Problem 40 - General Chemistry

Question

Aluminum metal reacts with bromine, a red-brown liquid with a noxious odor. The reaction is vigorous and produces aluminum bromide, a white crystalline substance. A sample of \(27.0 \mathrm{~g}\) of aluminum yields \(266.7 \mathrm{~g}\) of aluminum bromide. How many grams of bromine react with \(18.1 \mathrm{~g}\) of aluminum?

Step-by-Step Solution

Answer

160.69 g of bromine react with 18.1 g of aluminum.

Step 1: Identify the Chemical Equation

The chemical reaction between aluminum (Al) and bromine (Br_2) forms aluminum bromide (AlBr_3). The balanced equation is: \[ 2Al + 3Br_2 \rightarrow 2AlBr_3 \] This equation indicates that 2 moles of aluminum react with 3 moles of bromine to produce 2 moles of aluminum bromide.

Step 2: Calculate Moles of Aluminum

First, we need to find the molar mass of aluminum, which is \( 26.98 \text{ g/mol} \). Then, calculate the moles of aluminum used in the original reaction with 27 g of aluminum: \[ \text{moles of Al} = \frac{27.0 \text{ g}}{26.98 \text{ g/mol}} = 1.001 \text{ mol} \]

Step 3: Calculate Moles of Aluminum Bromide Produced

Find the molar mass of AlBr_3: \[ \text{Molar mass of AlBr}_3 = 26.98 \text{ (for Al)} + 3 \times 79.904 \text{ (for Br)} = 266.68 \text{ g/mol} \] Now, calculate the moles of aluminum bromide produced: \[ \text{moles of AlBr}_3 = \frac{266.7 \text{ g}}{266.68 \text{ g/mol}} = 0.9998 \text{ mol} \]

Step 4: Use Stoichiometry to Find Moles of Bromine

From the balanced equation, 2 moles of aluminum produce 2 moles of aluminum bromide, meaning 1 mole of aluminum produces 1 mole of aluminum bromide. Thus, 1.001 moles of aluminum will require \( \frac{3}{2} \times 1.001 = 1.5015 \) moles of bromine.

Step 5: Determine Mass of Bromine in Original Reaction

To find the mass of bromine that reacts, use the molar mass of bromine \( Br_2 \) which is \( 159.808 \text{ g/mol} \). Then, multiply by the moles of bromine: \[ \text{mass of Br}_2 = 1.5015 \text{ mol} \times 159.808 \text{ g/mol} = 239.94 \text{ g} \]

Step 6: Scale the Mass to 18.1 g of Aluminum

Now scale the amount of bromine for 18.1 g of aluminum. Since 27.0 g of aluminum requires 239.94 g bromine, use the proportion: \[ \frac{18.1 \text{ g Al}}{27.0 \text{ g Al}} \times 239.94 \text{ g Br}_2 = 160.69 \text{ g Br}_2 \]

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It is based on the conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction. By using the balanced chemical equation, stoichiometry allows us to predict the amounts of products made from given reactants and vice versa.

For example, in a reaction where aluminum reacts with bromine to form aluminum bromide, stoichiometry helps us determine how much bromine is needed to completely react with a certain mass of aluminum. The balanced equation gives us a ratio of how the moles of reactants and products relate to each other.

Understanding stoichiometry is crucial for solving chemical equations and predicting the outcomes of reactions, ensuring that reactions are conducted with the optimal amounts of reactants to avoid waste.

Molar Mass

Molar mass is an important property in chemistry. It allows us to relate the mass of a substance to the number of moles, which is essential in stoichiometric calculations. The molar mass of an element is its atomic mass in g/mol, and for a compound, it's the sum of the molar masses of its constituent atoms.

For instance, aluminum has a molar mass of 26.98 g/mol, while bromine, as a diatomic molecule ( ext{Br}_2), has a molar mass of 159.808 g/mol. When calculating the molar mass of aluminum bromide ( ext{AlBr}_3), we sum up the atomic masses of one aluminum atom and three bromine atoms. This helps in determining the amount of substance in moles from a given mass.

Having a solid grasp on molar masses aids in the conversion between mass and moles, facilitating a deeper understanding of the chemical dynamics of reactions.

Balanced Chemical Equation

A balanced chemical equation is central to understanding chemical reactions. It ensures that the number of atoms for each element is conserved on both sides of the equation, reflecting the conservation of mass.

In our example, the balanced equation for the reaction between aluminum and bromine to form aluminum bromide is: \[ 2Al + 3Br_2 \rightarrow 2AlBr_3 \]

This equation indicates that 2 moles of aluminum react with 3 moles of bromine to form 2 moles of aluminum bromide. Balancing equations involves adjusting the coefficients to achieve the same number of atoms on each side. This is crucial not only for calculating the correct stoichiometric ratios but also for understanding the molecular story of a reaction.

Learning to balance equations is a fundamental skill in chemistry, providing clarity and precision in chemical analysis and synthesis.

Aluminum Bromide

Aluminum bromide ( ext{AlBr}_3) is a chemical compound formed from the reaction between aluminum and bromine. It appears as a white crystalline substance and is known for its various applications in chemical synthesis and industrial processes.

Understanding the properties and formation of aluminum bromide starts with the balanced equation for its creation, where aluminum metal reacts vigorously with bromine liquid. The production of aluminum bromide illustrates the concept of stoichiometry, as we determine how much reactant is needed to produce a certain amount of the compound.

Studying compounds like aluminum bromide helps highlight the practical significance of chemical equations and stoichiometric calculations in everyday chemical processes, demonstrating how theoretical chemistry translates into real-world uses.