Problem 69 - Chemistry The Central Science In Si Units

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

(a) 0.00550 moles of aluminum were used. (b) 1.4678 grams of aluminum bromide were formed.

Step 1: Calculate the Volume of Aluminum

Calculate the volume of the aluminum foil using the formula for volume: \( V = ext{area} \times ext{thickness} \). The area given is \(1.00 \, \text{cm}^2\) and the thickness is \(0.550 \, \text{mm}\), which needs to be converted to cm: \(0.550 \, \text{mm} = 0.0550 \, \text{cm}\). Therefore, \( V = 1.00 \, \text{cm}^2 \times 0.0550 \, \text{cm} = 0.0550 \, \text{cm}^3 \).

Step 2: Calculate the Mass of Aluminum

Using the density \( \rho = 2.699 \, \text{g/cm}^3 \), calculate the mass \( m \) using the formula \( m = \rho \times V \). Substituting the calculated volume, \( m = 2.699 \, \text{g/cm}^3 \times 0.0550 \, \text{cm}^3 = 0.14845 \, \text{g} \).

Step 3: Convert Mass to Moles of Aluminum

Convert the mass of aluminum to moles using the molar mass of aluminum, which is \( 26.982 \, \text{g/mol} \). Calculate the moles: \( \text{moles} = \frac{0.14845 \, \text{g}}{26.982 \, \text{g/mol}} = 0.00550 \, \text{mol} \).

Step 4: Write the Chemical Reaction

The chemical reaction is \( 2 \text{Al} + 3 \text{Br}_2 \rightarrow 2 \text{AlBr}_3 \). This shows that 2 moles of aluminum react with bromine to form 2 moles of aluminum bromide.

Step 5: Calculate Moles of Aluminum Bromide Formed

According to the balanced equation, 1 mole of aluminum produces 1 mole of aluminum bromide, therefore \( 0.00550 \, \text{mol} \) of aluminum produces \( 0.00550 \, \text{mol} \) of \( \text{AlBr}_3 \).

Step 6: Convert Moles of Aluminum Bromide to Grams

Calculate the mass of \( \text{AlBr}_3 \) produced. The molar mass of \( \text{AlBr}_3 \) is \( 26.982 + 3 \times 79.904 = 266.694 \, \text{g/mol} \). Mass of \( \text{AlBr}_3 \) is \( 0.00550 \, \text{mol} \times 266.694 \, \text{g/mol} = 1.4678 \, \text{g} \).

Key Concepts

Understanding Molar Mass

The concept of molar mass is essential when dealing with chemical reactions. It refers to the mass of one mole of a substance. Molar mass is expressed in units of grams per mole (g/mol). To find the molar mass of any substance, you sum up the atomic masses of all the atoms in its chemical formula.

For example, for aluminum (Al), with an atomic mass of approximately 26.982 g/mol, the molar mass tells us how much one mole of aluminum weighs. This value serves as a conversion factor that allows you to convert between grams and moles, which is crucial in stoichiometric calculations.

In chemical calculations, knowing the molar mass of your reactants and products helps predict how much of each substance is needed or produced. In our problem, we used the molar mass of aluminum to convert the mass of a given aluminum piece to moles, moving us a step closer to solving the complete reaction equation.

Exploring Chemical Reactions

Chemical reactions provide the framework for understanding how substances transform during a chemical process. They involve reactants converting into products, illustrated by a balanced chemical equation.

Consider the reaction of aluminum with bromine to form aluminum bromide: \[ 2 \text{Al} + 3 \text{Br}_2 \rightarrow 2 \text{AlBr}_3 \]This equation depicts stoichiometry, showing the fixed ratio in which substances react. For every 2 moles of aluminum, 3 moles of bromine participate to yield 2 moles of aluminum bromide.

• This fixed ratio helps us calculate how much product we can create from a given amount of reactant.
• In our exercise, after determining the moles of aluminum used, the balanced equation allowed us to find that the same number of moles of aluminum bromide forms.

Understanding the balance lets us predict product amounts or needed reactants, ensuring efficient use of materials in chemical processes.

The Role of the Conversion Factor

Conversion factors are pivot points in stoichiometric problems, allowing us to translate one measure into another using fixed relationships.

• For instance, to find the number of moles from a given mass, use the conversion factor derived from the substance's molar mass.
• This method follows the formula: \[ \text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \]

This was applied to aluminum in our problem. We started with its mass and used its molar mass to find the number of moles present. Conversion factors simplify transformations between units in a balanced chemical equation, leading to accurate results.
By effectively utilizing these tools, we smoothly transition from mass to moles, then apply stoichiometric ratios to predict outcomes across chemical reactions.

Basics of Density Calculations

Density is a fundamental property defined as mass per unit volume, typically expressed in grams per cubic centimeter (g/cm³). Knowing the density of a material lets us find its mass if we know its volume, and vice versa.

In practice, calculating the mass of a piece of material starts with its volume and density. Use the formula:\[ \text{Mass} = \text{Density} \times \text{Volume} \]

• This was pivotal in our setup, where the density of aluminum (2.699 g/cm³) helped translate a thin aluminum foil's volume into its mass.
• The measured area and thickness gave us volume, then multiplied by density provided the mass needed for further stoichiometric analysis.

Density calculations are instrumental in connecting physical dimensions of materials to chemical properties, forming a bridge in stoichiometry ensuring correct material quantities in reactions.