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?

We are given the dimensions of the aluminum foil as 1.00 cm² in area and 0.550 mm in thickness. We can find the volume of the foil by multiplying the area and thickness, and then convert the volume to mass using the given density of aluminum. Volume of foil = Area x Thickness Convert the thickness from mm to cm (1 cm = 10 mm): Thickness = 0.550 mm × (1 cm / 10 mm) = 0.055 cm Volume of foil = 1.00 cm² × 0.055 cm = 0.0550 cm³ Now we use the density of aluminum (2.699 g/cm³) to find the mass: Mass of aluminum = Volume × Density Mass of aluminum = 0.0550 cm³ × 2.699 g/cm³ = 0.148 g

Now that we have the mass of aluminum (0.148 g), we can use the molar mass of aluminum to find the number of moles. The molar mass of aluminum is 26.98 g/mol. Moles of aluminum = Mass of aluminum / Molar mass of aluminum Moles of aluminum = 0.148 g / 26.98 g/mol = 0.00548 mol

The balanced chemical equation for the reaction between aluminum and bromine to form aluminum bromide is: 2 Al (s) + 3 Br2 (l) → 2 AlBr3 (s) From the stoichiometry of the reaction, 1 mole of aluminum reacts to form 1 mole of aluminum bromide. Since we have 0.00548 moles of aluminum, an equal number of moles of aluminum bromide will be formed. Moles of aluminum bromide = 0.00548 mol Now we need to convert moles of aluminum bromide to mass. The molar mass of aluminum bromide (AlBr3) is: 1 Al = 26.98 g/mol 3 Br = 3 × 79.90 g/mol = 239.7 g/mol Molar mass of AlBr3 = 26.98 + 239.7 = 266.68 g/mol Mass of aluminum bromide = Moles of aluminum bromide × Molar mass of aluminum bromide Mass of aluminum bromide = 0.00548 mol × 266.68 g/mol = 1.46 g Therefore, 1.46 g of aluminum bromide is formed assuming the aluminum reacts completely.