Problem 73 - Chemistry The Central Science

Question

Write a balanced equation for the reaction that occurs in each of the following cases: (a) Potassium metal burns in 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 is reacted with molten sulfur.

Step-by-Step Solution

Answer

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

Step 1: (a) Potassium metal and chlorine gas reaction

: 1. Identify reactants and products: The reactants are potassium metal (K) and chlorine gas (\(Cl_2\)). When a metal reacts with a halogen, it forms a metal halide. In this case, the product is potassium chloride (KCl). 2. Write the unbalanced equation: \(K + Cl_2 \rightarrow KCl\) 3. Balance the equation: To balance the equation, we must ensure that there are equal numbers of each atom on both sides of the equation. Adjusting the coefficients, we get the balanced equation: \(2K + Cl_2 \rightarrow 2KCl\).

Step 2: (b) Strontium oxide and water reaction

: 1. Identify reactants and products: The reactants are strontium oxide (SrO) and water (H_2O). When a metal oxide reacts with water, it forms a corresponding metal hydroxide. In this case, the product is strontium hydroxide (Sr(OH)_2). 2. Write the unbalanced equation: \(SrO + H_2O \rightarrow Sr(OH)_2\) 3. Balance the equation: The equation is already balanced, so we have \(SrO + H_2O \rightarrow Sr(OH)_2\).

Step 3: (c) Lithium metal and oxygen gas reaction

: 1. Identify reactants and products: The reactants are lithium metal (Li) and oxygen gas (\(O_2\)). When a metal reacts with oxygen, it forms a metal oxide. In this case, the product is lithium oxide (Li_2O). 2. Write the unbalanced equation: \(Li + O_2 \rightarrow Li_2O\) 3. Balance the equation: To balance the equation, we must ensure that there are equal numbers of each atom on both sides of the equation. Adjusting the coefficients, we get the balanced equation: \(4Li + O_2 \rightarrow 2Li_2O\).

Step 4: (d) Sodium metal and molten sulfur reaction

: 1. Identify reactants and products: The reactants are sodium metal (Na) and molten sulfur (S). When a metal reacts with a non-metal, it forms a metal sulfide. In this case, the product is sodium sulfide (Na_2S). 2. Write the unbalanced equation: \(Na + S \rightarrow Na_2S\) 3. Balance the equation: To balance the equation, we must ensure that there are equal numbers of each atom on both sides of the equation. Adjusting the coefficients, we get the balanced equation: \(2Na + S \rightarrow Na_2S\).

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental step in understanding chemical reactions. The goal is to ensure that the number of atoms for each element is the same on both sides of the equation. This aligns with the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.

• Start by writing the unbalanced equation with the reactants on the left and the products on the right.
• Count the number of atoms of each element on both sides of the equation.
• Adjust coefficients (the numbers before molecules) to balance the atoms. This might require trial and error, adjusting coefficients iteratively.
• Recheck all elements after each change to ensure balance is achieved.

An example is the reaction between potassium and chlorine. The unbalanced reaction is: \(K + Cl_2 \rightarrow KCl\). Balancing it results in: \(2K + Cl_2 \rightarrow 2KCl\). Both sides now have two potassium atoms and two chlorine atoms, making it a balanced equation.

Metal and Non-metal Reactions

When metals react with non-metals, they tend to form ionic compounds because metals tend to lose electrons while non-metals tend to gain them. This electron transfer leads to the formation of ions.

• An example reaction is between sodium (a metal) and sulfur (a non-metal).
• The reaction forms sodium sulfide, an ionic compound.
• The balanced chemical equation for this reaction is: \(2Na + S \rightarrow Na_2S\).

These reactions are often exothermic, releasing energy. Metals like sodium react vigorously with non-metals due to this energy release. The nature of the resulting compound, in this case sodium sulfide, is determined by the properties of the metal and non-metal involved.

Metal Oxides and Hydroxides

Metal oxides are compounds formed when metals react with oxygen. Adding water to metal oxides often leads to the formation of metal hydroxides. These hydroxides are bases, meaning they can neutralize acids.

• Take strontium oxide reacting with water to form strontium hydroxide as an example.
• The balanced equation: \(SrO + H_2O \rightarrow Sr(OH)_2\), begins the reaction with metal oxide and water.

This process showcases the transition from an oxide to a hydroxide. Metal oxides like SrO are often used in various industrial processes for producing metal hydroxides, which are vital in many applications ranging from everyday cleaning agents to complex chemical manufacturing.

Formation of Metal Halides

When metals react with halogens, the result is the formation of metal halides. Halogens are a group of elements that include fluorine, chlorine, bromine, iodine, and astatine. These elements are highly reactive, especially with metals.

• A common example is when potassium reacts with chlorine. The resulting compound is potassium chloride.
• The balanced chemical equation is \(2K + Cl_2 \rightarrow 2KCl\).

Metal halides are ionic compounds composed of a metal cation and a halogen anion. They have widespread uses in various fields, including street lighting and food flavoring, due to their stability and reactivity properties compared to pure elements. Understanding this formation helps in appreciating the compound's applications in real-world situations.