Problem 60 - Chemistry The Molecular Science

Question

Aluminum reacts with oxygen to give aluminum oxide. $$4 \mathrm{Al}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3(\mathrm{~s})$$ If you have \(6.0 \mathrm{~mol} \mathrm{Al},\) calculate the amount (mol) and mass (g) of \(\mathrm{O}_{2}\) needed for complete reaction. Calculate the mass of \(\mathrm{Al}_{2} \mathrm{O}_{3},\) in grams, that is produced.

Step-by-Step Solution

Answer

4.5 moles of \(\mathrm{O}_{2}\) are needed, mass is 144.0 g. Produces 305.88 g \(\mathrm{Al}_{2} \mathrm{O}_{3}\).

Step 1: Understand the Balanced Reaction

The balanced chemical reaction for the formation of aluminum oxide is \(4 \mathrm{Al} + 3 \mathrm{O}_{2} \rightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}\). This tells us that 4 moles of aluminum react with 3 moles of oxygen to form 2 moles of aluminum oxide.

Step 2: Calculate Amount of \(\mathrm{O}_{2}\) Needed

Given 6.0 moles of aluminum, use the stoichiometric ratios to find moles of oxygen: \[\text{Ratio from equation: } \frac{4}{3} = \frac{6.0}{x}\]Solving for \(x\), the moles of \(\mathrm{O}_{2}\) needed:\[x = \left( \frac{6.0 \times 3}{4} \right) = 4.5 \text{ moles of } \mathrm{O}_{2}\]

Step 3: Find Mass of \(\mathrm{O}_{2}\) Required

The molar mass of \(\mathrm{O}_{2}\) is approximately 32.00 g/mol. Calculate the mass using the formula:\[\text{Mass} = \text{moles} \times \text{molar mass} = 4.5 \text{ moles} \times 32.00 \text{ g/mol} = 144.0 \text{ g of } \mathrm{O}_{2}\]

Step 4: Calculate Amount of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) Produced

Using the stoichiometric ratio from the balanced equation: \[\frac{4}{2} = \frac{6.0}{y}\]Solving for \(y\), the moles of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) produced:\[y = \frac{6.0 \times 2}{4} = 3.0 \text{ moles of } \mathrm{Al}_{2} \mathrm{O}_{3}\]

Step 5: Find Mass of \(\mathrm{Al}_{2} \mathrm{O}_{3}\)

The molar mass of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) is calculated as:\[\text{Molar mass of } \mathrm{Al}_{2} \mathrm{O}_{3} = (2 \times 26.98) + (3 \times 16.00) = 101.96 \text{ g/mol}\]Using this, calculate the mass:\[\text{Mass} = 3.0 \text{ moles} \times 101.96 \text{ g/mol} = 305.88 \text{ g of } \mathrm{Al}_{2} \mathrm{O}_{3}\]

Key Concepts

Stoichiometry

Stoichiometry is like a recipe for chemical reactions. It provides the exact proportion of ingredients needed to make products. In our reaction, aluminum (Al) reacts with oxygen (\( \text{O}_2 \)) to form aluminum oxide (\( \text{Al}_2\text{O}_3 \)). Understanding stoichiometry means knowing how to use the balanced equation to find the right amounts of reactants and products.
For example, the balanced equation \( 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \) tells us that:

• 4 moles of Al react with 3 moles of \( \text{O}_2 \).
• 2 moles of \( \text{Al}_2\text{O}_3 \) are formed.

This concept allows us to calculate how much oxygen we need if we start with 6 moles of aluminum, and it predicts the amount of aluminum oxide produced.

Balancing Chemical Equations

Balancing chemical equations ensures that matter is conserved in a reaction. Each side of the equation must reflect the same number of atoms for each element. In our provided reaction:\( 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \), the coefficients before each compound show the balance of aluminum and oxygen atoms.
Balancing equations involves these steps:

• Count the number of atoms for each element in the reactants and products.
• Adjust coefficients to ensure equal numbers of each atom on both sides.
• This keeps the reaction's identity and integrity intact.

For example, you can see there are 4 aluminum atoms and 6 oxygen atoms on both sides of the equation. Balancing chemical equations doesn't change how much product forms, but it is crucial for using stoichiometry correctly.

Molar Mass Calculation

Molar mass helps us relate moles to grams and vice versa. It's the mass of one mole of a substance, measured in grams per mole (g/mol). Understanding molar mass is essential for turning mole calculations into practical laboratory information.
In our exercise:

• The molar mass of \( \text{O}_2 \) is 32.00 g/mol, calculated as 2 times the atomic mass of O, 16.00 g/mol each.
• The molar mass of \( \text{Al}_2\text{O}_3 \) is 101.96 g/mol, found by adding the atomic masses: - \( 2 \times 26.98 \ \text{(atomic mass of Al)} \)- \( 3 \times 16.00 \ \text{(atomic mass of O)} \)

Using these, you can convert between mass and moles, finally allowing us to calculate that 4.5 moles of \( \text{O}_2 \) corresponds to 144 g needed, and 3 moles of \( \text{Al}_2\text{O}_3 \) corresponds to 305.88 g produced.