Question

Write a balanced equation for the preparation of (a) \(\mathrm{N}_{2}\) from \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2}\) (b) \(\mathrm{O}_{2}\) from \(\mathrm{O}_{3}\) (c) \(\mathrm{S}\) from \(\mathrm{H}_{2} \mathrm{~S}\)

Short answer

Question: Write balanced chemical equations for the preparation of the following elements: (a) Nitrogen (N2) from Lead Azide (Pb(N3)2) (b) Oxygen (O2) from Ozone (O3) (c) Sulfur (S) from Hydrogen Sulfide (H2S). Answer: (a) 3Pb(N3)2 → 6N2 + 2Pb (b) 2O3 → 3O2 (c) 2H2S → S2 + 2H2

Step-by-step solution

Write unbalanced equation

Write the equation with the reactant and product: Pb(N3)2 -> N2

Balance the equation

Balancing the equation requires multiplying the reactants or products with coefficients as needed to ensure an equal number of atoms on both sides of the equation. 3Pb(N3)2 -> 6N2 + 2Pb

Finalize the balanced equation

The balanced equation for the preparation of N2 from Pb(N3)2 is: 3Pb(N3)2 → 6N2 + 2Pb (b) Preparation of O2 from O3

Write unbalanced equation

Write the equation with the reactant and product: O3 -> O2

Balance the equation

Balancing the equation requires multiplying the reactants or products with coefficients as needed to ensure an equal number of atoms on both sides of the equation. 2O3 -> 3O2

Finalize the balanced equation

The balanced equation for the preparation of O2 from O3 is: 2O3 → 3O2 (c) Preparation of S from H2S

Write unbalanced equation

Write the equation with the reactant and product: H2S -> S

Balance the equation

Balancing the equation requires multiplying the reactants or products with coefficients as needed to ensure an equal number of atoms on both sides of the equation. 2H2S -> S2 + 2H2

Finalize the balanced equation

The balanced equation for the preparation of S from H2S is: 2H2S → S2 + 2H2

Key concepts

Balancing Equations

Balancing chemical equations is a fundamental concept in chemistry. It ensures that the equation obeys the Law of Conservation of Mass. This law states that matter cannot be created or destroyed in a closed system. Therefore, a balanced equation has an equal number of each type of atom on both the reactant and product side.

To balance an equation, you start by writing the unbalanced chemical equation, also known as the skeleton equation. This equation shows the reactants and the products without any coefficients. For example, for the decomposition of lead azide to form nitrogen and lead, the unbalanced equation is written as:

• \[ \text{Pb(N}_3)_2 \rightarrow \text{N}_2 \]

Next, adjust the coefficients (the numbers in front of compounds) to ensure that the number of atoms for each element is the same on both sides of the equation. Use the smallest whole numbers possible to achieve this balance.

For the lead azide example, balancing achieves a diamond-check of atoms per element:

• \[3 \text{Pb(N}_3)_2 \rightarrow 6 \text{N}_2 + 2 \text{Pb}\]

By following these steps, you ensure chemical equations accurately represent real chemical reactions.

Stoichiometry

Stoichiometry is like the recipe for a chemical reaction. It involves calculating the amounts of reactants and products in a chemical reaction based on a balanced equation. It helps you determine how much of each substance you need or will produce, similar to measuring ingredients in a recipe.

Using stoichiometry, you can predict the amount of product formed in a reaction from given quantities of reactants, or determine the amount of reactants required to produce a desired quantity of product. This process involves using molar ratios, which are directly derived from the coefficients in a balanced chemical equation.

Example of Stoichiometry

Consider the preparation of \(\text{O}_2\) from \(\text{O}_3\). The balanced equation is:

• \[2\text{O}_3 \rightarrow 3\text{O}_2 \]

This tells us that 2 moles of ozone yield 3 moles of oxygen. If you start with 4 moles of ozone, you can calculate the moles of oxygen produced. Simply multiply the moles of ozone by the ratio of oxygen moles to ozone moles:

• \[ 4\text{ mol } \text{O}_3 \times \left(\frac{3\text{ mol } \text{O}_2}{2\text{ mol } \text{O}_3}\right) = 6\text{ mol } \text{O}_2\]

By understanding stoichiometry, you can better appreciate the quantitative aspect of chemistry and its calculations.

Chemical Reactions

Chemical reactions are processes where substances (reactants) transform into new substances (products). They are the cornerstone of chemistry, explaining how substances interact with each other to form new compounds.

Reactions can be classified into various types like synthesis, decomposition, single-replacement, and double-replacement. The examples provided demonstrate decomposition reactions, where a single compound breaks down into two or more simpler substances.

Decomposition Reaction Examples

• Nitrogen preparation: The decomposition of lead azide \(\text{Pb(N}_3)_2\) into nitrogen gas \(\text{N}_2\) and lead \(\text{Pb}\) represents a decomposition reaction.
• Oxygen preparation: Similarly, ozone \(\text{O}_3\) decomposes into oxygen gas \(\text{O}_2\).
• Sulfur preparation: The reaction of hydrogen sulfide \(\text{H}_2\text{S}\) decomposing to sulfur \(\text{S}_2\) and hydrogen gas \(\text{H}_2\).

Recognizing the type of reaction is crucial for predicting products in a reaction and understanding the changes occurring at the molecular level. A balanced equation is key to accurately representing these changes and adhering to the laws governing chemical reactions.