Problem 69 - Chemistry The Central Science

Question

A piece of aluminum foil \(1.00 \mathrm{~cm}\) square 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 \(\left.2.699 \mathrm{~g} / \mathrm{cm}^{3} .\right)\) (b) How many grams of aluminum bromide form, assuming the aluminum reacts completelv?

Step-by-Step Solution

Answer

(a) Moles of aluminum used: \(0.00549 \: mol\) (b) Mass of aluminum bromide formed: \(1.464 \: g\) (approximately)

Step 1: Calculate the mass of aluminum

Given the dimensions of the aluminum foil and its density, we can calculate the mass by finding the volume of the foil and then multiplying it with the density; Density = Mass/Volume => Mass = Density × Volume First, find the volume of the aluminum foil: Volume = Length × Width × Height = 1.00 cm × 1.00 cm × 0.0550 cm \(Volume = 0.0550 \mathrm{~cm}^3\) Now, using the given density of aluminum (2.699 g/cm³), find the mass: Mass = Density × Volume = 2.699 g/cm³ × 0.0550 cm³ \(Mass = 0.1484 \: g\) (approximately)

Step 2: Convert mass of aluminum to moles

To find the number of moles of aluminum used, we'll use the molar mass of aluminum (Al), which is 26.98 g/mol. Moles (Al) = Mass (Al) / Molar Mass (Al) = 0.1484 g / 26.98 g/mol \(Moles\: (Al) = 0.00549 \: mol\) (approximately)

Step 3: Write the balanced chemical equation

The chemical equation for the reaction between aluminum (Al) and bromine (Br₂) to form aluminum bromide (AlBr₃) is: \(2\: Al + 3\: Br_2 ⟶ 2\: AlBr_3\)

Step 4: Find the moles of aluminum bromide formed

According to the balanced reaction equation, 2 moles of aluminum react with 3 moles of bromine to produce 2 moles of aluminum bromide. From step 2, we calculated 0.00549 moles of aluminum reacting. Using the mole ratio from the equation, we can calculate the moles of aluminum bromide formed: \(\frac{Moles\: (Al)}{2} = \frac{Moles\: (AlBr_3)}{2}\) Moles (AlBr₃) = 0.00549 mol

Step 5: Calculate the mass of aluminum bromide formed

Now, we'll convert the moles of aluminum bromide to mass. First, we need to find the molar mass of AlBr₃, which is: Molar Mass (AlBr₃) = Molar Mass (Al) + 3 × Molar Mass (Br) = 26.98 g/mol + 3 × 79.90 g/mol \(Molar\: Mass\: (AlBr_3) = 266.68 \: g/mol\) Now, we can find the mass of AlBr₃ formed: Mass (AlBr₃) = Moles (AlBr₃) × Molar Mass (AlBr₃) = 0.00549 mol × 266.68 g/mol \(Mass\: (AlBr_3) = 1.464 \: g\) (approximately) So to summarize our results: (a) Moles of aluminum used: \(0.00549 \: mol\) (b) Mass of aluminum bromide formed: \(1.464 \: g\) (approximately)

Key Concepts

Chemical Reactions

Understanding chemical reactions is fundamental to grasping the basics of stoichiometry. A chemical reaction involves the transformation of one set of chemical substances into another through the rearrangement of atoms. It is governed by laws of conservation which state that mass and energy remain constant in an isolated system.

In the problem provided, aluminum reacts with bromine to produce aluminum bromide according to the balanced chemical equation:
\[2 Al + 3 Br_2 \rightarrow 2 AlBr_3\]
This reaction showcases how stoichiometry can predict the amounts of reactants consumed and products formed. The equation indicates that two moles of aluminum react with three moles of bromine to produce two moles of aluminum bromide. In the context of this problem, we can use the coefficients of the balanced equation to calculate the corresponding amounts of substances involved.

Molar Mass

Molar mass is a crucial concept in stoichiometry, which is defined as the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It serves as a conversion factor between the mass of a substance and the number of moles. In essence, the molar mass enables chemists to count atoms by weighing them since the mole is a unit that bridges the microscopic world of atoms to the macroscopic world we can measure.

In our exercise, the molar mass of aluminum (Al) is 26.98 g/mol. When calculating the mass of aluminum bromide (AlBr₃), we also need to use its molar mass. This is found by summing the molar masses of aluminum and bromine (taking into account the number of each in the compound), which totals 266.68 g/mol. Knowing the molar mass of aluminum and aluminum bromide allows us to calculate the number of moles and the mass of the products formed in the reaction.

Mole Concept

The mole concept is a fundamental principle in chemistry that defines the mole as the unit of measurement for the amount of substance in the International System of Units (SI). One mole contains exactly 6.022×1023 elementary entities (this number is known as Avogadro's number). Understanding the mole concept enables us to relate the mass of a substance to the number of atoms, molecules, or ions it contains.

In the solution to our problem, after calculating the mass of the aluminum foil, we converted this mass into moles using the molar mass of aluminum. Similarly, when we determined the amount of aluminum bromide produced, we first calculated the moles and then converted it to grams. The mole provides a direct way to calculate the number of particles involved in a reaction, and by extension, the various quantities associated with the reaction, whether it's the amount of reactants needed or the yield of products.