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Write a balanced equation for the reaction that occurs in each of the following cases: (a) Potassium metal is exposed to an atmosphere of chlorine gas. (b) Strontium oxide is added to water. (c) A fresh surface of lithium metal is exposed to oxygen gas. (d) Sodium metal reacts with molten sulfur.

Short Answer

Expert verified
(a) \(2K + Cl_2 -> 2KCl\) (b) \(SrO + H_2O -> Sr(OH)_2\) (c) \(4Li + O_2 -> 2Li_2O\) (d) \(2Na + S -> Na_2S\)

Step by step solution

01

(a) Potassium metal is exposed to an atmosphere of chlorine gas.

When potassium (K) reacts with chlorine gas (Cl2), the product is potassium chloride (KCl). To balance the equation, we need to make sure there are an equal number of potassium and chlorine atoms on both sides. The balanced chemical equation can be written as: 2K + Cl₂ -> 2KCl
02

(b) Strontium oxide is added to water.

When strontium oxide (SrO) reacts with water (H2O), the resulting products are strontium hydroxide (Sr(OH)2). To balance this equation, let's consider the number of atoms in the reactants and the products. The balanced chemical equation is: SrO + H₂O -> Sr(OH)₂
03

(c) A fresh surface of lithium metal is exposed to oxygen gas.

Lithium (Li) reacts with oxygen gas (O2) to form lithium oxide (Li2O). To balance this equation, we need to ensure there are equal numbers of lithium and oxygen atoms on both sides of the equation. The balanced chemical equation for this reaction is: 4Li + O₂ -> 2Li₂O
04

(d) Sodium metal reacts with molten sulfur.

When sodium (Na) reacts with sulfur (S), it forms sodium sulfide (Na2S). We need to balance the equation for this reaction by considering the number of sodium and sulfur atoms present. The balanced chemical equation for this reaction is: 2Na + S -> Na₂S

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Key Concepts

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

Chemical Reactions
The breadcrumbs of chemistry lie in the essence of chemical reactions, which are processes whereby substances, known as reactants, transform into new substances called products. Taking our textbook scenario as an example, potassium metal combining with chlorine gas results in potassium chloride. Here, potassium and chlorine are reactants that, when brought together, undergo a chemical change to form a product.

During these reactions, bonds between atoms break in the reactants and new bonds form in the products. Each reaction has its unique set of reactants and products which vary in properties. In the potassium and chlorine reaction (2K + Cl₂ -> 2KCl), we notice that two individual elements combine to form a compound, which is a typical synthesis reaction.

Understanding chemical reactions is pivotal, as it helps us comprehend how substances interact, predict the outcomes of such interactions, and allows us to use these reactions to our advantage in industrial, pharmaceutical, and other practical applications.
Reaction Stoichiometry
Reaction stoichiometry is like the cookbook for chemists, denoting precise measurements needed to get a perfect chemical dish without leftovers. It revolves around the proportional relationship between reactants and products in a chemical reaction, ensuring the law of conservation of mass is obeyed. When strontium oxide and water combine (SrO + H₂O -> Sr(OH)₂), stoichiometry tells us how much of each reactant is needed to produce a certain amount of product.

By utilizing coefficients, the numbers placed before molecules in a balanced equation, we quantify the number of reacting units. For example, in the reaction of lithium with oxygen (4Li + O₂ -> 2Li₂O), stoichiometry explains that four atoms of lithium react with one molecule of oxygen to form two molecules of lithium oxide.

Grasping reaction stoichiometry is essential for scientists and engineers to design reactions that maximize product yield and for students to predict the outcomes of lab experiments accurately.
Law of Conservation of Mass
The law of conservation of mass is the classical rule dictating that mass cannot be created or destroyed in a closed system, through a chemical reaction. It ensures the quantity of matter remains constant, merely changing forms. In balancing chemical equations, this law is the golden rule.

When we balance an equation, such as that of sodium metal with sulfur (2Na + S -> Na₂S), we are essentially accounting for every atom that participates in the reaction. This process ensures that the mass of the reactants equals the mass of the products, adhering to the law of conservation of mass.

It's the rigorous application of this law that allows us to predict amounts of products and reactants with precision in the lab and in industry, making it a cornerstone principle in the study of chemistry and a fundamental concept for all students to understand fully.

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Most popular questions from this chapter

Consider the following equation: $$\mathrm{Ca}^{+}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{Ca}(g)$$ Which of the following statements are true? (i) The energy change for this process is the electron affinity of the Ca' ion. (ii) The energy change for this process is the negative of the first ionization energy of the Ca atom. (ii) The energy change for this process is the negative of the electron affinity of the Ca atom.

Note from the following table that there is a significant increase in atomic radius upon moving from Y to La, whereas the radii of \(Z r\) to Hf are the same. Suggest an explanation for this effect.

Tungsten has the highest melting point of any metal in the periodic table: \(3422^{\circ} \mathrm{C}\) . The distance between \(\mathrm{W}\) atoms in tungsten metal is 2.74 A. (a) What is the atomic radius of atungsten atom in this environment? (This radius is called the metallic radius.) If you put tungsten metal under high pressure, predict what would happen to the distance between \(\mathrm{W}\) atoms.

Identify two ions that have the following ground-state electron configurations: \((\mathbf{a}) [\) Ar \(],(\mathbf{b})[\) Ar \(] 3 d^{5},(\mathbf{c})[\mathrm{Kr}] 5 s^{2} 4 d^{10}\).

Use electron configurations to explain the following observa tions: (a) The first ionization energy of phosphorus is greater than that of sulfur. (b) The electron afnity of nitrogen is lower (less negative) than those of both carbon and oxygen. (c) The second ionization energy of oxygen is greater than the first ionization energy of fluorine. (d) The third ionization energy of manganese is greater than those of both chromium and iron.

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