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?

Short answer

(a) The number of moles of aluminum used is \(0.00548~mol\). (b) The mass of aluminum bromide formed, assuming the aluminum reacts completely, is \(1.46~g\).

Step-by-step solution

Calculate the volume and mass of the aluminum foil

First, let's calculate the volume of the aluminum foil using the given dimensions: Volume = Area × Thickness Volume = \(1.00 cm^2 × 0.550 mm\) As we need to have consistent units, let's convert the thickness from mm to cm: \(0.550 mm = 0.0550 cm\) Now, calculate the volume: Volume = \(1.00 cm^2 × 0.0550 cm = 0.0550 cm^3\) Next, we can find the mass of the aluminum foil using the given density: Density = \(\frac{Mass}{Volume}\) Mass = Density × Volume Mass = \(2.699 \frac{g}{cm^3} × 0.0550 cm^3 = 0.148 g\)

Determine the moles of aluminum used

Now we need to find the moles of aluminum, using the molar mass of aluminum (Al) which is \(26.98 g/mol\): Moles of aluminum = \(\frac{Mass}{Molar~mass}\) Moles of aluminum = \(\frac{0.148~g}{26.98~g/mol} = 0.00548~mol\)

Calculate the moles and mass of aluminum bromide formed

According to the balanced chemical reaction, 2 moles of aluminum react with 3 moles of bromine to form 2 moles of aluminum bromide: \(2 Al (s) + 3 Br_2 (l) → 2 AlBr_3 (s)\) Using the stoichiometry of the reaction, we can determine the moles of aluminum bromide formed: Moles of aluminum bromide = Moles of aluminum × \(\frac{2~ moles~ of~AlBr_3}{2~moles~of~Al}\) Moles of aluminum bromide = \(0.00548 mol × 1 = 0.00548 mol\) Next, we need to find the mass of aluminum bromide formed, using its molar mass (which is \(26.98 g/mol~ for~ Al\) + \(3 × 79.9 g/mol~ for~ Br\) = \(266.7 g/mol\)): Mass of aluminum bromide = Moles × Molar mass Mass of aluminum bromide = \(0.00548 mol × 266.7 g/mol = 1.46 g\)

Answers

(a) The number of moles of aluminum used is \(0.00548 mol\). (b) The mass of aluminum bromide formed, assuming the aluminum reacts completely, is \(1.46 g\).

Key concepts

Density Calculations

Density is a fundamental concept in chemistry and physics and is used to determine how much matter is within a set volume. In the given problem, we're dealing with aluminum foil, and we use density to find out how much this material weighs given its volume. Density is mathematically represented as \[\text{Density} = \frac{\text{Mass}}{\text{Volume}}\]For our calculation, the density of aluminum is provided as \(2.699 \mathrm{~g/cm^3}\). Knowing the density helps us calculate the mass once we determine the volume. Volume, in this case, is calculated by multiplying the area of the foil by its thickness, ensuring all measurements use the same units. Once we have the volume, we multiply it by the density to get the mass of the aluminum foil. Therefore, understanding density calculations helps in bridging the volume and the mass of a substance.

Molar Mass

Molar mass is crucial when converting between mass and moles for any substance. It represents the mass of one mole of a substance, expressed in g/mol. In the aluminum foil problem, we use the molar mass of aluminum to convert from grams to moles.To find the molar mass, you can refer to the periodic table, where it is typically listed under the element's symbol. For instance, aluminum has a molar mass of \(26.98 \mathrm{~g/mol}\). Knowing the mass of aluminum, we can find the moles used through:\[\text{Moles of aluminum} = \frac{\text{Mass}}{\text{Molar mass}}\]This calculation is essential in stoichiometry, as it allows us to understand the proportion of macroscopic amounts (grams) of materials applied in a reaction to microscopic quantities (moles). This conversion is a pivotal step as it sets the stage to explore chemical reactions.

Chemical Reactions

Chemical reactions describe the process through which substances transform into new substances. In the given exercise, aluminum foil reacts with bromine to form aluminum bromide. Understanding the balanced chemical equation is foundational for determining the relationship between reactants and products. For this reaction, the balanced equation is:\[2 \text{Al} (s) + 3 \text{Br}_2 (l) \rightarrow 2 \text{AlBr}_3 (s)\]This equation indicates that 2 moles of aluminum react with 3 moles of bromine, yielding 2 moles of aluminum bromide. The stoichiometry, or the mole ratio, from this balanced equation is a key factor that enables us to determine the amount of product formed from a given amount of reactant. Grasping these stoichiometric coefficients is vital because they establish the exact proportions needed in reactions, ensuring you can predict how much of each substance is involved in any given reaction.

Volume and Mass Conversion

Volume and mass conversion can be a common stepping stone to many chemistry problems. It involves converting one form of measurement into another to facilitate calculations. In our problem, this begins with finding the volume of aluminum foil using its area and thickness, which requires unit consistency.For conversion, we initially changed the thickness from \(\mathrm{mm}\) to \(\mathrm{cm}\) to match the area units. With both dimensions in consistent units, the multiplication gives the volume in \(\mathrm{cm}^3\). Following the volume calculation, mass is calculated using the substance's density. The conversion from volume to mass is critical first-hand knowledge for measuring substances practically in chemistry lab settings. Mastering these conversions ensures accuracy when determining how much of a material is available or needed during experimental procedures.