{Al}(\mathrm{s}) \rightarrow\) (c) \(\mathrm{CaO}(\mathrm{s})+\mathrm{SiO}_{2}(\) s \() \rightarrow\)

(a) Arrange the hypohalous acids in order of increasing acid strength. (b) Saggest an interpretation of that order in terms of electronegativities.

The azide ion has an ionic radess of \(148 \mathrm{pm}\) and forms many ionic and covalent compounds that are similar to those of the halides. (a) Write the Lewis formula for the azide ion and predict the \(\mathrm{N}-\mathrm{N}-\mathrm{N}\) bond angle. (b) On the basis of its ionic radius, where in Ciroup 17/VII would you place the azide ion? (c) Compare the acidity of hydrazoic acid with those of the hydrohalic acids and explain any differences (for \(\mathrm{HN}_{3}, K_{2}=1.7 \times\) \(\left.10^{-5}\right\rangle\). (d) Write the formulas of three jonic or coralent azides.

Explain from structural considerations why the dipole moment of the \(\mathrm{NH}_{3}\) molecule is lower than that of the \(\mathrm{H}_{2} \mathrm{O}\) molecule.

a When Xe and Pt are treated wath fuorine gas under pressure at \(200^{0} \mathrm{C}\), the ionic solid \(\mathrm{XeF}_{5}+\mathrm{PtF}_{6}^{-}\)is formed. Write the Lewis structure for \(\mathrm{XeF}_{5}+\) and, from VSEPR theory, predict its shape. (b) When two molecules of XeF_ react with one molecule of AsF 5 , the resulting compound has the formala \(\mathrm{Xe}_{2} \mathrm{~F}_{3}{ }^{+} \mathrm{AsF}_{6}{ }^{-}\). Predict the Lewis structure of the \(\mathrm{Xe}_{2} \mathrm{~F}_{3}{ }^{*}\) ion.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free