A piece of aluminum foil 1.00 \(\mathrm{cm}^{2}\) and 0.550 -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?

Understanding the mole concept is fundamental in solving stoichiometry problems. A mole is a unit in chemistry that signifies a specific number of particles, atoms, ions, or molecules. This specific number, known as Avogadro's number, is approximately \(6.022 \times 10^{23}\) entities and is consistent for all substances.

For instance, in calculating the moles of aluminum used in the given problem, we first found the mass of the aluminum foil and then applied the mole concept. By dividing the mass of aluminum by its molar mass (the mass of one mole of aluminum), we were able to determine the number of moles present. This concept connects the microscopic particles to macroscopic measurements we can easily handle in the laboratory.

When applying the mole concept to chemical equations, it's crucial to consider the stoichiometry of the reaction, which tells us the proportional relationship between reactants and products in a balanced equation. It is through this concept that we can convert moles of one substance to moles of another, using the mole ratios derived from the balanced equation.

Chemical reactions involve the transformation of reactants into products through the breaking and forming of chemical bonds. Writing a balanced chemical equation is an essential step in solving stoichiometry problems. A balanced equation obeys the Law of Conservation of Mass, ensuring that the number of atoms for each element is the same on both sides of the equation.

In our problem, the balanced chemical equation for the reaction between aluminum and bromine to form aluminum bromide is:\[2 \text{Al (s)} + 3 \text{Br}_2 (l) \rightarrow 2 \text{AlBr}_3 (s)\]
This tells us that two moles of aluminum will react with three moles of bromine to produce two moles of aluminum bromide. Understanding the stoichiometry of this reaction is crucial as it allows us to predict the amounts of reactants and products involved. Once we know the moles of aluminum, the equation guides us to find the moles of aluminum bromide formed, given the mole-to-mole conversion between reactants and products.

The molar mass is the mass of one mole of a given substance, usually expressed in grams per mole (g/mol). It is a bridge between the atomic or molecular scale and the macroscopic scale, which we use to weigh substances. To find the molar mass of a compound, such as aluminum bromide (AlBr₃), you sum up the molar masses of the individual atoms within the molecule. Each element's molar mass is found on the periodic table and represents the mass of one mole of atoms of that element.

For example, to calculate the molar mass of AlBr₃, we added three times the molar mass of bromine to the molar mass of aluminum. Knowing the molar mass allowed us to convert between moles and grams, a key step in solving the problem. In stoichiometry problems, after determining the number of moles of a substance involved in a reaction, the molar mass is used to translate this amount into a measurable quantity, such as grams, which is more practical for laboratory work.