/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 134 Complete and balance the equatio... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Complete and balance the equations for the following reactions. a. \(\mathrm{Cs}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow\) b. \(\mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \rightarrow\)

Short Answer

Expert verified
The balanced equations for the given reactions are: a. \[\mathrm{2Cs}(s)+\mathrm{2H}_{2} \mathrm{O}(l) \rightarrow \mathrm{2CsOH}(aq)+\mathrm{H}_{2}(g)\] b. \[\mathrm{2Na}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{2NaCl}(s)\]

Step by step solution

01

a. Cesium and Water Reaction

In this reaction, cesium (Cs) reacts with water (Hâ‚‚O) to form cesium hydroxide (CsOH) and hydrogen gas (Hâ‚‚): Unbalanced Equation: \[\mathrm{Cs}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{CsOH}(aq)+\mathrm{H}_{2}(g)\] Now we have to balance the equation. We can do this by placing coefficients in front of the compounds to achieve an equal number of atoms on both sides of the equation. Balanced Equation: \[\mathrm{2Cs}(s)+\mathrm{2H}_{2} \mathrm{O}(l) \rightarrow \mathrm{2CsOH}(aq)+\mathrm{H}_{2}(g)\]
02

b. Sodium and Chlorine Reaction

In this reaction, sodium (Na) reacts with chlorine (Clâ‚‚) to form sodium chloride (NaCl): Unbalanced Equation: \[\mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{NaCl}(s)\] Now we have to balance the equation. We can do this by placing coefficients in front of the compounds to achieve an equal number of atoms on both sides of the equation. Balanced Equation: \[\mathrm{2Na}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{2NaCl}(s)\] #Summary# Now we have balanced the chemical equations for both reactions. Remember, the process involves identification and writing of the products of the reaction followed by balancing the equation by adjusting the coefficients to make the number of atoms of each element the same on both sides. The final balanced equations are: a. \[\mathrm{2Cs}(s)+\mathrm{2H}_{2} \mathrm{O}(l) \rightarrow \mathrm{2CsOH}(aq)+\mathrm{H}_{2}(g)\] b. \[\mathrm{2Na}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{2NaCl}(s)\]

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 91Ó°ÊÓ!

Key Concepts

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

Cesium and Water Reaction
When cesium, which is a highly reactive alkali metal, comes into contact with water, a vigorous reaction occurs. During this process, cesium displaces hydrogen from water to form cesium hydroxide (\text{CsOH}), which is a strong base, and releases hydrogen gas (\text{H}_2). The reaction is exothermic, releasing a significant amount of heat, and can be explosive.

The basic approach to balance this reaction starts with writing the unbalanced chemical formula:\[\mathrm{Cs}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{CsOH}(aq)+\mathrm{H}_{2}(g)\]
In order to balance it, it's necessary to ensure that there are equal numbers of each type of atom on both sides of the equation. By adding coefficients, we obtain the balanced chemical equation:\[\mathrm{2Cs}(s)+\mathrm{2H}_{2} \mathrm{O}(l) \rightarrow \mathrm{2CsOH}(aq)+\mathrm{H}_{2}(g)\]
The coefficients 2 in front of cesium and water indicate that two atoms of cesium react with two molecules of water to produce two units of cesium hydroxide and one molecule of hydrogen gas. It's crucial for students to understand not only how to balance the equation but also the implications of the coefficients in terms of the quantities of reactants and products involved in the reaction.
Sodium and Chlorine Reaction
The reaction between sodium (Na) and chlorine (Cl_2) to form sodium chloride (NaCl), commonly known as table salt, is a classic example of a redox reaction. Here, sodium, an alkali metal, donates an electron to chlorine, a halogen, resulting in the formation of sodium ions (\text{Na}^+) and chloride ions (\text{Cl}^-), which combine to form the ionic compound sodium chloride.

To balance this reaction, we start with the unbalanced equation:\[\mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{NaCl}(s)\]
Balancing requires assessing the number of atoms or ions of each element on both the reactant and product sides. By adjusting the coefficients, we get:\[\mathrm{2Na}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{2NaCl}(s)\]
The coefficient 2 in front of sodium and sodium chloride shows that two atoms of sodium are needed to react completely with one molecule of chlorine gas. The balancing process underscores the importance of conservation of mass and charge in chemical reactions and helps students understand the mole-to-mole ratios that dictate the quantities of reactants and products.
Chemical Reaction Stoichiometry
Chemical reaction stoichiometry is the quantitative study of the relationships between the amounts of reactants used and products formed in a chemical reaction. It is fundamental to understanding chemical reactions because it allows scientists and students to predict the quantities required and generated in any given reaction. Stoichiometry is based on the law of conservation of mass and the concept of the mole, which is the standard unit for amount of substance in chemistry.

To perform stoichiometric calculations, one must first write a balanced chemical equation. A balanced equation ensures that the same number of each type of atom is present on both sides of the equation, adhering to the conservation of mass. From there, one can use the coefficients to determine the mole ratios between reactants and products, which are essential when calculating amounts in grams, liters, or particles like atoms, molecules, or ions.

The stoichiometric coefficients, as seen in the balancing of cesium and water and sodium and chlorine reactions, inform us about these mole ratios and are used to calculate how much of each substance is involved in the reaction. Understanding these concepts helps students to perform precise calculations and predict the outcomes of reactions in terms of the amount of products formed from given reactants.
Chemical Education
Chemical education is the process of learning and teaching chemistry in schools, colleges, and universities. It involves not only memorizing facts and equations but also understanding underlying concepts and developing critical thinking and problem-solving skills. Effective chemical education empowers students to comprehend and predict the behavior of substances at the molecular level.

When approaching topics such as the reactions of cesium with water and sodium with chlorine, chemical education focuses on the principles of reaction balancing, stoichiometry, and the safe handling of chemicals due to their potential hazards. Educators often employ a variety of tools, including visual aids, lab experiments, digital simulations, and interactive problem-solving sessions to enhance understanding.

Furthermore, in the context of a digital age, chemical education extends into online platforms, which provide innovative ways for students to engage with chemical concepts and receive personalized instruction. These resources can offer immediate feedback, additional exercises, and visual representations that can greatly aid in the learning process. A strong foundation in the chemical sciences forms the base for future innovation and the responsible application of chemical knowledge in real-world situations.

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

The four most abundant elements by mass in the human body are oxygen, carbon, hydrogen, and nitrogen. These four elements make up about \(96 \%\) of the human body. The next four most abundant elements are calcium, phosphorus, magnesium, and potassium. Write the expected ground-state electron configurations for these eight most abundant elements in the human body.

An ion having a \(4+\) charge and a mass of \(49.9\) amu has 2 electrons with principal quantum number \(n=1,8\) electrons with \(n=2\), and 10 electrons with \(n=3\). Supply as many of the properties for the ion as possible from the information given. (Hint: In forming ions for this species, the \(4 s\) electrons are lost before the \(3 d\) electrons.) a. the atomic number b. total number of \(s\) electrons c. total number of \(p\) electrons d. total number of \(d\) electrons e. the number of neutrons in the nucleus f. the ground-state electron configuration of the neutral atom

Which of the following orbital designations are incorrect: \(1 s\), \(1 p, 7 d, 9 s, 3 f, 4 f, 2 d ?\)

In each of the following sets, which atom or ion has the smallest ionization energy? a. \(\mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}\) d. \(\mathrm{S}^{2-}, \mathrm{S}, \mathrm{S}^{2+}\) b. \(\mathrm{K}, \mathrm{Mn}, \mathrm{Ga}\) e. \(\mathrm{Cs}, \mathrm{Ge}\), Ar c. \(\mathrm{N}, \mathrm{O}, \mathrm{F}\)

For each of the following pairs of elements \((\mathrm{C}\) and \(\mathrm{N})\) \((\mathrm{Ar}\) and \(\mathrm{Br})\) pick the atom with a. more favorable (exothermic) electron affinity. b. higher ionization energy. c. larger size.

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.