Problem 45 - Chemistry Matter and Change

Question

One carbon (C) and one ozone (O \(_{3} )\) react to form one carbon monoxide (CO) and one oxygen gas \(\left(\mathrm{O}_{2}\right)\) particle. How many ozone particles are needed to form 24 particles of oxygen gas (O \(_{2} ) ?\)

Step-by-Step Solution

Answer

24 ozone particles are needed to form 24 oxygen gas particles.

Step 1: Write the balanced chemical equation

First, we will write the unbalanced chemical equation: C + O₃ → CO + O₂ Now, we'll check if the equation is balanced. On both sides of the equation, there are: - 1 carbon atom - 3 oxygen atoms Since the number of each type of element is equal on both sides of the equation, the reaction is balanced as it stands.

Step 2: Calculate the stoichiometric ratio between O₃ and O₂

Based on the balanced chemical equation (C + O₃ → CO + O₂), we can see that one O₃ molecule forms one O₂ molecule. This means that the ratio between O₃ and O₂ is 1:1.

Step 3: Determine the required number of O₃ particles

Using the stoichiometric ratio we found in step 2 (1:1), we can determine the number of O₃ particles needed to produce 24 O₂ particles. Since the ratio is 1:1, the number of O₃ particles needed will also be 24. Therefore, 24 ozone particles are needed to form 24 oxygen gas particles.

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is an essential skill in chemistry. It ensures that the same number of each type of atom appears on both sides of the equation. In our exercise, we start with the equation: \( \text{C} + \text{O}_3 \rightarrow \text{CO} + \text{O}_2 \).

To balance it, we count the atoms: one carbon and three oxygen atoms on both sides. Since they match, the equation is balanced. This step is crucial for accurately predicting the amounts of reactants and products.

Remember, balance by adjusting coefficients, never changing subscripts in the chemical formulas.
Balance ensures conservation of mass and helps in understanding chemical transformations.

Stoichiometry

Stoichiometry involves calculating the relative quantities of reactants and products. It uses the balanced equation to make these calculations. In our balanced equation \( \text{C} + \text{O}_3 \rightarrow \text{CO} + \text{O}_2 \), we see the direct relation of components.

Stoichiometry helps determine how much of each substance is needed or produced. It's like a recipe in cooking – ensuring the ingredients are in perfect proportions. This mathematical approach is key for solving many chemical problems, such as determining theoretical yields in reactions.

With stoichiometry, you're applying core principles of conservation within chemical reactions, ensuring everything adds up as nature intends.

Chemical Reactions

Chemical reactions represent the transformation of substances. In our problem, carbon reacts with ozone to form carbon monoxide and oxygen gas. The equation \( \text{C} + \text{O}_3 \rightarrow \text{CO} + \text{O}_2 \) illustrates this transformation.

Reactions involve the breaking and forming of bonds, leading to new products. Understanding the types of reactions, such as synthesis, decomposition, or combustion, helps predict their behavior.

Involving energy changes, chemical reactions can be endothermic (absorbing heat) or exothermic (releasing heat). Knowing how reactions behave helps in fields like energy production, pharmaceuticals, and materials science.

Reaction Stoichiometry

Reaction stoichiometry delves into the quantitative relationships among reactants and products. It's crucial when scaling reactions from particles to grams or liters.

In our example, the balanced equation shows a 1:1 ratio between ozone and oxygen gas. This means one particle of \( \text{O}_3 \) produces one particle of \( \text{O}_2 \). If we need 24 particles of \( \text{O}_2 \), we use 24 particles of \( \text{O}_3 \).

This concept is vital for designing experiments and industrial processes, ensuring the right amounts of materials are used for desired outcomes. Reaction stoichiometry is about precision – you rely on it to make feasible predictions in any chemical study.