Write a balanced equation for the reaction that occurs in each of the following cases: (a) Cesium is added to water. (b) Strontium is added to water. (c) Sodium reacts with oxygen. (d) Calcium reacts with iodine.

Short Answer

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(a) Cs + H2O → CsOH + 1/2 H2 (b) Sr + 2H2O → Sr(OH)2 + H2 (c) 4Na + O2 → 2Na2O (d) Ca + I2 → CaI2

Step by step solution

01

(a) Cesium is added to water.

The reactants in this reaction are cesium (Cs) and water (H2O). When cesium reacts with water, it produces cesium hydroxide (CsOH) and hydrogen gas (H2). Unbalanced equation: Cs + H2O → CsOH + H2 To balance this equation, we only need to adjust the coefficients of cesium hydroxide: Balanced equation: Cs + H2O → CsOH + 1/2 H2
02

(b) Strontium is added to water.

The reactants in this reaction are strontium (Sr) and water (H2O). When strontium reacts with water, it produces strontium hydroxide (Sr(OH)2) and hydrogen gas (H2). Unbalanced equation: Sr + H2O → Sr(OH)2 + H2 To balance this equation, we need to adjust the coefficients of water and strontium hydroxide: Balanced equation: Sr + 2H2O → Sr(OH)2 + H2
03

(c) Sodium reacts with oxygen.

The reactants in this reaction are sodium (Na) and oxygen (O2). When sodium reacts with oxygen, it produces sodium oxide (Na2O). Unbalanced equation: Na + O2 → Na2O To balance this equation, we need to adjust the coefficients of sodium and oxygen: Balanced equation: 4Na + O2 → 2Na2O
04

(d) Calcium reacts with iodine.

The reactants in this reaction are calcium (Ca) and iodine (I2). When calcium reacts with iodine, it produces calcium iodide (CaI2). Unbalanced equation: Ca + I2 → CaI2 The equation is already balanced: Balanced equation: Ca + I2 → CaI2

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Key Concepts

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

Chemical Reactions
Chemical reactions are processes where reactants transform into products through breaking and forming of chemical bonds. This transformation follows the law of conservation of mass, meaning that atoms are neither created nor destroyed during a chemical reaction.

For instance, when cesium reacts with water, cesium atoms and water molecules rearrange to form new substances: cesium hydroxide and hydrogen gas. A similar transformation occurs when strontium reacts with water, producing strontium hydroxide and hydrogen gas. These reactions not only show the change from reactants to products but also the importance of balancing the equation to reflect the mass conservation law.
Reactivity of Metals
The reactivity of metals plays a significant role in determining how vigorously a metal reacts with other substances. Metals like cesium and sodium are highly reactive with water and oxygen due to their position in the reactivity series, a list of metals arranged according to their ability to displace other metals in a compound.

For example, cesium displaces hydrogen from water with such vigor because it is a highly reactive metal from the alkali metals group. On the other hand, calcium's reaction with iodine is less vigorous, as calcium is less reactive than cesium or sodium. Understanding the reactivity of metals helps predict the outcome of chemical reactions and the safety precautions necessary during those reactions.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. By following the basic principle that matter cannot be created or destroyed, stoichiometry involves calculating the amounts of reactants required or products formed in a reaction.

For instance, in the reaction of sodium with oxygen, stoichiometry dictates that four sodium atoms react with one molecule of oxygen to produce two formula units of sodium oxide. Here, stoichiometry ensures the balanced equation complies with the law of conservation of mass. Students can use stoichiometric calculations to predict the amount of each substance involved, ensuring the equation is quantitatively accurate and allowing for the proper scaling of reactions for various applications.

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Most popular questions from this chapter

Discussing this chapter, a classmate says, "An element that commonly forms a cation is a metal." Do you agree or disagree?

Write the electron configurations for the following ions, and determine which have noble-gas configurations: \((\mathbf{a})\mathrm{Ru}^{3+}\) \((\mathbf{b}) \mathrm{As}^{3-},(\mathbf{c}) \mathrm{Y}^{3+},(\mathbf{d}) \mathrm{Pd}^{2+},(\mathbf{e}) \mathrm{Pb}^{2+},(\mathbf{f}) \mathrm{Au}^{3+}.\)

Write balanced equations for the following reactions: (a) potassium oxide with water, (b) diphosphorus trioxide with water, (c) chromium(III) oxide with dilute hydrochloric acid, (d) selenium dioxide with aqueous potassium hydroxide.

Which of the following chemical equations is connected to the definitions of (a) the first ionization energy of oxygen, (b) the second ionization energy of oxygen, and (c) the electron affinity of oxygen? \((\mathbf{i})\mathrm{O}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{o}^{-}(g) \quad\) \((\mathbf{ii})\mathrm{O}(g) \longrightarrow \mathrm{o}^{+}(g)+\mathrm{e}^{-}\) \((\mathbf{iii})\mathrm{O}(g)+2 \mathrm{e}^{-} \longrightarrow \mathrm{O}^{2-}(g) \quad(\mathbf{i v}) \mathrm{O}(g) \longrightarrow \mathrm{O}^{2+}(g)+2 \mathrm{e}^{-}\) \((\mathbf{v}) \mathrm{O}^{+}(g) \longrightarrow \mathrm{O}^{2+}(g)+\mathrm{e}^{-}\)

(a) Which ion is smaller, \(\mathrm{Co}^{3+}\) or \(\mathrm{Co}^{4+} ?(\mathbf{b})\) In a lithium-ion battery that is discharging to power a device, for every \(\mathrm{Li}^{+}\) that inserts into the lithium cobalt oxide electrode, a \(\mathrm{Co}^{4+}\) ion must be reduced to a \(\mathrm{Co}^{3+}\) ion to balance charge. Using the CRC Handbook of Chemistry and Physics or other standard reference, find the ionic radii of \(\mathrm{Li}^{+}, \mathrm{Co}^{3+},\) and \(\mathrm{Co}^{4+} .\) Order these ions from smallest to largest. (c) Will the lithium cobalt oxide cathode expand or contract as lithium ions are inserted? (d) Lithium is not nearly as abundant as sodium. If sodium ion batteries were developed that function in the same manner as lithium ion batteries, do you think "sodium cobalt oxide" would still work as the electrode material? Explain. (e) If you don’t think cobalt would work as the redox-active partner ion in the sodium version of the electrode, suggest an alternative metal ion and explain your reasoning.

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