Problem 69 - Chemistry: The Central Science in SI Units, Global Edition

Question

A piece of aluminum foil \(1.00 \mathrm{~cm}^{2}\) and \(0.550-\mathrm{mm}\) thick is allowed to react with bromine to form aluminum bromide. (a) How many moles of aluminum were used? (The density of aluminum is \(2.699 \mathrm{~g} / \mathrm{cm}^{3} .\) ) (b) How many grams of aluminum bromide form, assuming the aluminum reacts completely?

Step-by-Step Solution

Answer

The mass of aluminum in the foil is 0.148 g, which is equal to 0.00548 moles. The mass of aluminum bromide formed from this reaction, assuming complete reaction, is 1.46 g.

Step 1: Find the mass of aluminum in the foil

We are given the dimensions of the aluminum foil as 1.00 cm² in area and 0.550 mm in thickness. We can find the volume of the foil by multiplying the area and thickness, and then convert the volume to mass using the given density of aluminum. Volume of foil = Area x Thickness Convert the thickness from mm to cm (1 cm = 10 mm): Thickness = 0.550 mm × (1 cm / 10 mm) = 0.055 cm Volume of foil = 1.00 cm² × 0.055 cm = 0.0550 cm³ Now we use the density of aluminum (2.699 g/cm³) to find the mass: Mass of aluminum = Volume × Density Mass of aluminum = 0.0550 cm³ × 2.699 g/cm³ = 0.148 g

Step 2: Calculate the moles of aluminum

Now that we have the mass of aluminum (0.148 g), we can use the molar mass of aluminum to find the number of moles. The molar mass of aluminum is 26.98 g/mol. Moles of aluminum = Mass of aluminum / Molar mass of aluminum Moles of aluminum = 0.148 g / 26.98 g/mol = 0.00548 mol

Step 3: Find the mass of aluminum bromide formed

The balanced chemical equation for the reaction between aluminum and bromine to form aluminum bromide is: 2 Al (s) + 3 Br2 (l) → 2 AlBr3 (s) From the stoichiometry of the reaction, 1 mole of aluminum reacts to form 1 mole of aluminum bromide. Since we have 0.00548 moles of aluminum, an equal number of moles of aluminum bromide will be formed. Moles of aluminum bromide = 0.00548 mol Now we need to convert moles of aluminum bromide to mass. The molar mass of aluminum bromide (AlBr3) is: 1 Al = 26.98 g/mol 3 Br = 3 × 79.90 g/mol = 239.7 g/mol Molar mass of AlBr3 = 26.98 + 239.7 = 266.68 g/mol Mass of aluminum bromide = Moles of aluminum bromide × Molar mass of aluminum bromide Mass of aluminum bromide = 0.00548 mol × 266.68 g/mol = 1.46 g Therefore, 1.46 g of aluminum bromide is formed assuming the aluminum reacts completely.

Key Concepts

Aluminum Bromide Formation

When aluminum foil reacts with bromine, it forms aluminum bromide, a compound consisting of aluminum and bromine atoms. This reaction happens because aluminum can easily combine with other elements, like bromine, to form a stable compound.

The chemical reaction can be described by the equation:

• \( 2 \text{Al (s)} + 3 \text{Br}_2 \text{(l)} \rightarrow 2 \text{AlBr}_3 \text{(s)} \)

This equation tells us that **two moles of aluminum** react with **three moles of bromine** to produce **two moles of aluminum bromide**. This is an example of synthesis, where simple elements combine to form a more complex compound.

Chemical Reaction Stoichiometry

Stoichiometry is the part of chemistry that helps us understand the proportions of elements needed for a chemical reaction. In the reaction to form aluminum bromide, the stoichiometry from the chemical equation tells us the ratio of reactants to products.

Here's how it applies:

• **2 moles of aluminum** react with **3 moles of bromine**.
• This produces **2 moles of aluminum bromide**.

Because the ratio of aluminum to aluminum bromide is 1:1, every mole of aluminum will yield one mole of aluminum bromide. This relationship helps chemists calculate how much of each substance will be consumed or produced in a reaction.

Molar Mass Calculation

The concept of molar mass is like a bridge between the mass of a substance and the amount in moles. Every element has a specific molar mass, typically measured in grams per mole (g/mol), based on its atomic weight.

For aluminum, the molar mass is **26.98 g/mol**. This means that 26.98 grams of aluminum is equal to one mole.

For aluminum bromide (AlBr₃), we calculate its molar mass by adding:

• One mole of aluminum: **26.98 g/mol**
• Three moles of bromine: **3 × 79.90 g/mol = 239.7 g/mol**

This gives a molar mass of **266.68 g/mol** for AlBr₃, allowing us to convert moles to grams for practical use in experiments.

Density and Volume Relationship

Density is the measure of how much mass is contained in a given volume. When working with solids like aluminum foil, knowing the density helps us calculate its mass from its volume.

The given density of aluminum is **2.699 g/cm³**. We find the volume by multiplying the area and thickness of the foil:

• **Volume** = Area × Thickness = \(1.00 \text{ cm}^2 \times 0.055 \text{ cm} = 0.0550 \text{ cm}^3\)

The mass can then be calculated by multiplying this volume by the density:

• **Mass** = Volume × Density = \(0.0550 \text{ cm}^3 \times 2.699 \text{ g/cm}^3 = 0.148 \text{ g}\)

Understanding this relationship is crucial for moving between properties like volume and mass.