Problem 74 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

The balanced chemical equations for the given reactions are: (a) \(Cs + 2H2O \to 2CsOH + H2\) (b) \(Sr + 2H2O \to Sr(OH)2 + H2\) (c) \(4Na + O2 \to 2Na2O\) (d) \(Ca + I2 \to CaI2\)

Step 1: Identify the products of the reaction

When cesium reacts with water, it forms cesium hydroxide and hydrogen gas. So, we can write the initial equation as: Cs + H2O -> CsOH + H2

Step 2: Balance the chemical equation

To balance the equation, we start by checking the number of atoms of each element on both sides: -Cs: 1 (left), 1 (right) -O: 1 (left), 1 (right) -H: 2 (left), 3 (right) The number of hydrogen atoms is not balanced. To balance it, we can multiply the cesium hydroxide molecule by 2: Cs + 2H2O -> 2CsOH + H2 The balanced chemical equation for the reaction of cesium with water is: \(Cs + 2H2O \to 2CsOH + H2\) (b) Reaction of Strontium with water

Step 1: Identify the products of the reaction

When strontium reacts with water, it forms strontium hydroxide and hydrogen gas. So, we can write the initial equation as: Sr + H2O -> Sr(OH)2 + H2

Step 2: Balance the chemical equation

To balance the equation, we check the number of atoms of each element on both sides: -Sr: 1 (left), 1 (right) -O: 1 (left), 2 (right) -H: 2 (left), 4 (right) Here, both oxygen and hydrogen are not balanced. To balance the equation, we can multiply the water molecule by 2: Sr + 2H2O -> Sr(OH)2 + H2 The balanced chemical equation for the reaction of strontium with water is: \(Sr + 2H2O \to Sr(OH)2 + H2\) (c) Reaction of Sodium with oxygen

Step 1: Identify the products of the reaction

When sodium reacts with oxygen, it forms sodium oxide. So, we can write the initial equation as: Na + O2 -> Na2O

Step 2: Balance the chemical equation

To balance the equation, we check the number of atoms of each element on both sides: -Na: 1 (left), 2 (right) -O: 2 (left), 1 (right) To balance the equation, we can multiply the sodium atom by 4 and the sodium oxide molecule by 2: 4Na + O2 -> 2Na2O The balanced chemical equation for the reaction of sodium with oxygen is: \(4Na + O2 \to 2Na2O\) (d) Reaction of Calcium with iodine

Step 1: Identify the products of the reaction

When calcium reacts with iodine, it forms calcium iodide. So, we can write the initial equation as: Ca + I2 -> CaI2

Step 2: Balance the chemical equation

The equation is already balanced because there is one calcium atom and two iodine atoms on both sides of the equation. The balanced chemical equation for the reaction of calcium with iodine is: \(Ca + I2 \to CaI2\)

Key Concepts

Balancing Chemical Equations

In chemical reactions, the number of atoms for each element must be the same on both the reactant and product sides of the equation. This is known as balancing a chemical equation. Balancing is crucial because it respects the Law of Conservation of Mass, which states that mass cannot be created or destroyed in a chemical reaction. Here's a simple approach to balancing equations:

• Write the unbalanced equation using the chemical formulas for each reactant and product.
• Count the number of atoms for each element on both sides of the equation.
• Add coefficients to balance the number of atoms for each element, starting with the one that appears in the largest number of molecules.
• Ensure all coefficients are in the lowest possible whole number ratio.

For example, in the reaction of cesium with water: Initially, it's written as \(Cs + H_2O \to CsOH + H_2\). Upon balancing, more water is added to account for the hydrogen imbalance, resulting in: \(Cs + 2H_2O \to 2CsOH + H_2\). This equation now has equal numbers of each type of atom on both sides.

Types of Chemical Reactions

Chemical reactions can be classified into several types, each with its distinct characteristics. Understanding these types helps predict the outcomes of chemical interactions.Some common types include:

• Synthesis Reaction: Two or more simple substances combine to form a more complex substance. For example, calcium reacting with iodine to form calcium iodide: \(Ca + I_2 \to CaI_2\).
• Decomposition Reaction: A complex substance breaks down into simpler substances. No example from the given reactions represents this type, as all involve synthesis or replacement.
• Single Replacement Reaction: An element replaces another in a compound. For instance, sodium reacts with oxygen where sodium replaces some oxygens: \(4Na + O2 \to 2Na2O\).
• Double Replacement Reaction: Two compounds exchange partners to form two new compounds. This type does not appear in the provided exercise.

Identifying the type of reaction provides insight into how reactants transform into products.

Chemical Reaction Products

Identifying products is a critical step in predicting the outcome of a chemical reaction. The products are the substances that are formed as a result of the reaction. In the given exercises:

• When cesium reacts with water, the products are cesium hydroxide \(CsOH\) and hydrogen gas \(H_2\).
• In the reaction of strontium with water, strontium hydroxide \(Sr(OH)_2\) and hydrogen gas \(H_2\) are produced.
• The combination of sodium and oxygen forms sodium oxide \(Na_2O\).
• Calcium reacting with iodine yields calcium iodide \(CaI_2\).

Each chemical reaction leads to the formation of specific products, which are determined by the nature of the reactants and the type of chemical reaction taking place. Predicting products involves understanding the reactivity of the elements involved and the principles governing chemical changes.