Problem 44 - Chemistry Matter and Change

Question

Interpret the following equation in terms of particles, moles, and mass. $$4 \mathrm{Al}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})$$

Step-by-Step Solution

Answer

In the given reaction, 4 moles of aluminum (Al) react with 3 moles of oxygen (O₂) to produce 2 moles of aluminum oxide (Al₂O₃). In terms of particles, it is \( (4 \times 6.022 \times 10^{23}) \: Al + (3 \times 6.022 \times 10^{23}) \: O_2 \rightarrow (2 \times 6.022 \times 10^{23}) \: Al_2O_3 \). In terms of mass, the reaction can be written as 107.92 g Al + 96 g O₂ -> 203.92 g Al₂O₃.

Step 1: Determine the number of atoms and molecules

The reaction can be broken down as: 4 Al(s) + 3 O₂(g) -> 2 Al₂O₃(s) Here, 4 moles of aluminum react with 3 moles of oxygen to form 2 moles of aluminium oxide. For each mole, there are Avogadro's number of particles. So, let's calculate the number of particles or atoms present.

Step 2: Conversion to particles using Avogadro's number

Using Avogadro's number (6.022 x 10²³), the given equation can be interpreted as: \( (4 \times 6.022 \times 10^{23}) \: Al + (3 \times 6.022 \times 10^{23}) \: O_2 \rightarrow (2 \times 6.022 \times 10^{23}) \: Al_2O_3 \)

Step 3: Calculate the number of moles

Now, we will express the equation in terms of moles. The balanced equation already gives us this information: 4 moles Al + 3 moles O₂ -> 2 moles Al₂O₃

Step 4: Calculate the mass of reactants and products

Finally, we need to calculate the mass of the reactants and products. To do this, we must consider the molar mass of each element and compound: Molar mass of Al = 26.98 g/mol Molar mass of O₂ = 32.00 g/mol Molar mass of Al₂O₃ = (2 × 26.98) + (3 × 16.00) = 101.96 g/mol Now, multiply the moles by the molar mass to get the mass: \( (4 \times 26.98) \:g \: Al + (3 \times 32.00) \:g \: O_2 \rightarrow (2 \times 101.96) \:g \: Al_2O_3 \) The reaction can be written in terms of mass as: 107.92 g Al + 96 g O₂ -> 203.92 g Al₂O₃

Key Concepts

Moles

When we work with chemical reactions, such as the one involving aluminum and oxygen in our example, understanding the concept of "moles" is crucial. A "mole" is a unit that measures the amount of any substance. It's similar to how a "dozen" measures the quantity of twelve items.
In chemistry, one mole of any substance contains an exact number of particles, such as atoms or molecules. This is why moles are the bridge between the atomic world and measurable quantities that we can see.
Let's take our reaction: 4 moles of Aluminum (\( ext{Al}\) ) means we have a large but specific number of aluminum atoms ready to react. When we say 3 moles of Oxygen (\( ext{O}_2\) ), we imply that there are countless oxygen molecules waiting to interact. The result is that 2 moles of aluminum oxide (\( ext{Al}_2 ext{O}_3\) ), a compound formed from these elements in the reaction.

Molar Mass

Each element in the periodic table has a unique molar mass, which is the mass of one mole of its atoms. Molar mass makes it possible to calculate the mass of substances involved in chemical equations.
For our equation, the molar mass of Aluminum (\( ext{Al}\) ) is 26.98 g/mol. This means one mole of aluminum weighs 26.98 grams. Similarly, the molar mass of oxygen (\( ext{O}_2\)) is 32.00 g/mol. Therefore, one mole of oxygen molecules weighs 32.00 grams.
To calculate the molar mass of aluminum oxide (\( ext{Al}_2 ext{O}_3\)), we depend on its formula: (2 \( imes\) 26.98) + (3 \( imes\) 16.00) = 101.96 g/mol.
This calculation helps us determine that when 4 moles of \( ext{Al}\) react with 3 moles of \( ext{O}_2\), 203.92 grams of aluminum oxide will be produced.

Avogadro's Number

Named after the scientist Amedeo Avogadro, Avogadro’s number is a fundamental concept in chemistry. It is defined as approximately 6.022 \( imes\) 10^{23}, representing the number of atoms or molecules in one mole of any substance. This large number helps us convert moles into actual particles when working with chemical equations.
In our example, 4 moles of aluminum convert to \(4 \ imes 6.022 \ imes 10^{23}\) aluminum atoms, and 3 moles of oxygen \( ightarrow\) \(3 \ imes 6.022 \ imes 10^{23}\) oxygen molecules.
Knowing Avogadro's number allows us to go from the theoretical world of chemical equations to practical calculations. This powerful tool is pivotal for chemists, enabling tangible analysis of reactions at the atomic and molecular levels.

Chemical Equations

Chemical equations are a shorthand way to describe chemical reactions. By using symbols for elements and chemical formulas for compounds, they indicate what substances react and what products are formed.
For our equation, \(4 \ ext{Al} \) + \(3 \ ext{O}_2 \rightarrow 2 \ ext{Al}_2 ext{O}_3\), we understand that aluminum reacts with oxygen to produce aluminum oxide. The coefficients indicate the ratios needed for a balanced reaction.
Balancing equations is critical because it follows the law of conservation of mass—matter cannot be created or destroyed. This law means that the number of atoms of each element on the reactant side must match the number on the product side.

• Reactants: Aluminum and Oxygen
• Products: Aluminum Oxide
• Balance: Atoms of Al and O are equal on both sides of the equation

Chemical equations are essential in predicting the quantities of reactants required and products formed, guiding chemists in practical applications.